Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
x³ = 64
x = \(\sqrt[3]{4 \times 4 \times 4}\)
x = 4 cm
∴ side of cube = 4 cm.

When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]
= 2 [8 × 4 + 4 × 4 + 4 × 8]
= 2 [32 + 16 + 32]
= 2 [80]
∴ Surface area of cuboid = 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere ¡s 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13 – 7) = 6 cm
Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere
= 2πRH + 2πR²
= 2πR [H + R]
= 2 × \(\frac{22}{7}\) × 7(16 + 7)
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm²

Question 3.
A toy is ¡n the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Slant height of cone = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\)

= \(\sqrt{(3.5)^{2}+(12)^{2}}\)

= \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\)
Slain height conk (l) = 12.5 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
= πRL + 2πR²
= πR[L + 2R]
= \(\frac{22}{7}\) × 3.5 [12.5 + 2 (3.5)1 cm²
= \(\frac{22}{7}\) × 3.5 [19.5] cm²
= \(\frac{1501.5}{7}\) = 214.5 cm²
∴ Total surface area of toy = 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?
Find the surface area of the solid.
Solution:
Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
2R = 7
R = \(\frac{7}{2}\) cm

Surface area of solid = (surface area of the cube) – (area of base of hemisphere) + curved surface area of hemisphere)
= 6l² – πR² + 2πR²
= 6l² + πR²
= 6(7)² + \(\frac{22}{7}\) \(\frac{7}{2}\)²
= [6 × 49 + 11 × \(\frac{7}{2}\)]cm²
= 294 + 38.5 = 332.5 cm²

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let each side of cube = a
∴ Diameter of hemisphere = Side of cube
2R = a
R = \(\frac{a}{2}\)

Surface area of remaining solid = Total surface area of euboid – Area of the top of cube + Inner curved Surface area of hemisphere
= 6 (side)² – πR² + 2πR²
= 6(a)² + πR²
= 6(a)² + π \(\frac{a}{2}\)²
= 6a² + π \(\frac{a^{2}}{4}\)
= a² 6 + \(\frac{\pi}{4}\) cm²

Question 6.
A medicine capsule is ¡n the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm
∴ 2R = 5 mm
R = \(\frac{5}{2}\) mm

Length of entire capsule = 14 mm
Height of cylinderical part = (14 – \(\frac{5}{2}\) – \(\frac{5}{2}\)) mm
= (14 – 5) mm
H = 9 mm
Surface area of capsule = Surface area of cylinder + 2 Surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πRH + 4πR²
= 2πR [H + 2R]
= 2 × \(\frac{22}{7} \times \frac{5}{2}\left[9+2\left(\frac{5}{2}\right)\right]\)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) [9 + 5]
= \(\frac{22}{7}\) × 5 × 14
= 220 mm²
Hence, Surface area of capsule = 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per (Note that the base of the tent will not be
covered with canvas.)
Solution:
Diameter of cone = Diameter of cylinder
2R = 4
R = 2 m

Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 in
Slant height of cone (L) = 2.8 m
Curved surface area of tent = Curved surface of cylinder + Curved surface of conical part
= 2πRH + πRL
= πR [2H + L]
= \(\frac{22}{7}\) × 2[2(2.1) + 2.8]
= \(\frac{22}{7}\) × 2[4.2 + 2.8]
= \(\frac{22}{7}\) × 2 × 7
= 44 m²
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = 44 × 500 = ₹ 22000
Hence, Total cost of canvas = ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Diameter of cylinder (D) = 1.4 cm = Diameter of cone
∴ Radius of cylinder = Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm

As we know, L² = R² + H² + (2.4)²
L = \(\sqrt{(0.7)^{2}+(2.4)^{2}}\)
= \(\sqrt{0.49+5.76}\) = \(\sqrt{6.25}\)
L = 2.5 cm
Total surface area of remaining solid = curved surface area of cylinder + Area of base of cylinder + Surface area of cone
= 2πRH + πR² + πRL
= πR [2R +R + L]
= \(\frac{22}{7}\) × 0.7 [2(2.4) + 0.7 + 2.5]

=\(\frac{22}{7}\) × \(\frac{7}{10}\) [4.8 + 3.2]

= \(\frac{22}{10}\) [8]

= \(\frac{176}{10}\) = 17.6 cm²

Hence, Total surface area remaining solid to nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πR [H + 2R]
= 2 × \(\frac{22}{7}\) × 3.5 [10 + 2(3.5)]
= \(\frac{44}{7}\) × \(\frac{35}{10}\) [10 + 7]
= 44 × \(\frac{5}{10}\) × 17
= 44 × \(\frac{1}{2}\) × 17
= 22 × 17 = 374 cm²
Hence, total surface area of article = 374 cm².

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in Fig., If PQ = 24 cm, PR =7 cm and O is the centre of the circle.

Solution:
PQ = 24 cm
PR = 7 cm
RQ is diameter of circle
∠RPQ = 90° Angle in semi circle
In ∆PQR,
QR² = RP² + PQ²
QR = \(\sqrt{(7)^{2}+(24)^{2}}=\sqrt{49+576}\)
= \(\sqrt{625}\)
QR = 25 cm
∴ Diameter of circle (QR) = 25 cm
Radius of circle (R) = \(\frac{25}{2}\) cm
Area of shaded region = Area of the semicircle – Area of ∆RPQ
= \(\frac{1}{2} \pi \mathrm{R}^{2}-\frac{1}{2} \mathrm{RP} \times \mathrm{PQ}\)

= \(\left[\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 7 \times 24\right]\) cm²

= \(\left[\frac{6875}{28}-84\right]\)
= 245.53 – 84 = 161.53 cm²
∴ Area of shaded region = 161.53 cm².

Question 2.
Find the area of the shaded region in Fig., if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
Radius of smaller circle (r) = 7 cm
Radius of bigger circle (R) = 14 cm
Central angle ∠AOC (θ) = 40°
Area of shaded region = Area of bigger sector OAC – Area of smaller sector OBD
= \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}-\frac{\pi r^{2} \theta}{360^{\circ}}\)

= \(\frac{\pi \theta}{360^{\circ}}\) [R² – r²]

= \(\frac{22}{7} \times \frac{40}{360}\) × [14² – 7²]

= \(\frac{22}{63}\) [196 – 49]

= \(\frac{22}{63}\) × 147 = 51.33 cm²
∴ Shaded Region = 51.33 cm².

Question 3.
Find the area of the shaded region in fig., if ABCD is a square of side 14 cm and APD and BPC are semi circles.

Solution:
Side of square = 14 cm
Diameter of semicircle (AB = BC) = 14 cm
Radius of semi circle (R) = 7 cm
Area of square = (Side)²
= 14 × 14 = 196cm²
Area of a semi circles = \(\frac{1}{2}\) πR²
= \(\frac{1}{2} \times \frac{22}{7}\) × 7 × 7
= 77 cm²
Area of two semi circle = 2(77) = 154 cm²
Area of shaded region = Area of square ABCD – Area of two semi circles
= (196 – 154) = 42 cm²
Area of shaded region = 42 cm².

Question 4.
Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn iith vertex O of an equilateral triangle OAB of side 12 cm as centre.

[Each angle of equilateral triangle is 60°]
Area of major sector of circle = Area of circle – Area of sector
= πR² – \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7}\) × 6 × 6 – \(\frac{22}{7}\) × 6 × 6 × \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 6 × 6 1 – \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 36 1 – \(\frac{1}{6}\)

= \(\frac{22}{7}\) × 36 × \(\frac{5}{6}\)
= 94.28 cm²
∴ Area of major sector of circle = 94.28 cm²
Area of equilateral triangle OAB = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{1.73}{4}\) × 12 × 12
= 1.73 × 36 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAB + Area of major sector of circle
= 62.28 + 94.28 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAR + Area of major sectoç of circle
= 62.28 + 94.28 = 156.56
Shaded Area = 156.56 cm².

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in fig. Find the area of the remaining portion of the square.
Solution:
Side of square = 4 cm
Radius of each semi circle cut out (r) = 1 cm

Diameter of circle (R) = 2 cm
.. Radius of circle (R) = 1 cm
Area of square = (Side)²
= (4)² = 16 cm²
Area of 4 quadrants = 4\(\left[\frac{\pi^{2} \theta}{360^{\circ}}\right]\)

= \(\frac{4 \times 90}{360} \times \frac{22}{7} 1 \times 1\)

= 1 × \(\frac{22}{7}\) × 1 × 1 = 3.14 cm²
Area of circle = πR²
= \(\frac{22}{7}\) × 1 × 1
Area of circle = 3.14 cm²
Required area = Area of square – Area of 4 quadrants – Area of circle
= (16 – 3.14 – 3.14) cm² = 9.72 cm²
Required Area = 9.72 cm².

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as
shown in fig. Find the area of the design (shaded region).

Solution:
Radius of table cover (R) = 32 cm
OA = OB = OC = 32 cm

∆ABC is equilateral triangle with AB = AC = BC
∠AOB = ∠BOC = ∠COA = 120°
Now, in ∆BOC,
From O draw, angle bisector of ∠BOC as well as perpendicular bisector 0M of BC.
∴ BM = MC = \(\frac{1}{2}\) BC
Also, OB = OC [radii of the circle]
∴ ∠B = ∠C
∴ ∠O + ∠B + ∠C = 180°
120° + 2∠B = 180°
∠B = 30°
and ∠B = ∠C = 30°
Also, ∠BOM = ∠COM = 60°
∆OMB ≅ ∆OMC [RHS Cong.]
∴ In ∠OMB,
∠OBM = 30° [∠O = 60° and ∠M = 90°]
∴ \(\frac{\mathrm{BM}}{\mathrm{OB}}\) = cos 30°

\(\frac{\mathrm{BM}}{32}=\frac{\sqrt{3}}{2}\)

BM = 16√3 cm.

∴ BC = 2 MB = 32√3 cm
Area of circle = πR² = \(\frac{22}{7}\) × (32)²
= \(\frac{22}{7}\) × 32 × 32 = 3218.28 cm²

Area of ∆ABC = 4 (side)²
= \(\frac{1.73}{4}\) × 32√3 × 32√3 = 1328.64 cm²

∴ Required Area = Area of circle – Area of ∆ABC
= 3218.28 – 1328.64 = 1889.64 cm²

Question 7.
In fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Side of square ABCD = 14 cm
Radius of circle (R) = 7 cm
Sector angle (θ) = 90° [Each angle of square 90°]
Area of square = (side)²
= 14 × 14 = 196 cm²
Area of four quadrants = 4 \(\left[\frac{\pi R^{2} \theta}{360}\right]\)
= 4 × \(\frac{22}{7} \times \frac{7 \times 7 \times 90}{360}\)
= 22 × 7 = 154 cm²
∴ Required shaded area = Area of square – Area of 4 quadrants
= 196 – 154 = 42 cm².

 

Question 8.
Fig. depicts a racing track whose left and right ends are semicircular. The distahce between the two inner parallel line segments is 60 m and they are each 106 m long. 1f the track is 10 m wide, find
(i) the distance around the track along its inner edge.
(ii) the area of the track.

Solution:

(i) Here AB = DC = 106 m
AF = BE = CG = HD = 10m
Diameter of inner semicircle (APD and BRC) =60 m
∴ Radius of inner semicircle (APD (r) = 30 m
Radius of outer semicircle (R) = r + 10 = 30 + 10 = 40 m
Distance around the track along inner edge = AB + circumference of semi circle BRC + CD + circumference of semi circle DPA
= 2 AB + 2 [circumference of semi circle BRC]
= 2 (106) + 2(\(\left(\frac{2 \pi r}{2}\right)\))
= 212 + 2πr
= 212 + 2 × \(\frac{22}{7}\) × 30
= 212 + \(\frac{60 \times 22}{7}\)
= 212 + 188.57 = 400.57 m.
∴ Distance around the track along its inner edge = 400.57 m

(ii) Area of track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + area of region.
= 2 Area of rectangle ABCD + 2 Area of region (II)
= 2 (AB × AF) + 2
[Area of semi circle with Radius 60 cm – Area of semi circle with radius 30 cm]
= 2 [106 × 10] + 2 [latex]\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}[/latex]
= 2 × 1060 + \(\frac{2 \pi}{2}\) [R² – r²]
= 2120 + \(\frac{22}{7}\) (40² – 30²)
= 2120 + \(\frac{22}{7}\) [1600 – 900]
= 2120 + \(\frac{22}{7}\) [700]
= 2120 + 2200 = 4320 m²
Area of track = 4320 m²

Question 9.
In Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Diameter of circle = 14 cm
Radius of circle = 7 cm
Diameter of smaller circle = 7 cm
∴ Radius of smaller circle = \(\frac{7}{2}\) cm
Since AB and CD are to perpendicular the diameters of a circle,
∴ AO ⊥ CD
Area of bigger circle = πR² × 7 × 7 = 154 cm²
Area of bigger semicircle = \(\frac{154}{2}\) = 77 cm²
Area of smaller circle = πr²
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= 38.50 cm²

Area of ∆ABC = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 14 × 7 = 49 cm²
∴ Shaded Area = Area of bigger semi circle + Area of smaller circle – Area of triangle
= (77 – 49 + 38.5) cm² = 66.5 cm²

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half of the length of the side of the triangle (see Fig.). Find the area of the shaded region.
(Use n = 3.14 and ,√3 = 1.73205)

Solution:
Area of equilateral triangle ABC = 17320.5 cm²

\(\frac{\sqrt{3}}{4}\) (side)² = 17320.5

(side)² = \(\frac{17320.5 \times 4}{1.73205}\)

(side)² = \(\frac{173205}{10} \times \frac{100000 \times 4}{173205}\)

side = \(\sqrt{4 \times 100 \times 100}\)
side = 2 × 100 = 200 cm
AB = BC = AC
Radius of circle (R) = \(\frac{A B}{2}=\frac{200}{2}\) = 100 cm
Sector angle, θ = 60°
Area of sector APN = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{3.14 \times 100 \times 100 \times 60}{360}\)

= \(\frac{15700}{3}\)

Area of three sector = 3 × \(\frac{15700}{3}\) cm²
∴ Required shaded Area = Area of triangle – Area of three sectors
= 17320.5 – 15700 = 1620.5 cm²
∴ Hence, Required shaded Area = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find the area of the remaining portion of the handkerchief.

Solution:
Radius of circle (R) = 7 cm
Diameter of circle = 2 × R = 2 × 7 = 14 cm
Since there are three circles along a side of square
∴ side of squrae = 3 [14] = 42 cm
Total area of handkerchief = Area of square = (side)²
= (42)² = 1764 cm².
Area of 9 circular designs = 9πR²
= 9 × \(\frac{22}{7}\) × (7)2
= 9 × \(\frac{22}{7}\) × 7 × 7
= 9 × 154 = 1386 cm²
∴ Required area of remaining portion = Area of square – Area of 9 circular designs
= 1764 – 1386 = 378 cm²
∴ Required area of remaining portion = 378 cm².

Question 12.
In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,
(ii) shaded region.

Solution:

Radius of quadrant (R) = 3.5 cm
Angle of sector (θ) = 90°
OD = 2 cm.

(i) Area of quadrant OACB = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{3.5 \times 3.5 \times 90}{360}\) = 9.625 cm².

(ii) Area of ODB = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 3.5 × 2 = 3.5 cm²

∴ Shaded Area = Area of quadrant OACB – Area of ∆ODB
= 9.625 – 3.5 = 6.125 cm²
∴ Hence, Shaded Area = 6.125 cm².

Question 13.
In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Solution:
Side of square ABCO = 20 cm
∠AOC = 90°
AB = OA

OB² = OA² + AB²

OB = \(\sqrt{(20)^{2}+(20)^{2}}\)

= \(\sqrt{400+400}\)

= \(\sqrt{800}=\sqrt{400 \times 2}\)
OB = 20√2 cm
Area of square OABC = (side)² = (20)²
∴ Area of square = 400 cm²
Radius of quadrant (R) = 20√2 cm
Sector angle (θ) = 90°
∴ Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 20 \sqrt{2} \times 20 \sqrt{2} \times 90}{360}\)
= 2 × 314 cm² = 628 cm²
Required shaded Area = Area of sector – Area of square
= (628 – 400) cm² = 228 cm²

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ZAOB = 30°, find the area of the shaded region.

Solution:
Radius of sector OBA (R) =21 cm
Radius of sector ODC (r) 7 cm
Sector angle (θ) = 30°
Area of bigger sector (OAB) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{21 \times 21 \times 30}{360}\) = 115.5 cm²

Area of smaller sector (ODC) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 30}{360}\) = 12.83 cm²

Area of smaller sector (ODC) = 12.83 cm²
Now, Shaded Area = Area of bigger sector OAB – Area of smaller sector OCD
= 115.5 – 12.83 = 102.66
Hence, Shaded Area = 102.66 cm².

Question 15.
In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of quadrant ACPB (r) = 14 cm
Sector angle (θ) = 90°
AB = AC = 14 cm

Area of triangle = \(\frac{1}{2}\) AB × AC
= \(\frac{1}{2}\) × 14 × 14
= 98 cm²

Area of sector ACPB = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{14 \times 14 \times 30}{360}\) = 154 cm²

∴ Area of BOCPB = Area of sector ABPC – Area of \ABC
= 154 cm² – 98 cm² = 56 cm²
In ∆BAC, AB² + AC² = BC²
(14)² + (14)² = BC²
BC = \(\sqrt{196+196}=\sqrt{2(196)}\) = 14√2

∴ Radius of semi circle BOCR = \(\frac{14 \sqrt{2}}{2}\) = 7√2

Area of semi circle = \(\frac{\pi \mathrm{R}^{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \sqrt{2} \times 7 \sqrt{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 2}{2}\)
= 154 cm²

Required Area = Area of semi circle – [Area of sector – Area of ∆BAC]
= 154 – [154 – 98]
= (154 – 56) cm² = 98 cm²
Hence, Shaded Area = 98 cm².

Question 16.
Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

Solution:
Side of square = 8 cm
Area of square = (8)² = 64 cm²
Line BD divides square ABCD into the equal parts
Area of ∆ABD = ar of ∆BDC
Sector angle θ = 90°
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{8 \times 8 \times 90}{360}\)

Area of sector = 50.28 cm²
Area of ∆ABD = \(\frac{1}{2}\) × AB × AD
= \(\frac{1}{2}\) × 8 × 8
= 32 cm²

∴ Area of segment DMBPD = Area of sector ∆BPD – Area of ∆ABD
= 50.28 – 32 = 18.28 cm²
Hence, Shaded area = 2 area of segment DMBPD = 2 (18.28) = 36.56 cm²

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:

Radius of first circle (r₁) = 19 cm
Radius of second circle (r₂) = 9 cm
Let radius of third circle be R cm
According to condition
circumference of first circle + circumference of second circle = circumference of third circle
2πr₁ + 2πr₂ = 2πR
2π (r₁ + r₂] = 2πR
19 + 9 = R
∴ R = 28
∴ Radius of third circle (R) = 28 cm.

Question 2.
The Radii of two circles are 8 cm and 6 cm respectively. Find radius of circle which is having area equal to sum of the area of two circles.
Solution:
Radius of first circle (r₁) = 8 cm
Radius of second circle (r₂) = 6 cm
Let radius of third circle be R cm
According to question
Area of third circle = Area of first circle + Area of second circle
πR² = πr₁² + πr₂²
πR² = π[r₁² + r₂²]
R² = (8)² + (6)²
R = \(\sqrt{64+36}=\sqrt{100}\)
R = 10 cm
∴ Radius of required circle (R) = 10 cm.

Question 3.
Fig. depicts an archery target marLed with its five scoring areas from the centre ‘utwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score ¡s 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:
Diameter of Gold region = 21 cm
Radius of Gold region (R₁) = 10.5 cm
∴ Area of gold region = πR₁²
= \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}=\frac{690}{2}\) cm²
= 346.5 cm²
Width of each band = 10.5 cm
∴ Radius of Red and Gold region (R₂) = (10.5 + 10.5) = 21 cm
Combined radius of Blue, Red and Gold region (R₃) = R₂ + 10.5 cm
= 21 cm + 10.5 cm = 31.5 cm
Combined radius of Black, Blue, Red and Gold (R₄) = R₃ + 10.5
= 31.5 + 10.5 = 42cm
Area of circle having black radius = (Combined area of Gold, Red, Blue and Black radius) – (Combined Area of Gold, Red and Blue radius)
= πr₄² – πr₃²
= π [(42)² – (31.5)²]
= \(\frac{22}{7}\) [1764 – 992.25]
= \(\frac{22}{7}\) [771.75] = 2425.5 cm²

Combined Radius of white, black, blue, red, gold region (R₅) = R₄ + 10.5
R₅ = 42 + 10.5 = 52.5 cm
Combined radius of black, blue, red and gold = (R₄) = 42 cm.
Area of circle white scoring region = (Combined area of white, bLack, red, blue, gold region) – (Combined Area of Black, blue and gold region)
= πR₅² – πR₄²
= π[R₅² – πR₄²]
= \(\frac{22}{7}\) × [(52.5)² – (42)²]
= \(\frac{22}{7}\) [2756.25 – 1764]
= \(\frac{22 \times 992.25}{7}=\frac{21829.5}{7}\)
= 3118.5 cm²
∴ Area of white scoring region = 3118.5 cm²

∴ Area of red region = Area of red and gold region – Area of gold region
= πR₂² – πR₁²
= π [(21)² – (\(\frac{21}{2}\))²]
= \(\frac{22}{7}\) × 441 [1 – \(\frac{1}{4}\)]
= 22 × 63 \(\frac{3}{4}\)
= \(\frac{11 \times 189}{4}=\frac{2079}{4} \mathrm{~cm}^{2}\)
= 1039.5 cm²

∴ Area of Red region = 1039.5 cm²
Combined Radius of Gold, Red and Blue region R₃ (10.5 + 10.5 + 10.5) = 31.5 cm

Area of blue scoring region = (Combined area of red, blue and gold region) – (Combined area of Gold and red region)
= πR₃² – πR₂²
= π[R₃² – πR₂²]
= \(\frac{22}{7}\) × [(31.5)² – (21)²]
= \(\frac{22}{7}\) [992.25 – 441]
= \(\frac{22}{7}\) × 551.25 = \(\frac{121275}{7}\)
= 1732.5 cm²

Hence, area of gold ring; red ring; blue ring : black ring; white ring are 3465 cm² ; 1039.5 cm²; 1732.5 cm²; 24255 cm² ; 3118.5 cm² respectively.

Question 4.
The wheels of a ca are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a spel of 66 km per hour?
Solution:
Diameter of wheel = 80 cm
Radius of wheel (R) = 40 cm
Circumference of whed = 2πr
= 2 × \(\frac{22}{7}\) × 0.04
= \(\frac{22}{7}\) × 0.08 m
Let us suppose wheel of cr complete n revolutions of the wheel in 10 minutes = n[0.08 × \(\frac{22}{7}\)]
Speed of car = 66 km/hr. = 66 × 1000 m
Distance covered in 60 minutes = \(\frac{66 \times 1000}{60} \times 10\) = 11000 m
According to question.
∴ n[\(\frac{22}{7}\) × 0.08] = 11000
n = \(\frac{11000}{0.08} \times \frac{7}{22}\)
n = 4375
Hence, number of complete revolutions made by wheel in 10 minutes = 4375.

Question 5.
Tick the correct answer in the following and justify your choice : If the perimeter and area of a clrde are numerically equal, then the radius of the circle Is
(A) 2 units
(B) π units
(C) 4 units
(D) n units
Solution:
Perimeter of circle = Area of circle
2πR = πR²
2R = R²
⇒ R = 2
∴ Correct option A is (R) = 2 unit.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of the construction.

Question 1.
Draw a circle of radius 6 cm from a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:

Steps of construction:
1. Draw a circle (1) of radius 6 cm.
2. Take a point ‘P’ at a distance of 10 m. from the centre of the circle. Join OP.
3. Draw perpendicular bisector of OP. Let ‘M’ be the mid point OP.
4. With ‘M’ as centre and radius MO, draw a circle (II) which intersects the circle (I) at T and T’.
5. Then FT and PT’ are two required tangents.

Justification of construction:
We know that tangent at a point is always perpendicular to the radius at the point. Now
we have to prove that ∠PTO = ∠PT’O = 90°.
OT is joined.
Now, PMO is the diameter of circle (II) and ∠PTO is in the semicircle.
∴ ∠PTO = 90° [Angle in semicircle is a right angle].
Similarly, ∠PT’O = 90°
∴ PT and PT’ has to be the tangents to the circle at T and T’.
(On measuring, the lengths of tangents
i.e., PT = 8.1 cm
PT’ = 8.1 cm.
Co-centric circles. Two or more circles having same centre but different radii are called CO-CENTRIC circles.

Question 2.
Construct a tangent to a circle of radivs 4 cm from a point on the co-centric circle of radIus 6 cm and measure its length.
Also, erify the measurement by actual calculation.
Solution: Steps of construction:
STEPS OF CONSTRUCTION:
1. Draw a circle with cente O’ and radius 4 cm. Mark it as 1
2. Draw another circle with same centre ‘O’ and radius 6 cm and mark it as II.
3. Take any point ‘P’ on circle II. Join OP.
4. Draw pependicu1ar bisector of OP. Let it intersects ‘OP’ at M.
5. With M is centre and radius MO’ or ‘MP’, draw a circie III which intersects the circle ‘1’ at T and T’.
6. Join PT.
PT is the required tangent.

Justification of the construction :
Join OT.
Now OP is the diameter of the circle III.
∠OTP is in the semicircle.
∴ ∠OTP = 90° …………….(1)
[∴ Angle in a semicircle isa right angle]
Now OT ⊥ PT [using (I)]
∵ A line which makes an angle of 900 with radius at any point on the circle, the line is tangent to the circle.
∴ PT is tangent to the circle ‘I’
i.e. PT is tangent to the circle of radius 4.5 cm.

To calculate the length of tangent:
Consider ∆OTP,
∠OTP = 90° [using (i)]
∴ ∆OTP is a right angled triangle.
OT = 4 cm [Radius of I circle (given)]
OP = 6 cm [Radius of the II circle (given)]
PT = ? [to be calculated]
In rt. triangle ∆OTP,
By Pythagoras theorem
OP² = OT² + PT²
[(Hyp)² = (Base)² + (Perp.)²]
or PT² = OP² – OT²
= 6² – 4²
= 36 – 16 = 20
PT = \(\sqrt{20}\) cm
= 2√5 = 2 × 2.24 = 4.48 cm.
So, length of tangent by actual calculation = 4.48 cm = 4.5 cm.
Length of tangent by measurement = 4.5 cm
Hence, the length of tangent ‘PT’ is verified.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points ‘P’ and ‘Q’.
Solution:
Steps of construction:
1. Draw a circle of radius 3 cm and centre ‘O’.

2. Draw its diameter ‘AB’ and extend it in both directions as OX and OX’.
3. Take a point P’ on OX” direction and ‘Q’ on OX’ direction such that OP = OQ = 7 cm.
4. Draw perpendicular bisectors of OP and OQ which intersects OP and OQ at ‘M’ and ‘M” respectively.
5. With ‘M’ as centre and radius = ‘MO’ or MP, draw a circle ‘II’ which intersects the circle ‘I’ at T and T’.
6. Similarly with ‘M’’ as centre and radius = M’O or MQ, draw a circle (III) which intersects the circle ‘I’ at S’ and ‘S’’.
7. Join PT, PT’ and QS and QS’.

Justification of construction :
Join OT’ and ‘OT” and ‘OS’ and OS’.
To prove ‘PT & PT’ tangents to the circle
we will prove that ∠PTO = ∠PT’O = 90°.
Now ‘OP’ acts as the diameter of circle ‘II’ and ∠OTP is in the semicircle.
∴ ∠OTP = 90° …………….(1) [∵ Angle in semicircle is 90°]
But ‘OT’ is the radius of circle ‘I’ and line ‘PT’ touches the circle at T’.
∵ The line which touches the circle at a point and makes an angle of 90° with radius at that point, is tangent to the circle.
∴ PT is tangent to the circle I at point T through point ‘P.
Similarly PT’, QS and QS’ are tangents to the circle I.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°
Solution:
Steps of construction:
1. Draw the rough sketch of required figure.

∵ the tangents make an angle of 60° with each other.
∠OTP = ∠OQT = 90°
[Tangent is perpendicular to the radius of circle]
1. To find inclination of radii with each other
∠TOQ + ∠OTP + ∠OQT + ∠TPQ = 360° [Angle sum property of quad.]
or ∠TOQ + 90° + 90° + 60 = 360°
or ∠TOQ = 360 – 90° – 90° – 60° = 120°
2. Draw a circle of radius 5 cm.
3. Draw two radii of circle which make an angle of 120° with each other.
4. The radii intersect the circle at ‘A’ and
5. Make an angle of 90° at each point A and B, which intersect each other at ‘P’.

6. PA and PB are the required tangents.

Question 5.
Draw a line segment AB of length 8 cm. Taking ‘A’ as centre, draw a circle of radius 4 cm and taking ‘B’ as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:

Steps of construction :
1. Draw a line segment AB = 8 cm.
2. With ‘A’ as centre and radius 4 cm, draw a circle (I)
3. With ‘B’ as centre and radius 3 cm, draw a circle ‘I’.
4. Draw the perpendicular bisector of line segment AB which inersects ‘AB’ at ‘M’.
5. With ‘M as centre and radius MA or MB. draw a circle (III) which intersects the circle (I) at ‘S’ and ‘T’ and circle (II) at ‘P’ and ‘Q’.
6. Join ‘AP’ and AQ’. These are required tangents to the circle with radius 3 cm. from point ‘A’.
7. Join ‘BS’ and ‘BT’. These are required tangents to the circle with radius 4 cm from point ‘B’.

Justification of Construction:
In circle (III), AB acts as diameter then ∠ASB and ∠BPA are in semicircle.
∴ ∠ASB = 90° ………………(1) [Angle in semicircle]
and ∠BPA = 90° .
But ∠ASB is angle between radius of circle (I) and line segment BS’ and ∠BPA is angle between radius of circle (II) and line segment ‘AP’.
∵ Line segment which is perpendicular to the radius of circle, is tangent to the circle through that point.
∴ BS is tangent to circle (I) at point ‘S’ and AP is tangent to circle (II) at point ‘P’.
Similarly AQ and BT are tangents to the circle (II) and (I) respectively.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 5 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from ‘A’ to this circle.
Solution:

Steps of construction:
1. Construct rt. angled triangle. ABC according to given conditions and measurements.
2. Draw BD ⊥ AC.
3. Take mid point of side BC take it as
4. Take ‘M’ as centre and BC as diameter,
draw a circle through B. C, D using property, angle in semicircle is 90° (∠BDC 90°). Take this circle as I.
5. Now join ‘A’ and ‘M.
6. Draw perpendicular bisector of AM intersecting AM in point N. Now with ‘N’ as centre and ‘NA or ‘NM’ as radius, draw a circle (II) which intersects the circle (I) at ‘B’ and ‘P’.
7. Join AP.
8 AP and AB are the required tangents.

Justification of construction:
Line segment AM’ is diameter of circle (II)
∠APM is in semicircle
∴ ∠APM = 90° [Angle in semicircle]
i.e., MP ⊥ AP
But ‘MP’ is the radius of circle (I)
∴ AP is tangent to the circle (II)
[∵ Any line ⊥ to radius of circle at any point on the circle is tangent to the circle.]
Similarly AB is tangent to circle (I).

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this poiñt
to the circle.
Solution:
To draw circle with bangle means the centre of circle is unknown. First find the centre of circle.

Steps of construction:
1. Draw a circle. using a bangle (I).
2. Take any two chords AB and CD (non parallel) on circle.
3. Draw the perpendicular bisectors of chords AB and CD. The perpendicular bisectors intersect each other
[∵ any point lying on perpendicular bisector of line segment is equidistant from its end points
[∵ ‘O’ lies on ⊥ bisector of AH and CD]
∴ OA = OB and OC = OD
∴ OA = OB = OC = OD (Radii of circle)
∴ ‘O’ is the centre of circle.
4. Take any point ‘P’ out side the circle.
5. Join OP.
6. Draw the perpcndicular bisector of OP let ‘M’ the mid point of OP.
7. With ‘M’ as centre and radius ‘MP’ or ‘MO’, draw a circle II which intersects the circle (I) at T and T’.
8. Join PT and PT’, which is required pair of tangents.

Justification of construction:
Tangent at a point is always perpendicular to the radius at the point. Now, we have to prove
that ∠PTO = PT’O = 90°
Join OT.
Now ∠PTO is in the semicircle I.
∵ ∠PTO = 90° [Angle in semicircle is a right angle]
Similarly ∠PT’O = 90°
∴ PT and PT’ has to be the tangents to the circle at T and T’.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

In each of the questions, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide ¡tin the ratio 5 : 8. Measure the two parts.
Solution:
Given: A line segment of length of 7.6 cm.
Steps of construction:
1. Take a line segment AB = 7.6 cm.
2. Draw any ray AX, making an acute angle ∠BAX.
3. Locate 5 + 8 = 13 (given ratio 5: 8) points A₁, A₂, A₃, A₄, A₅, ………….., A₁₀, A₁₁, A₁₂, A₁₃ on ray AX such that A₁A₂ = A₂A₃ = A₃A₄ = …………. = A₁₁A₁₂ = A₁₂ A₁₃.
4. Join BA₁₃.
5. Through point A5, draw a line A₅C || A₁₃B (by making an angle equal to ∠A₁₃B) at A₅ intersecting AB at ‘C’. Then AC : CB = 5 : 8;

Justification:
Let us see how this method gives us the required division.
In ∆AA₁₃B,
Since A₅C || A₁₃B
∴ By Basic Proportionality Theorem
\(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}=\frac{\mathrm{AC}}{\mathrm{CB}}\)

By construction, \(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}=\frac{5}{8}\)

∴ \(\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{5}{8}\)
This shows that ‘C’ divides AB in the ratio 5 : 8.
On measuring the two parts, AC = 2.9 cm and CB = 4.7 cm.

Alternative Method:
Steps of construction:
1. Take a line segment AB = 7.6 cm
2. Draw any acute angle ∠BAX
3. Draw angle ∠ABY such that ∠ABY = ∠BAX.
4. Locate the points A₁, A₂, A₃, A₄, A₅ on ray AX such that A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
5. Locate the points B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈ on ray BY such that B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈
6. Join A₅B₈ let it intersects AB at point Then AC : CB = 5 : 8.

justification:
In ∆ACA₅ and ∆BCB₈,
∠ACA₅ = ∠BCB₈ [vertically opp. ∠s]
∠BAA₅ = ∠ABB₈ [construction]
∴ AACA₅ ~ ABCB₈ [AA-similarity cond.]
∴ Their corresponding sides must be in the same ratio. ,
\(\frac{A C}{B C}=\frac{C A_{5}}{C B_{8}}=\frac{A_{5} A}{B_{8} B}\)
(I)(II) (III)
From I and III, \(\frac{A C}{B C}=\frac{A_{5} A}{B_{8} B}\)

But, \(\frac{A_{5} A}{B_{8} B}=\frac{5}{8}\) [construction]

\(\frac{A C}{C B}=\frac{5}{8}\).

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of corresponding sides of the first triangle.
Solution:
Steps of construction:
1. Construct a triangle ABC with given measurements. AB = 5 cm, AC = 4 cm and BC = 6 cm.
2. Make any acute angle ∠CBX below the side BC.

3. Locate three points (greater of 2 and 3 in \(\frac{2}{3}\))B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃C.
5. Through B₂ (smaller of 2 and 3 in \(\frac{2}{3}\) draw a line parallel to B₃C, which intersect BC in C’.
6. Through C’, draw a line parallel to CA meeting BA is A’.
Thus ∆A’BC’ is the required triangle whose sides are of corresponding sides of ∆ABC.

Justification of construction :
First we will show that first triangle and constructed triangle are similar.
i.e. ∆A’BC’ ~ ∆ABC.
Consider ∆A’BC’ and ∆ABC.
∠B = ∠B [Common]
∠A’C’B= ∠ACB [By construction]
∆A’C’B ~ ∆ACB [AA – similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) …………….(1)
Now, consider ∆B₂BC’ and ∆B₃BC,
∠B = ∠B [common]
∠B₂C’B = ∠B₂CB [construction]
∴ ∆B₂BC’ ~ ∆B₃BC [AA -similarity]
∴ Their corresponding sides must be in the same ratio.

\(\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}_{2}}{\mathrm{CB}_{3}}\)

I II III

Taking (I) and (II).
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}\)

But, \(\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{2}{3}\) [construction]

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{2}{3}\) ……………(2)

From (1) & (2),
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{2}{3}\)

⇒ A’B = \(\frac{2}{3}\) AB and BC’ = \(\frac{2}{3}\) BC; C’A’ = \(\frac{2}{3}\) CA.
Hence, the construction is Justified.

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.
Solution:

Steps of construction :
1. Construct a triangle ABC in which AB = 7 cm, BC 6 cm and AC =5 cm.
2. Make any acute angle ∠BAX below the base AB.
3. Locate seven points A₁, A₂, A₃, A₄, A₅, A₆, A₇ on the ray AX such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
4. Join BA₅.
5. Through A₇, draw a line parallel A₅B. Let it meets AB at B’ on being produced such that AB’= \(\frac{7}{5}\) AB.
6. Through B’, draw a line parallel to BC which meets AC at C’ on being produced.
∆AB’C’ is the required triangle.

Justification of the construction.
In ∆ABC and ∆AB’C’,
∠A = ∠A [common]
∠ABC = ∠AB’C’ [corresponding ∠s]
∴ ∠ABC – ∠AB’C’ [AA-similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{CA}}{\mathrm{C}^{\prime} \mathrm{A}}\) ……………..(1)

Again, in ∆AA₅B and AA₇B’
∠A = ∠A [common]
∠AA5B = ∠AA7 B’ [corresponding ∠s]
∴ ∆AA5B ~ ∆AA7B’ [AA – similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}=\frac{\mathrm{A}_{5} \mathrm{~B}}{\mathrm{~A}_{7} \mathrm{~B}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}\)

⇒ \(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}\) [construction]

But, \(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{5}{7}\) …………….(2)

From (1) and (2),

\(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{CA}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{5}{7}\)

or \(\frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{C^{\prime} A}{C A}=\frac{7}{5}\)

⇒ AB’ = \(\frac{7}{5}\) AB; B’C’ = \(\frac{7}{5}\) BC and C’A’ = \(\frac{7}{5}\) CA

Hence, the sides of ∆AB’C’ are \(\frac{4}{4}\) of ∆ABC.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 12 times
the corresponding sides of the isosceles triangle.
Solution:
Given: Base of isosceles triangle is 8 cm and Altitude = 4 cm
To construct: A triangle whose sides are times the sides of isosceles triangle.
Steps of construction:
1. Take base AB = 8 cm.
2. Draw perpendicular bisector of AB. Let it intersect AB at ‘M’.
3. With M as centre and radius = 4 cm, draw an arc which intersects the perpendicular bisector at ‘C’
4. Join CA and CB.
5. ∆ABC is an isosceles with CA = CB.
6. Make any acute angle ∠BAX below the side BC.
7. Locate three (greater of ‘2’ & ‘3’ in 1\(\frac{1}{2}\) or \(\frac{3}{2}\))
A₁, A₂, A₃ on ‘AX’ such that A A₁ = A₁ A₂ = A₂ A₃.

8. Join A₂ (2nd point smaller of ‘2 and ‘3’ in ) and B.
9. Through A₃, draw a line parallel to A₂B meet AB is B’ cm being produced.
10. Through B’, draw a line parallel to BC which meets AC in C’ on being produced. ∆AB’C’ is the required triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of ∆ABC.

Justification of construction :
First we will prove ∆AB’C’ are ∆ABC and similar.
Consider ∆ AB’C’ and ∆ ABC
∠A = ∠A [Common]
∠AB’C’ = ∠ABC [By construction]
∠AB’C’ ~ ∠ABC [By AA – similarityj
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}}{\mathrm{CA}}\) ……………(1)

Now consider ∆ A₃AB’ and ∆ A,AB
∠A = ∠A [common]
∠B’A₃A = ∠BA₂A [By construction]
∴∆ A₃A B’ – ∆A₂AB [AA – similarity]
∴ Their corresponding sides must be in the same ratio
\(\frac{A_{3} A}{A_{2} A}=\frac{A B^{\prime}}{A B}=\frac{B^{\prime} A_{3}}{B A_{2}}\)
I II III
Taking (I) & (II),
\(\frac{A B^{\prime}}{A B}=\frac{A_{3} A}{A_{2} A}\)

But, \(\frac{A_{3} A}{A_{2} A}=\frac{3}{2}\) [construction]
⇒ \(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{3}{2}\) ……………..(2)
From (1) & (2)m
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}}{\mathrm{CA}}=\frac{3}{2}\left(1 \frac{1}{2}\right)\)

⇒ AB’ = 1\(\frac{1}{2}\) (AB); B’C’ = 1\(\frac{1}{2}\) BC and C’A’ = 1\(\frac{1}{2}\) (CA)
Hence, given result is justified.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB =5 cm and ¿ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
1. Take a line segment BC = 6 cm
2. Construct an angle of measure 60° at point B. i.e., ∠CBX = 60°.
3. With B as centre and radius 5 cm draw an arc intersecting BX at ‘A’

4. Join A and C.
5. At B, make any acute angle ∠CBY below the side BC.
6. Locate four points (greater of 3 and 4 in \(\frac{3}{4}\)) B₁, B₂, B₃, B₄ on BY such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄. .
7. Join B₄ and C.
8. Draw a line through B₃ (smaller of 3 and 4 in ) parallel to B₄C making corresponding angles. Let the line through B₃ intersects BC in C’.
9. Through C’, draw a line parallel to CA which intersects BA at A’.
The ∆A’BC’ is the required triangle whose sides are \(\frac{3}{4}\) of sides of ∆ABC.

Justification of the construction:
Consider ∆A’BC’ and ∆ABC
∠B = ∠B [commoni
∠A’C’B = ∠ACB [corresponding ∠s]
∴ ∆A’BC’ ~ ∆ABC [AA – similarity]
∴ Their corresponding sides must be in the same ratio.

∴ \(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ……………..(1)

Now consider ∆B₃B C’ and ∆B₄BC.
∠B = ∠B [common]
∠ C’ B₃B = ∠CB₄B [corresponding ∠s]
∆B₃BC’ ~ ∆B₄BC [AA – similarity con.]
Their corresponding sides must be in the same ratio.
\(\frac{B_{3} B}{B_{4} B}=\frac{B C^{\prime}}{B C}=\frac{C^{\prime} B_{3}}{C B_{4}}\)
(I) (II) (III)

From (I) and (II),
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{3} \mathrm{~B}}{\mathrm{~B}_{4} \mathrm{~B}}\)

But, = \(\frac{\mathrm{B}_{3} \mathrm{~B}}{\mathrm{~B}_{4} \mathrm{~B}}=\frac{3}{4}\) [construction]

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{3}{4}\) ………..(3)

From (1) and (3)
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{3}{4}\)

and C’A’= \(\frac{3}{4}\) CA.
∆A’BC’ is the required triangle whose sides are \(\frac{3}{4}\) sides of ∆ABC.

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°. ∠A = 105°. Then construct a triangle whose sides are j- times
the corresponding sides of ∆ABC.
Solution:
Steps of construction:
1. Construct the triangle ABC with the given measurements.
BC = 7 cm; ∠B = 45, ∠A = 105°
By angle sum property of triangle
∠A + ∠B + ∠C= 180°
105° + 45° + ∠C = 180°
∠C = 180 – 150° = 30°
2. Make any acute angle ∠CBX at point B, below the sides BC.

3. Locate four points (greater of 3 and 4 in \(\frac{4}{3}\)) B₁, B₂, B₃, B₄ on ‘BX’ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
4. Join B₃C (smaller of 3 and 4 in \(\frac{4}{3}\)).
5. Through B₄, draw a line parallel to B₃C meeting BC in C’ on being produced.
6. Through C’, draw another line parallel to CA meeting BA in A’ on being produced.
7. ∆A’BC’ is the required triangle whose sides are times the triangle ABC.

Justification of construction:
Consider the ∆ A’BC’ and ∆ ABC,
∠B = ∠B [common]
∠A’C’B = ∠ACB [construction]
∴ ∆A’BC’ ~ ∆ABC [AA – similarity]
∴ Their corresponding sides must be in the same ratio

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ………….(1)

Again, consider ∆B₄B C’ and ∆B₃BC,
∠B = ∠B [common]
∠C’B4B = ∠CB3B [By consiruction]
∴ BB C’ AB3BC [AA-si niilarity]
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{B}_{4} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}_{4}}{\mathrm{CB}_{3}}\)
I II III

Taking I and II members.

\(\frac{B C^{\prime}}{B C}=\frac{B_{4} B}{B_{3} B}\)

But, \(\frac{B_{4} B}{B_{3} B}=\frac{4}{3}\) (construction)

or \(\frac{B C^{\prime}}{B C}=\frac{4}{3}\) ………….(2)

From (1) and (2),

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{4}{3}\)

⇒ A’B = \(\frac{4}{3}\) AB; BC’ = \(\frac{4}{3}\) BC and C’A’ = \(\frac{4}{3}\) CA
Hence the construction is justified.

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of construction:
1. Draw a right triangle using given conditions. Consider the triangle as ABC in which BC = 4 cm; AB = 3 cm and
∠B = 90°.
2. Make any acute angle ∠CBX below the line BC.
3. Locate five points (greater of 5 and 3 in \(\frac{5}{3}\)) B₁, B₂, B₃, B₄. B₅ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
4. Join B₃ (smaller of ‘5’ and ‘3’ in \(\frac{5}{3}\)) and ‘C’.

5. Through B₅. draw a line parallel to BC meeting BC is C’ on being produced.
6. Again draw a line through C’ parallel to CA meeting BA in A’ on being produced.
∆A’BC’ is the required triangle whose sides are \(\frac{5}{3}\) times the sides of ∆ABC.

Justification of construction :
Consider ∆A’BC’ and ∆ABC
∠B = ∠B [common]
∠A’C’B = ∠ACB [By construction]
∴ ∆A’BC’ ~ ∆ABC [AA-similarity condition]
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ……………..(1)

Again, in ∆B₅C’B and ∆XB₃CB,
∠B = ∠B [common]
∠C’B₅B = ∠CB₃B [By construction]
∴ ∆B₅C’B ~ ∆B₃CB [AA-similarityj
∴ Their corresponding sidcs must be in the same ratio.

\(\frac{\mathrm{B}_{5} \mathrm{C}^{\prime}}{\mathrm{B}_{3} \mathrm{C}}=\frac{\mathrm{C}^{\prime} \mathrm{B}}{\mathrm{CB}}=\frac{\mathrm{BB}_{5}}{\mathrm{BB}_{3}}\)

I II III

Taking II and III members.
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{5} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}\)

But, \(\frac{B_{5} B}{B_{3} B}=\frac{5}{3}\) [construction]

\(\frac{B C^{\prime}}{B C}=\frac{5}{3}\) ……………(2)
From (1) and (2),

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{5}{3}\)

⇒ A’B = \(\frac{5}{3}\) AB; BC’ = \(\frac{5}{3}\) BC and C’A’ = \(\frac{5}{3}\) CA
Hence the construction is justified.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

In Question 1 to 3, choose the corred option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
A circle with centre O from a point, Q the length of the tangent to a circle is 24 cm and distance of Q from the centre is 25 cm.

∴ ∠QPO = 90°
Now, in right angled ∠OPQ,
OQ² = PQ² + OP²
(25)² = (24)² + OP²
Or 625 = 576 + OP²
Or OP² = 625 – 576
Or OP² = 49 = (7)²
Or OP = 7 cm
∴ Option (A) is correct.

Question 2.
In Fig., if TP and TQ and tangents to a circle with centre O so ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
In figure, OP is radius and PT is tangent to circle.
∠OPT = 90°
Similarly, ∠OQT = 90° and ∠POQ = 110° (Given)
Now, POQT is a Quadrilateral.
∴ ∠POQ + ∠OQT + ∠PTQ + ∠TPO = 360°
110° + 90° + ∠PTQ + 90° = 360°
Or ∠PTQ + 290° = 360°
Or ∠PTQ = 360° – 290°
Or ∠PTQ = 70°
∴ Option (B) is correct.

Question 3.
In tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then LPOA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
In given firgure, OA is radius and AP is a tangent to the circle.

∴ ∠OAP = 90°
Similarly, ∠OBP = 90°
Now, in right angled ∆PAO and ∆PBO
∠PAO = ∠PBO = 90°
OP = OP (Common side)
OA = OB (radii of same Circle)
∴ ∆PAO ≅ ∆PBO [RHS congruence]
∴ ∠AOP = ∠BOP [CPCT]
Or ∠AOP =∠BOP = \(\frac{1}{2}\) ∠AOB
Also, In Quad. OAPB,
∠OBP + ∠BPA + ∠PAO + ∠AOP = 360°
90° +80° +90° + ∠AOB = 360°
∠AOB = 360° – 260°
∠AOB = 100°
Form (1) and (2), we get
∠AOP = ∠BOP = \(\frac{1}{2}\) × 100° = 50°

∴ Option (A) is correct.

Question 4.
Prove that, the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Given: A circle with center O and AB as its diameter l and m are tangents at points A and B.
To Prove: l || m
Proof: OA is the radius and l is the tangent to the circle.

∴ ∠1 = 90°
Similarly, ∠2 = 90°
Or ∠1 = ∠2 = 90°
But these are alternate angles between two lines, when one transversal cuts them.
∴ l || m
Hence, tangents drawn at the ends of a diameter of a circle are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Given. A circle with centre O. AB its tangent meet circle at P.
i.e., P is the point of contact.

To Prove: Perpendicular at the point of contact to the tangent to a circle passes through the centre.
Construction: Join OP.
Proof: The perpendicular to a tangent line AB through the point of contact passes through the centre of the circle because only one perpendicular, OP can be drawn to the line AB through the point P.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at a distance 5 cm. from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
A circle with centre ‘O’ A is any point outside the circle at a distance of 5 cm from the centre.

Length of tangent = PA = 4 cm
Since, OP is the radius and PA is the tangent to the circle.
∠OPA = 90°
Now, in right angled ∠OPA.
Using Pythagoras Theorem.
OA² = OP² + PA²
(5)² = OP² + (4)²
Or OP² = 25 – 16
Or OP² = 9 = (3)²
Or OP = 3 cm.
Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of Ihe chord of the larger circle which touches the smaller circle.
Solution:
Two concentric circles having same centre O, and radii 5 cm and 3 cm respectively. Let PQ be the chord of larger circle
but tangent to the smaller circle.

Since, OM be the radius of smaller circle and PMQ be the tangent.

∴ ∠OMP = ∠OMQ = 90°
Consider, right angled triangles OMP and OMQ,
∠OMP = ∠OMQ = 90°
OP = OQ [radii of same circle]
OM = OM [common side]
∴ ∆OMP ≅ ∆OMQ [RHS congurence]
∴ PM = MQ [CPCT]
Or PQ = 2PM = 2MQ
Now, in right angled ∆OMQ.
Using Pythagoras Theorem,
OQ² = OM² + MQ²
(5)² = (3)² + (MQ)²
Or MQ = 25 – 9
Or MQ² = 16 = (4)²
Or MQ = 4 cm
Length of chord PQ = 2 MQ = 2 (4) cm = 8 cm
Hence, length of required chord is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to the circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

Solution:
Given: A Quadrilateral ABCD is drawn to circumscribe a circle.
To Prove: AB + CD = AD + BC
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside the circle and BP : BQ are tangents to the circle.
∴ BP = BQ ……………(1)
Similarly, AP = AS ………….(2)
and CR = CQ …………..(3)
Also, DR = DS ………….(4)
Adding (1), (2), (3) and (4), we get
(BP + AP) + (CR + DR) = (BQ + CQ) + (AS + DS)
AB + CD = BC + AD
is the required result.

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Given: A circle with centre O having two parallel tangents XY and X’Y’ and= another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B.
To Prove: ∠AOB = 90°
Contruction: Join OC, OA and OB.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal.
Now, A is any point outside the circle from two tangents PA and AC are drawn to the circle.
∴ PA = AC
Also, in ∆ POA and ∆ AOC,
PA = AC (Proved)
OA = OA (common side)
OP = OC (radii of same circle)
∴ ∆POA ≅ ∆AOC [SSS congruence]
and ∠PAO = ∠CAO [CPCT]
Or ∠PAC = 2 ∠PAO = 2 ∠CAO ……………(1)
Similarly, ∠QBC = 2∠OBC = 2 ∠OBQ ………………(2)
Now, ∠PAC + ∠QBC = 180°
[Sum of the interior angles on the same side of transversal is 180°]
Or 2∠CAO + 2∠OBC = 180° [Using (1) & (2)]
Or ∠CAO + ∠OBC = 180 = 90° …(3)
Now, in ∆OAB,
∠CAO + ∠OBC + ∠AOB = 180°
90°+ ∠AOB = 180° [Using (3)]
Or ∠AOB = 180° – 90° = 90°
Hence, ∠AOB = 90°.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. [Pb. 20191
Solution:
Given. A circle with centre O. P is any point outside the circle PQ and PR are the tangents to the given circle from point P.

To Prove. ∠ROQ + ∠QPR = 180°
Proof. OQ is the radius and PQ is tangent from point P to the given circle.
∴ ∠OQP = 90° ………..(1)
[∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact]
Similarly, ∠ORP = 90°
Now, in quadrilateral ROQP,
∠ROQ + ∠PRO + ∠OQP + ∠QPR = 360°
Or ∠ROQ + 90° + 90° + ∠QPR = 360° [Using (1) & (2)]
Or ∠ROQ + ∠QPR + 180 = 360°
Or ∠ROQ + ∠QPR = 360° – 180°
Or ∠ROQ + ∠QPR = 180°
Hence, the angle between the two tangents drawn from and external point to a circle is supplementary to angle subtended by the line segment joining the pnts of contact at the centre.

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Given: A parallelogram ABCD circumscribed a circle with centre O.
To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thé circle and BE; BF are tangents to the circle.

To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thè circle and BE; BF are tangents to the circle.
∴ BE = BF
Similarly AE = AH ……………(2)
and CG = CF
Also DG = DH
Adding (1), (2), (3) and (4), we get
(BE + AE) + (CG + DG) = (BF + CF) – (AH + DH)
Or AB + CD = BC +AD ……….(5)
Now, ABCD is a parallelogram, (Given)
∴ AB = CD and BC = AD …………(6)
From (5) and (6), we get
AB + AB = BC + BC
Or 2AB = 2BC or AB = BC
Or AB = BC = CD = AD
∴ ABCD is a rhombus.
Hence, parallelogram circumscribing a circle is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig). Find the sides AB and AC.

Solution:
A triangle ABC is drawn to circumscribe a circle of radius 4 cm and the sides BC, CA, AB of ∆ABC touch the circle at
D, E, F respectively. Since the lengths of tangents drawn from an external point to a circle are equal.
∴ AE = AF = x cm(say)
CE = CD = 6 cm (Given)
and BF = BD = 5 cm
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC; OE ⊥ AC and OF ⊥ AB.
Also, OE = OD = OF = 4 cm.
Consider, ∆ABC
a = AC = (x + 6) cm ;
b = CB = (6 + 8) cm = 14 cm
c = BA = (8 + x) cm
S = \(\frac{a+b+c}{2}\)
∴ S = \(\frac{x+6+14+8+x}{2}\) = \(\frac{2 x+28}{2}\) = (x + 14)

area (∆ABC)= \(\sqrt{\mathrm{S}(\mathrm{S}-a)(\mathrm{S}-b)(\mathrm{S}-c)}\)

= \(\sqrt{\begin{array}{r}
(x+14)(x+14-\overline{x+6}) \\
(x+14-14)(x+14-\overline{8+x})
\end{array}}\)

= \(\sqrt{(x+14)(8)(x)(6)}\)

= \(\sqrt{48 x^{2}+672 x}\) cm² ………………(1)

area (∆OBC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 14 × 4 = 28cm² …………….(2)

area(∆BOA)= \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (8 + x) × 4 = 28cm²…………….(3)

area (∆AOC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (6 + x) × 4 = 28cm² …………….(4)

From the figure, by addition of areas, we have
Or (∆ABC) = ar (∆OBC) + ar (∆BOA) + ar (∆AOC)
\(\sqrt{48 x^{2}+672 x}\) = 28 + 16 + 2x + 12 + 2x
Or \(\sqrt{48 x^{2}+672 x}\) = 4x + 56
Or 48x² + 672x =4[x+ 14]
Squaring both sides, we get
Or 48x² + 672x = 16 (x + 14)²
Or 48x (x + 14) = 16(x + 14)²
Or 3x = x + 14
Or 2x = 14
Or x = \(\frac{14}{2}\) = 7
∴ AC = (x + 6) cm = (7 + 6) cm= 13 cm
and AB = (x + 8) cm = (7 + 8) cm = 15 cm
Hence, AB = 15 cm and AC = 13 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Given:
A quadrilateral PQRS circumscribing a circle having centre O. Sides PQ, QR, RS and SP touches the circles at L, M, N, T
respectively.

To Prove:
∠POQ + ∠SOR = 180°
and ∠SOP + ∠ROQ = 180°
Construction:
Join OP, OL, OQ, 0M. OR, ON, OS, OT
Proof: Since the two tangents drawn from an external point subtend equal angles at the centre.
∴ ∠2 = ∠3; ∠4 = ∠5 ; ∠6 = ∠7; ∠8 = ∠1
But, sum of all angles around a point is 360°
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
Or ∠1 + ∠2 + ∠2 + ∠5 +∠5 + ∠6 + ∠6 + ∠1 = 360°
Or 2(∠1 + ∠2 + ∠5 + ∠6) = 360°
Or (∠1 + ∠2) + (∠5 + ∠6) = \(\frac{360^{\circ}}{2}\) = 180°
Or ∠POQ + ∠SOR = 180°
Similarly, ∠SOP + ∠ROQ = 180°
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
Since at any point on a circle, there can be one and only one tangeni. But circle is a collection of infinite points, so we can draw infinite number of tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………………. point(s).
Solution:
one

(ii) A line intersecting a circle in two points is called a ………………..
Solution:
secant.

(iii) A circle can have ……………. parallel tangents at the most.
Solution:
two

(iv) The common point of a lingent h, circle and the circit is called ………………
Solution:
point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution:
According to given information we draw the figure such that,

OP = 5 cm and OQ = 12 cm
∵ PQ is a tangent and OP is the radius
∵ ∠OPQ = 90°
Now, In right angled ∆OPQ.
By Pythagoras Theorem,
OQ² = OP² + QP²
Or (12)² = (5)² + QP²
Or QP² = (12)² – (5)²
Or QP² = 144 – 25 = 119
Or QP = \(\sqrt{119}\) cm.
Hence, option (D) is correct.

Question 4.
Draw a circle and two lines parallel to given line such that one is a tangent and other a secant to the circle.
Solution:
According to thc given information we draw a circle having O as centre and l is the given line.

Now, m and n be two lines parallel to a given line l such that m is tangent as well as parallel to l and n is secant to the circle as well as parallel to l.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing 220 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, ¡f the angle made by the rope with the ground level is 30° (see fig.).

Solution:
Let AB be the heignt of pole;
AC = 20 m be the length of rope.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in figure.

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = sin 30°

or \(\frac{\mathrm{AB}}{20}=\frac{1}{2}\)

or AB = \(\frac{1}{2}\) × 20 = 10
Hence, height of pole is 10 m.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 300 with it. The distance between the foot of the tree to the point where the top touches the ground is 8 rn Find the height of the tree.
Solution:
Let BD be length of tree before storm.
After storm AD = AC = length of broken part of tree.
The angle of elevation in this situation is 30°.
Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h_{1}}{8}=\frac{1}{\sqrt{3}}\)
or h₁ = \(\frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{8}{3} \sqrt{3}\) m ……….(1)

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = cos 30°

or \(\frac{8}{h_{2}}=\frac{\sqrt{3}}{2}\)

or \(h_{2}=\frac{8 \times 2}{\sqrt{3}}=\frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

h₂ = \(\frac{16}{3}\) √3 …………..(2)

Total height of the tree = h₁ + h₂
= \(\frac{8}{3}\) √3 + \(\frac{16}{3}\) √3 [Using (1) & (2)]

= \(\left(\frac{8+16}{3}\right) \sqrt{3}=\frac{24}{3} \sqrt{3}\) = 8√3 m.
Hence, height of the tree is 8√3 m.

Question 3.
A contractor plants to install two slides for the children to play in a park. For the children below the age of 5 years, she
prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Case I:
For children below 5 years.
Let AC = l₁ m denote the length of slide and BC = 1.5 m be the height of slide. The angle of elevation is 30°.
Various arrangements are shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 30°

or \(\frac{1 \cdot 5}{l_{1}}=\frac{1}{2}\)

or l₁ = 1.5 × 2 = 3 m.

Case II:
For Elder children
Let AC = 12 m represent the length of slide and BC = 3 m be the height of slide. The angle of elevation is 60°. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 60°

or \(\frac{3}{l_{2}}=\frac{\sqrt{3}}{2}\)

or l₂ = \(\frac{3 \times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}}\)

= \(\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6 \sqrt{3}}{3}\)

= 2√3 m.

Hence, length of slides for children below 5 years and elder children are 3 m and 2 m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
Let BC = h m be the height of tower and AB = 30 m be the distance at ground level. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 30°

or \(\frac{h}{30}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}\)

= 10√3 = 10 × 1.732
h = 17.32 (approx).
Hence, height of tower is 17.32 m.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
Let us suppose position of the kite is at point CAC = l m be length of string with which kite is attached. The angle of elevation for this situation be 60°. Various arrangements are as shown in the figure.

In right angled ∆ABC,

\(\frac{\mathrm{CB}}{\mathrm{AB}}\) = sin 60°

or \(\frac{60}{l}=\frac{\sqrt{3}}{2}\)

or l = \(\frac{60 \times 2}{\sqrt{3}}=\frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{120 \sqrt{3}}{3}\) = 40√3 m.
Hence, length of the string be 40√3 m.

Question 6.
A 15 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution. Let ED = 30 m be the height of building and EC = l5 m be the height of boy.
The angle of elevation at different situation are 30° and 60° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{DC}}{\mathrm{AC}}\) = tan 30°

or \(\frac{28 \cdot 5}{x+y}=\frac{1}{\sqrt{3}}\)

or x + y = 28.5 × √3 m ………………(1)

Now, in right angled ∆BCD,

\(\frac{\mathrm{DC}}{\mathrm{BC}}\) = tan 60°

or \(\frac{28 \cdot 5}{y}=\sqrt{3}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}}\)

or y = \(\frac{28 \cdot 5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{28 \cdot 5 \times \sqrt{3}}{3}\) ……….(2)

Distance covered towards building = x = (x + y) – y
= (28.5 × √3) – (\(\frac{28.5}{3}\) × √3) m [sing (1) and (2)]

= 28.5 (1 – \(\frac{1}{3}\)) √3 m

= 28.5 (\(\frac{3-1}{4}\)) √3 m

= [28.5 × \(\frac{2}{3}\)]√3 m = 19√3 m.

Hence, distance covered by boy towards the building is 19√3 m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
Let BC = 20 m be the height of building and DC = h m be the height of transmission tower. The angle of elevation of
the bottom and top of a transmission tower are 45° and 60° respectively.
Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{\mathrm{AB}}{20}\) = 1
or AB = 20 m ………………..(1)
Also, in right angled ∆ABD,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{20+h}=\frac{1}{\sqrt{3}}\)

AB = \(\frac{(20+h)}{\sqrt{3}}\) ………….(2)

From (1) and (2), we get

20 = \(\frac{(20+h)}{\sqrt{3}}\)
or 20√3 = 20 + h
or h = 20√3 – 20
or h = 20 (√3 – 1) m
= 20 (1.732 – 1) m
= 20 × 0.732 = 14.64 m.

Hence, height of the tower is 14.64 m.

Question 8.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution. Let BC = h m be the height of Pedestal and CD = 1.6 m be the height of statue.
The angle of elevation of top of statue and top of pedestal are 60° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 45°

or \(\frac{A B}{h}\) = 1

or AB = h m ………….(1)

In right angled ∆ABC,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = cot 60°

or \(\frac{\mathrm{AB}}{h+1.6}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{h+1.6}{\sqrt{3}}\) ……….(2)

From (1) and (2), we get
h = \(\frac{h+1.6}{\sqrt{3}}\)
or √3h = h + 1.6
or (√3 – 1) h = 1.6
or (1.732 – 1) h = 16
or (0.732) h = 1.6
or h = \(\frac{1.6}{0.732}\) = 2.1857923
= 2.20 m (approx.)
Hence, height of pedestal is 2.20 m.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the
building.
Solution:
Let BC = 50 m be height of tower and AD = h m be height of building. The angle of elevation of the top of a building from the foot of tower and top of tower from foot of the building are 30° and 60° respectively. Various arrangement are as shown in figure.

In right angled ∆ABC,
\(\frac{A B}{B C}\) = cot 60°

or \(\frac{\mathrm{AB}}{50}=\frac{1}{\sqrt{3}}\)

or AB = \(\frac{50}{\sqrt{3}}\) …………(1)

Also, in right angled ∆DAB,
\(\frac{\mathrm{AB}}{\mathrm{DA}}\) = cot 30°

or \(\frac{A B}{h}\) = √3
or AB = h√3 ……………(2)

From (1) and (2), we get
\(\frac{50}{\sqrt{3}}\) = h√3

or \(\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\) = h

or h = \(\frac{50}{3}\) = 16.6666

or h = 16.70 m (approx).
Hence, height of building is 16.70 m.

Question 10.
Two poles of equal heights are tanding opposite each other on either side of he road, which is 80 m wide. From a point
between them on the road the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
Let BC = DE = h m he height of two equal poles and point A be the required position where the angle of elevations of top of two poles are 30° and 60° respectively. Various arrangement are as shown in the figure.

In right angled ∆ADE,

\(\frac{E D}{D A}\) = tan 30°

or \(\frac{h}{x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{x}{\sqrt{3}}\) ……………(1)

In right angled ∆ABC,

\(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan 60°

or \(\frac{h}{80-x}\) = √3

or h = (80 – x) √3 …………(2)

From (1) and (2), we get
\(\frac{x}{\sqrt{3}}\) = (80 – x)
or x = (80 – x) √3 × √3
or x = (80 – x) 3
or x = 240 – 3x
or 4x = 240
or x = \(\frac{240}{4}\) = 60
Substitute this value of x in (I), we get
h = \(\frac{60}{\sqrt{3}}=\frac{60^{\circ}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{60 \sqrt{3}}{3}=20 \sqrt{3}\)

= (20 × 1.732) m = 34.64 m
DA = x = 60 m
and AB = 80 – x = (80 – 60) m = 20 m.
Hence, heigth of the poles are 3464 m and the distances of the point from the poles are 20 m and 60 m respectively.

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 30° (see fig.). Find the height of the tower and the width of the canal.

Solution:
Let BC = x m be the width of canal and CD = h m be height of TV tower. The angles of elevation of top of tower at different position are 30° and 60° respectively. Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60°

or \(\frac{h}{x}\) = √3
or h = √3x …………..(1)

Also, in right angled ∆ABD,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = tan 30°

or \(\frac{h}{20+x}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{20+x}{\sqrt{3}}\) ……………….(2)

From (1) and (2), we get

√3x = \(\frac{20+x}{\sqrt{3}}\)
or √3(√3x) = 20 + x
or 3x = 20 + x
or 2x = 20
or x = \(\frac{20}{2}\) = 10

Substitute this value of x in (1), we get
h = 10(√3)
= 10 × 1.732
h = 17.32 m
Hence, height of TV tower is 17.32 m and. width of the canal is 10 m.

Question 12.
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let BD = hm be the height of cable tower and AE = 7 m be the height of building. The angle of elevation of the top of a cable tower and angle of depression of its foot from top of a building are 60° and 45° respectively.
Various arrangements are as shown in figure.

In right angled ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AE}}\) = cot 45°

or \(\frac{\mathrm{AB}}{7}\) = 1

or AB = 7 m. ……………..(1)

Also, in right angled ∆DCE,

\(\) = cot 60°
or \(\frac{\mathrm{EC}}{h-7}=\frac{1}{\sqrt{3}}\)

or EC = \(\frac{h-7}{\sqrt{3}}\) ……………..(2)

But AB = EC ………….(Given)
7 = \(\frac{h-7}{\sqrt{3}}\) [Using (1) and (2)]
or 7√3 = h – 7
h = 7√3 + 7 = 7 (√3 + 1)
or h = 7 (1.732 + 1) = 7(2.732)
or h = 19.124
or h = 19.20 m (approx.)
Hence, height of the tower is 19.20 m.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:

Let CD = 75 m be the height of light house and point D be top of light house from w’here angles of depression of two ships are 30° and 45° respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD,
\(\frac{\mathrm{BC}}{\mathrm{CD}}\) = cot 45°

or \(\frac{y}{75}\) = 1
or y = 75 m ……………(1)

Also, in right angled ∆ACD
\(\frac{\mathrm{AC}}{\mathrm{CD}}\) = cot 30°

or \(\frac{x+y}{75}\) = √3
or x + y = 75√3
or x + 75 = 75√3 [using (1)]
or x = 75√3 – 75
= 75 (√3 – 1)
= 75( 1.732 – 1)
= 75 (0.732)
or x = 54.90
Hence, distance between the two ships is 54.90 m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant ¡s 60°. After some time, the angle of elevation reduces to 30° (see fig.). Find the distance travelled by the balloon during the interval.

Solution:
Let ‘AB’ be the position of 1.2 m tall girl, at the point of the angles of elevation of balloon at
different distances are 30° and 60° respectively. Various arrangements are as shwon in th figure.
According to question,
FG = ED = CE – CD
= 88.2 m – 1.2 m
= 87 m

In right angled ∆AGF,
\(\frac{A G}{G F}\) = cot 60°

or \(\frac{x}{87}=\frac{1}{\sqrt{3}}\)

or x = \(\frac{87}{\sqrt{3}}\) m.

Also, in right angled ∆ADE,
\(\frac{A D}{E D}\) = cot 30°

or \(\frac{x+y}{87}\) = √3

or x + y = 87√3
or \(\frac{87}{\sqrt{3}}\) + y = 87√3
or y = 87√3 – \(\frac{87}{\sqrt{3}}\)

or y = 87√3 – \(\frac{1}{\sqrt{3}}\)

or y = 87 \(\frac{3-1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

or y = \(\frac{87 \times 2 \times \sqrt{3}}{3}\)

or y = 58√3
or y = 58(1.732) = 100.456
or y = 100.456 m.
Hence, distance travelled by the balloon during the interval is 100.46 m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the further t(me taken by the car to reach the foot of the tower.
Solution:
Let CD = h m. be the tower of height.
Let A be initial position of the car and after six seconds the car be at 13. The angles of depression at A and B are 30° and 60° respectively. Various arrangements are as shown in figure.

Let speed of the car be υ metre per second using formula, Distance = Speed x Time
AB = Distance covered by car in 6 seconds
AB = 6υ metre
Also, time taken by car to reach the tower be ‘n’ seconds.
∴ BC = nυ metre
In right angled ∆ACD.
\(\frac{\mathrm{CD}}{\mathrm{AC}}\) = tan 30°

or \(\frac{h}{6 v+n v}=\frac{1}{\sqrt{3}}\)

or h = \(\frac{6 v+n v}{\sqrt{3}}\) ……………….(1)

Also, in right angled ∆BCD,
\(\frac{C D}{B C}\) = tan 60°

or \(\frac{h}{n v}\) = √3
h = nv (√3) ……….(2)

From (1) and (2), we get
\(\frac{6 v+n v}{\sqrt{3}}\) = nυ(√3)
or 6υ + nυ = nυ(√3)
or 6υ + nυ = 3nυ
or 6υ = 2nυ
or n = \(\frac{6 v}{2 v}\) = 3
Hence, time taken by car to reach the foot of tower is 3 seconds.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let CD = h m be the height of tower and B ; A be the required points which are at a distance of 4 m and 9 m from the tower respectively. Various arrangements are as shown in the figure.

In right angled ∆BCD
\(\frac{\mathrm{CD}}{\mathrm{BC}}\) = tan θ

or \(\frac{h}{4}\) = tan θ ………….(1)

Also, in right angled ∆ACD,
\(\frac{C D}{A C}\) = tan (90 – θ)

or \(\frac{h}{9}\) = cot θ

Multiplying (1) and (2), we get
\(\frac{h}{4} \times \frac{h}{9}\) = tan θ cot θ

or \(\frac{h^{2}}{36}=\tan \theta \times \frac{1}{\tan \theta}\)

or h² = 36 = (6)²
or h = 6
Hence, height of the tower is 6 m.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios of sin A, sec A and tan A in terms of cot A.
Solution:
By using Identity,
cosec² A – cot² A = 1
⇒ cosec² A = 1 + cot² A
⇒ (cosec A)² = cot² A + 1
⇒ \(\left(\frac{1}{\sin A}\right)^{2}\) = cot² A + 1
⇒ (sin A)² = \(\frac{1}{\cot ^{2} \mathrm{~A}+1}\)
⇒ sin A = ± \(\frac{1}{\sqrt{\cot ^{2} \mathrm{~A}+1}}\)
We reject negative values of sin A for acute angle A.
Therefore, sin A = \(\frac{1}{\sqrt{\cot ^{2} A+1}}\)
By using identity,
sec² A – tan² A = 1
⇒ sec² A = 1 + tan² A
= 1 + \(\frac{1}{\cot ^{2} A}\)
= \(\frac{\cot ^{2} A+1}{\cot ^{2} A}\)

⇒ sec A = \(\sqrt{\frac{\cot ^{2} A+1}{\cot ^{2} A}}\)

tan A = \(\frac{1}{\cot A}\).

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
By using Identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\frac{1}{\sec ^{2} \cdot A}\) = \(\frac{\sec ^{2} A-1}{\sec ^{2} A}\)

⇒ (sin A)² = \(\frac{\sec ^{2} A-1}{\sec ^{2} A}\)

⇒ sin A = ± \(\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}\)

[Reject – ve sign for acute angle A]
⇒ sin A = ± \(\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}\)
cos A = \(\frac{1}{\sec A}\)
1 + tan² A = sec² A
tan² A = sec² A – 1
(tan A)² = sec² A – 1
⇒ tan A = ± \(\sqrt{\sec ^{2} A-1}\)
[Reject – ve sign for acute angle A]
i.e., tan A = \(\sqrt{\sec ^{2} A-1}\)
cosec A = \(\frac{1}{\sin A}=\frac{1}{\sqrt{\sec ^{2} A-1}}\)

= \(\frac{\sec A}{\sqrt{\sec ^{2} A-1}}\)

cot A = \(\frac{1}{\tan A}=\frac{1}{\sqrt{\sec ^{2} A-1}}\).

Question 3.
Evaluate:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

(ii) sin 25° cos 65° + cos 25° sin 65°.
Solution:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

= \(\frac{\left\{\sin \left(90^{\circ}-27^{\circ}\right)+\sin ^{2} 27^{\circ}\right\}}{\cos ^{2} 17^{\circ}+\left\{\cos \left(90^{\circ}-17^{\circ}\right)\right\}^{2}}\)
[∵ sin(90 – θ) = cos θ and cos (90 – θ) = sin θ]

= \(\frac{\left\{\cos 27^{\circ}\right\}^{2}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\left\{\sin 17^{\circ}\right\}^{2}}\)

= \(\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}\)
= \(\frac{1}{1}\) = 1.

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° × cos (90° – 25°)
+ cos 25° × sin (90° – 25°)
[∵ cos (90° – θ) = sin θ
sin(90° – θ) = cos θ].
= sin 25° × sin 25° + cos 25° × cos 25°
= sin² 25° + cos² 25° = 1.

Question 4.
Choose the correct option. Justify your choice:
(i) 9 sec² A – 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) θ
(B) 1
(C) 2
(D) – 1.

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A.

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =
(A) sec²A
(B) – 1
(C) cot² A
(D) tan² A.

Solution:
(i) Consider, 9 sec² A – 9 tan² A
= 9 (sec² A – tan² A)
= 9 × 1 = 9.
Option (B) is correct.

(ii) Consider, (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= \(\left\{1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right\} \times\left\{1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right\}\)

= \(\left\{\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right\} \times\left\{\frac{\sin \theta+\cos \theta+1}{\sin \theta}\right\}\)

= \(\begin{array}{r}
\{(\cos \theta+\sin \theta)+1\} \\
\times\{(\cos \theta+\sin \theta)-1\} \\
\hline \cos \theta \times \sin \theta
\end{array}\)

= \(\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \times \sin \theta}\)

[∵ (a + b) (a – b) = a² – b²]

= \(\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}{\cos \theta \times \sin \theta}\)

= \(\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}\) = 2.

Option (C) is correct.

(iii) Consider, (sec A + tan A) (1 – sin A)
= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\) × (1 – sin A)

= \(\frac{(1+\sin A)}{\cos A}\) × (1 – sin A)

= \(\frac{(1+\sin A)(1-\sin A)}{\cos A}\)

= \(\frac{(1)^{2}-(\sin A)^{2}}{\cos A}=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}\)
[∵ cos² A = 1 – sin² A]
= cos A.
Option (D) is correct.

(iv) Consider, \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)

= tan² A.
Option (D) is correct.

Question 5.
Prove the following Identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ) = \(\frac{1-\cos \theta}{1+\cos \theta}\)

(ii) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 sec A.

(iii) \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]

(iv) \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
[Hint: Simplify L.H.S. and R.H.S. separately]

(v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A = 1 + cot² A.

(vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A

(vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
[Hint : Simplify L.H.S. and R.H.S. separately]

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A.

Solution:
(i) L.H.S. = (cosec θ – cot θ)²
= \(\left\{\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right\}^{2}\)

= \(\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}\)
Using identity, sin² θ + cos² θ = 1
⇒ sin² θ = 1 – cos² θ
= \(\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}\)
= \(\)
[∵ a² – b² = (a + b) (a – b)]

= \(\)

∴ L.H.S. = R.H.S.
Hence, (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)

(ii) L.H.S. = \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\)

= \(\frac{2}{\cos A}\) = cos A
∴L.H.S. = R.H.S.
Hence, \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 sec A.

(iii) L.H.S. = \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

= \(\frac{\left(\frac{\sin \theta}{\cos \theta}\right)}{\left(1-\frac{\cos \theta}{\sin \theta}\right)}+\frac{\left(\frac{\cos \theta}{\sin \theta}\right)}{\left(1-\frac{\sin \theta}{\cos \theta}\right)}\)

= \(\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}=\frac{1}{\cos \theta \sin \theta}+1\)

= 1 + \(\left(\frac{1}{\cos \theta}\right)\left(\frac{1}{\sin \theta}\right)\) = 1 + sec θ cosec θ
∴L.H.S. = R.H.S.
Hence, \(\frac{\tan \theta}{1+\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

(iv) L.H.S. = \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
= \(\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}\)
= 1 + cos A …………….(1)
R.H.S = \(\frac{\sin ^{2} A}{1-\cos A}\)
(∵ 1 – cos² A = sin² A.)
= \(\frac{1-\cos ^{2} A}{1-\cos A}\)

= \(\frac{(1+\cos A)(1-\cos A)}{(1-\cos A)}\)

= 1 + cos A. …………….(2)
From (1) and (2) it is clear that
∴ L.H.S. = R.H.S.
Hence, \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\).

(v) L.H.S. = \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A

= cosec A + cot A
= R.H.S
∴ L.H.S. = R.H.S.
Hence, \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) using the identity cosec² A = 1 + cot² A.

(vi) L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)

= \(\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}\)

= \(\sqrt{\frac{(1+\sin A)^{2}}{(1)^{2}-(\sin A)^{2}}}\)

= \(\sqrt{\frac{(1+\sin A)^{2}}{1-\sin ^{2} A}}=\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)

= \(\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A
∴ L.H.S. = R.H.S.
Hence, \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A.

(vii) L.H.S. = \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\)

∴ L.H.S. = R.H.S.
Hence, \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ

(viii) L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A × cosec A) + {cos² A + sec² A
+ 2 cos A × sec A)
= [sin² A + co² A + 2sin A × \(\frac{1}{\sin A}\)] + [cos² A + sec² A + 2 cosA × \(\frac{1}{\cos A}\)]
= (sin² A + cosec² A + 2) + (cos² A + sec² A + 2)
= 2 + 2 + (sin² A + cos² A) + sec² A + cosec² A
= 2 + 2 + 1 + 1 + tan² A + 1 + cot² A
= 7 tan² A + cot² A
∴ L.H.S. = R.H.S.
Hence, (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

(ix) L.H.S. = (cosec A – sin A) (sec A – cos A)

From (1) and (2), it is clear that
L.H.S. = R.H.S.
Hence, (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)\)
(∵ 1 + tan² A = sec² A
and 1 + cot² A = cosec² A)

From (1) and (2), it is clear that
LH.S. = R.H.S.
Hence, \(\left(\frac{1+\tan ^{2} A}{1+\cot A^{2}}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution.
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= \(\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}\)
= \(\frac{\sin 18^{\circ}}{\sin 18^{\circ}}\) = 1
[∵ cos (90° – θ) = sin θ]

(ii) \(\frac{\tan 26^{\circ}}{\cos 64^{\circ}}=\frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)}\)
= \(\frac{\tan 26^{\circ}}{\tan 26^{\circ}}\) = 1
[∵ cot (90°- θ) = tan θ]

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
[∵ cos (90° – 0) = sin O]
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59°
=cosec 31° – sec (90° – 31°)
= cosec 31° – cosec 31°
[∵ sec (90° – θ) = cosec θ].

Question 2.
Show that:
(i) tan 4 tan 230 tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) L.H.S.
= tan 48° tan 23° tan 42° tan 67°
= tan 48° × tan 23° × tan (90° – 48°) × tan (90° – 23°)
= tan48° × tan 23° × cot48° × cot 23°
= tan 48C × tan 23° × \(\frac{1}{\tan 48^{\circ}}\) × \(\frac{1}{\tan 23^{\circ}}\) = 1
∴ L.H.S. = R.H.S.

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°
= cos 38° × cos (90 – 38°) – sin 38° × sin (90° – 38°)
= cos 38° × sin 38° – sin 38° × cos 38
= 0.
∴ L.H.S. = RH.S.

Question 3.
If tan 2A = cot (A – 18°) where 2A is an acute angle, find the value of A.
Solution:
Given: tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°)
[cot (90° – θ) = tan θ]
⇒ 90°- 2A = A – 18°
⇒ 3A = 108°
⇒A = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
Given that: tan A = cot B
⇒ tan A = tan(90° – B)
[∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B.
⇒ A + B = 90°..

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Given that: sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A — 20°)
[∵ cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20°
⇒ 5A = 110°
⇒ A = 22°.

Question 6.
If A, B and C interior angles of a triangle ABC, then show that: \(\sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)\)
Solution:
Since, A, B and C are interior angles of a triangle
∴ A + B + C = 180°
[Sum of three angles of a triangle is 180°]
⇒ B + C = 180° – A
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}\)
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
Taking sin on both sides, we get
⇒ \(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
[∵ sin (90° – θ) = cos θ].

Question 7.
Express sin 67° + cos 75° in terms of Trigonometric ratios of angles between 0° and 45°.
Solution:
Given that: sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
[∵ sin(90° – θ) = cos θ and cos (90° – θ) = sin θ].