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Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.
Solution:
(i) Let ∆ABC, with AB = 7 cm BC = 24 cm, AC = 25 cm
AB² + BC² = (7)² + (24)²
= 49 + 576 = 625
AC² = (25)² = 625
Now AB² + BC² = AC²
∴ ∆ABC is right angled triangle. Hyp. AC = 25cm.

(ii) Let ∆PQR with PQ = 3 cm, QR = 8 cm PR = 6 cm
PQ² + PR² = (3)² + (6)²
= 9 + 36 = 45
QR² = (8)² = 64.
Here PQ² + PR² ≠ QR²
∴ ∆PQR is not right angled triangle.

(iii) Let ∆MNP, with MN =50 cm, NP = 80 cm, MP = 100 cm
MN ²+ NP² = (50)² + (80)²
= 2500 + 6400 = 8900
MP² = (100)² = 10000
Here MP² ≠ MN² + NP².
∴ ∆MNP is not right angled triangle.

(iv) Let ∆ABC, AB = 13 cm, BC = 12 cm, AC = 5 cm
BC² + AC² = (12)² + (5)²
= 144 + 25 = 169
AB² = (13)² = 169
∴ AB² = BC² + AC²
∆ABC is right angled triangle.
Hyp. AB = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Solution:
Given: ∆PQR is right angled at P and M is a point on QR such that PM ⊥ QR.
To prove : PM² = QM × MR

Proof: ∠P = 90° (Given)
∴ ∠1 + ∠2 = 90°
∠M = 900 (Given)
In ∆PMQ,
∠1 + ∠3 + ∠5 = 180°
=> ∠1 + ∠3 = 90° [Angle Sum Property] ………….(2) [∠5 = 90°]
From (1) and (2),
∠1 + ∠2 = ∠1 + ∠3
∠2 = ∠3
In ∆QPM and ∆RPM,
∠3 = ∠2 (Proved)
∠5 = ∠6 (Each 90°)
∴ ∆QMP ~ ∆PMR [AA similarity]
\(\frac{{ar} .(\Delta \mathrm{QMP})}{{ar} .(\Delta \mathrm{PMR})}=\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)

[If two triangles are similar, ratio o their areas is equal to square of corresponding sides]
\(\frac{\frac{1}{2} \mathrm{QM} \times \mathrm{PM}}{\frac{1}{2} \mathrm{RM} \times \mathrm{PM}}=\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)

[area of ∆ = \(\frac{1}{2}\) Base × Altitude]

\(\frac{\mathrm{QM}}{\mathrm{RM}}=\frac{\mathrm{PM}^{2}}{\mathrm{RM}^{2}}\)

PM² = QM × RM Hence proved.

Question 3.
In fig., ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC.BD
(ii) AC² = BC.DC
(üi) AD² = BD.CD.
Solution:
Given. A right angled ∆ABD in which right angled at A and AC ⊥ BD.
To Prove:
(i) AB² = BC.BD
(ii) AC² = BC.DC .
(iii) AD² = BD.ÇD .
Proof. In ∆DAB and ∆DCA,
∠D = ∠D (common)
∠A = ∠C (each 90°)
∴ ∆DAB ~ ∆DCA [AA similarity]
In ∆DAB and ∆ACB,
∠B = ∠B (common)
∠A = ∠C . (each 90°)
∴ ∆DAB ~ ∆ACB, .
From (1) and (2),
∆DAB ~ ∆ACB ~ ∆DCA.
(i) ∆ACB ~ ∆DAB (proved)
∴ \(\frac{{ar} .(\Delta \mathrm{ACB})}{{ar} .(\Delta \mathrm{DAB})}=\frac{\mathrm{AB}^{2}}{\mathrm{DB}^{2}}\)

[If two triangles are similar corresponding sides are proportional]

\(\frac{\frac{1}{2} \mathrm{BC} \times \mathrm{AC}}{\frac{1}{2} \mathrm{DB} \times \mathrm{AC}}=\frac{\mathrm{AB}^{2}}{\mathrm{DB}^{2}}\)
[Area of triangle = \(\frac{1}{2}\) Base × Altitude]
BC = \(\frac{\mathrm{AB}^{2}}{\mathrm{BD}}\)
AB² = BC × BD.

(iii) ∆ACB ~ ∆DCA (proved)
\(\frac{{ar} .(\Delta \mathrm{DAB})}{{ar} .(\Delta \mathrm{DCA})}=\frac{\mathrm{DA}^{2}}{\mathrm{DB}^{2}}\)
[If two triangles are similar corresponding sidec are proportional]

\(\frac{\frac{1}{2} \mathrm{CD} \times \mathrm{AC}}{\frac{1}{2} \mathrm{BD} \times \mathrm{AC}}=\frac{\mathrm{AD}^{2}}{\mathrm{BD}^{2}}\)

CD = \(\frac{\mathrm{AD}^{2}}{\mathrm{BD}}\)
⇒ AD² = BD × CD.

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given: ABC is an isosceles triangle right angled at C.
To prove : AB² = 2AC².
Proof: In ∆ACB, ∠C = 90° & AC = BC (given)

AB² = AC² + BC²
[By using Pythagoras Theorem]
=AC² + AC² [BC = AC]
AB² = 2AC²
Hence proved.

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is right triangle.
Solution:
Given: ∆ABC is an isosceles triangle AC = BC
To prove: ∆ABC is a right triangle.

Proof: AB² = 2AC² (given)
AB² = AC² + AC²
AB² = AC² + BC² [AC = BC]
∴ By Converse of Pythagoras Theorem,
∆ABC is right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
∆ABC is equilateral triangle with each side 2a
AD ⊥ BC
AB = AC = BC = 2a
∆ADB ≅ ∆ADC [By RHS Cong.]
∴ BD = DC = a [c.p.c.t]

In right angled ∆ADB
AB² = AD² + BD²
(2a)² = AD² + (a)²
4a² – a² = AD².
AD² = 3a²
AD = √3a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [Pb. 2019]
Solution:
Given: Rhombus, ABCD diagonal AC and BD intersect each other at O.
To prove:
AB² + BC² + CD² + AD² = AC² + BD²
Proof:The diagonals of a rhombus bisect each other at right angles.

∴ AO = CO, BO = DO
∴ ∠s at O are rt. ∠s
In ∆AOB, ∠AOB = 90°
∴ AB² = AO² + BO² [By Pythagoras Theorem] …………..(1)
Similarly, BC² = CO² + BO² ……………..(2)
CD² = CO² + DO² ……………(3)
and DA² = DO² + AO² ……………….(4)
Adding. (1), (2), (3) and (4), we get
AB² + BC² + CD² + DA² = 2AO² + 2CO² + 2BO² + 2DO²
= 4AO² + 4BO²
[∵ AO = CO and BO = DO]
= (2AO)² + (2BO)² = AC² + BD².

Question 8.
In fig., O is a point In the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²

(ii)AF² + BD² + CE² = AE² + CD² + BF².

Solution:
Given: A ∆ABC in which OD ⊥ BC, 0E ⊥ AC and OF ⊥ AB.

To prove:
(i) AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²
(ii) AF² + BD² + CE² = AE² + CD² + BF².
Construction: Join OB, OC and OA.
Proof: (i) In rt. ∠d ∆AFO, we have
OA² = OF² + AF² [By Pythagoras Theorem]
or AF² = OA² – OF² …………..(1)

In rt. ∠d ∆BDO, we have:
OB² = BD²+ OD² [By Pythagoras Theorem]
⇒ BD² = OB² – OD² …………..(2)

In rt. ∠d ∆CEO, we have:
OC² = CE² + OE² [By Pythagoras Theorem]

⇒ CE² = OC² – OE² ……………(3)

∴ AF² + BD² + CE² = OA² – OF² + OB² – OD² + OC² – 0E²
[On adding (1), (2) and (3)]
= OA² + OB² + OC² – OD² – OE² – OF²
which proves part (1).
Again, AF² + BD² + CE² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)
= AE² + CD² + BF²
[∵AE² = AO² – OE²
CD² = OC² – OD²
BF² = OB² – OF²].

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Height of window from ground (AB) = 8m.

Length of ladder (AC) = 10 m
Distance between foot of ladder and foot of wall (BC) = ?
In ∆ABC,
AB² + BC² = AC² [By Pythagoras Theorem]
(8)² + (BC)² = (10)²
64 + BC² = 100
BC² = 100 – 64
BC = √36
BC = 6 cm.
∴ Distance between fóot of ladder and foot of wall = 6 cm.

Question 10.
A guy wire attached to a vertical pole of height 18 m Is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is height of pole (AB) = 18 m
AC is length of wire = 24 m

C is position of stake AB at ground level.
In right angle triangle ABC,
AB² + BC² = AC² [By Pythagoras Theorem]
(18)² + (BC)² = (24)²
324 + (BC)² = 576
BC² = 576 – 324
BC = \(\sqrt{252}=\sqrt{36 \times 7}\)
BC = 6√7 m.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two pLanes after 1\(\frac{1}{2}\) hours?
Solution:
Speed of first aeroplane = 1000km/hr.

Distance covered by first aeroplane due north in 1\(\frac{1}{2}\) hours =1000 × \(\frac{3}{2}\)
OA = 1500 km
Speed of second aeroplane = 1200 km/hr.
Distance covered by second aeroplane in 1\(\frac{1}{2}\) hours = 1200 × \(\frac{3}{2}\)
OB = 1800 km.
In right angle ∆AOB
AB² = AO² + OB² [By Phyrhagoras Theorem]
AB² = (1500)² + (1800)²
AB = \(\sqrt{2250000+3240000}\)
= \(\sqrt{5490000}\)
= \(\sqrt{61 \times 90000}\)
AB = 300√61 km.
Hence, Distance between two aeroplanes = 300√61 km.

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the
distance between their tops.
Solution:
Height of pole AB = 11 m
Height of pole (CD) = 6 m

Distance between foot of pole = 12 m
from C draw CE ⊥ AB. such that
BE = DC = 6 m
AE = AB – BE = (11 – 6) m = 5 m.
and CE = DB = 12 m.
In rt. ∠d ∆AEC,
AC² = AE² + FC²
[By Phythagoras Theorem)
AC = \(\sqrt{(5)^{2}+(12)^{2}}\)
= \(\sqrt{25+144}\)
= \(\sqrt{169}\) = 13.
Hence, Distance between their top = 13m.

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
Prove that AE² + BD² = AB² + DE².
Solution:
Given: In right angled ∆ABC, ∠C = 90° ;
D and E are points on sides CA & CB respectively.
To prove: AE² + BD² = AB² + DE²
Proof: In rt. ∠d ∆BCA,
AB² = BC² + CA² …………..(1) [By Pythagoras Theorem]

In rt. ∠d ∆ECD,
DE² = EC² + DC² ……………….(2) [By Pythagoras Theorem]
In right angled triangle ∆ACE,
AE² = AC² + CE² ……………….(3)
In right angled triangle ∆BCD
BD² = BC² + CD² ……………….(4)
Adding (3) and (4),
AE² + BD² = AC² + CE² + BC² + CD²
= [AC² + CB²] + [CE² + DC²]
= AB² + DE²
[From (1) and (2)]
Hence 2 + BD² = AB² + DE².
Which is the required result.

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB² = 2AC² + BC².

Solution:
Given: ∆ABC, AD ⊥ BC
BD = 3CD.
To prove: 2AB² = 2AC² + BC².

Proof: In rt. ∠d triangles ADB and ADC, we have
AB² = AD² + BD²;
AC² = AD² + DC² [By Pythagoras Theorem]
∴ AB² – AC² = BD² – DC²
= 9 CD² – CD²; [∵ BD = 3CD]
= 8CD² = 8 (\(\frac{\mathrm{BC}}{4}\))²
[∵ BC = DB + CD = 3 CD + CD = 4 CD]
∴ CD = \(\frac{1}{4}\) BC
∴ AB² – AC² = \(\frac{\mathrm{BC}^{2}}{2}\)
⇒ 2(AB² – AC²) = BC²
⇒ 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC².
Which is the required result.

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9 AD² = 7 AB².
Solution:
Given: Equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC.
To prove: 9AD² = 7 AB².
Construction: AB ⊥ BC.
Proof: ∆AMB ≅ ∆AMC [By R.HS. Rule since AM = AM and AB = AC]

∴ BM = MC = \(\frac{1}{2}\) BC [c.p.c.t.]
Again BD = \(\frac{1}{3}\) BC and DC = \(\frac{1}{3}\) BC (∵ BC is trisected at D)
Now in ∆ADC, ∠C is acute
∴ AD² = 2AC² + DC² – 2 DC × MC
= AC² + \(\left[\frac{2}{3} \mathrm{BC}\right]^{2}\) – 2 \(\left[\frac{2}{3} \mathrm{BC}\right] \frac{1}{2} \mathrm{BC}\)

[∵ DC = \(\frac{2}{3}\) BC and MC = \(\frac{1}{2}\) BC]
= AB² + \(\frac{4}{9}\) AB² – \(\frac{2}{3}\) AB²
[∵ AC = BC = AB]
= (1 + \(\frac{4}{9}\) – \(\frac{2}{3}\)) AB²

= \(\left(\frac{9+4-6}{9}\right) \mathrm{AB}^{2}=\frac{7}{9} \mathrm{AB}^{2}\)

∴ AD² = \(\frac{7}{9}\) AB²
⇒ 9 AD² = 7 AB².

Question 16.
In an equilateral triangle, prove that three times the square of one side Ls equal to four times the square of one of its
altitudes.
Solution:
Given:
ABC is equilateral ∆ in which AB = BC = AC

To prove: 3 AB² = 4 AD²
Proof: In right angled ∆ABD,
AB² = AD² + BD² (Py. theorem)
AB² = A BD² (Py. theorem)

AD² = \(\frac{3}{4}\) AB²
⇒ 4 AD² = 3 AB²
Hence, the result.

Question 17.
Tick the correct answer and justify: In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6√3 cm. [The angles of B are respectively
(A) 120°
(B) 64°
(C) 90°
(D) 45°
Solution.
AC = 12 cm
AB = 6√3 cm
BC = 6 cm
AC² = (12)² = 144 cm
AB² + BC² = (6√3)² + (6)²
= 108 + 36
AB√3 + BC√3 = 144
∴ AB√3 + BC√3 = AC√3
Hence by converse of pythagoras theorem ∆ABC is right angred triangle right angle at B
∴ ∠B = 90°
∴ correct option is (C).

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.4

1. Write all the names of the following triangles in all orders:

Solution:
(i) ∆ABC, ∆ACB, ∆BAC, ∆BCA, ∆CAB, ∆CBA.
(ii) ∆XYZ, ∆XZY, ∆YZX, ∆YXZ, ∆ZXY, ∆ZYX
(iii) ∆LMN, ∆LNM, ∆MNL, ∆MLN, ∆NML, ∆NLM.

2. Write the name of vertices, sides and angles of the following triangles:

Solution:

	(i)	(ii)	(iii)
Vertices	P, R, Q	D, E, F	T, P, S
Sides	PR, QR, PQ	DE, EF, DF	TP, PS, TS
Angles	\(\angle \mathrm{P}, \angle \mathrm{R}, \angle \mathrm{Q}\)	\(\angle \mathrm{D}, \angle \mathrm{E}, \angle \mathrm{F}\)	\(\angle \mathrm{T}, \angle \mathrm{P}, \angle \mathrm{S}\)

3. In the given figure, name the points that lie:

Question (i)
On the boundary of ∆GEM
Solution:
Points on the boundary of AGEM are: G, A, E, C, M

Question (ii)
In the interior of ∆GEM
Solution:
Points in the interior of AGEM are : P, X, D

Question (iii)
In the exterior of ∆GEM.
Solution:
Points in the exterior of AGEM are : Y, B.

4. In the given figure, write the name of:

Question (i)
All different triangles
Solution:
All different triangles are :
∆AOD, ∆DOC, ∆BOC, ∆AOB, ∆ABD, ∆BCD, ∆ACD, ∆ABC

Question (ii)
Triangles having O as the vertex
Solution:
Triangles having O as the vertex are :
∆AOB, ∆BOC, ∆COD, ∆AOD

Question (iii)
Triangles having A as the vertex.
Solution:
Triangles having A as the vertex are:
∆AOB, ∆AOD, ∆ABD, ∆ABC, ∆ACD.

5. Fill in the blanks of the following:

Question (i)
A triangle has …………… vertices.
Solution:
3

Question (ii)
A triangle has …………… angles.
Solution:
3

Question (iii)
A triangle has ……………. sides.
Solution:
3

Question (iv)
A triangle divide the plane into ……………. parts.
Solution:
3

Question (v)
A triangle has …………… parts.
Solution:
6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 8 Basic Geometrical Concepts Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 8 Basic Geometrical Concepts Ex 8.3

1. Name the given angles in all ways:

Solution:
(i) \(\angle \mathrm{DEF}, \angle \mathrm{FED}, \angle \mathrm{E}, \angle a\)
(ii) \(\angle \mathrm{XOY}, \angle \mathrm{YOX}, \angle \mathrm{O}, \angle 1\)
(iii) \(\angle N O M, \angle M O N, \angle O, \angle x\)

2. Name the vertex and the arms of given angles:

Solution:

	(i)	(ii)	(iii)
Vertex	B	Q	o
Arm	\(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{BA}}\)	\(\overrightarrow{\mathrm{QP}}, \overrightarrow{\mathrm{QR}}\)	\(\overrightarrow{\mathrm{OS}}, \overrightarrow{\mathrm{OP}}\)

3. Name all the angles of the given figure:

Solution:
(i) \(\angle \mathrm{X}, \angle \mathrm{Y}, \angle \mathrm{Z}\)
(ii) \(\angle \mathrm{P}, \angle \mathrm{Q}, \angle \mathrm{R}, \angle \mathrm{S}\)
(iii) \(\angle \mathrm{AOB}, \angle \mathrm{BOC}, \angle \mathrm{AOC}\)

4. In the given figure, name the points that lie:

Question (i)
In the interior of \(\angle \mathrm{DOE}\)
Solution:
Points in the interior of \(\angle \mathrm{DOE}\) are :
A, X, M

Question (ii)
In the exterior of \(\angle \mathrm{DOE}\)
Solution:
Points in the exterior of \(\angle \mathrm{DOE}\) are :
H, L

Question (iii)
On the \(\angle \mathrm{DOE}\)
Solution:
Points on the \(\angle \mathrm{DOE}\) are :
D, B, O, E.

5. In the given figure, write another name for the following angles :

Question (i)
\(\angle \mathrm{1}\)
Solution:
\(\angle S \text { or } \angle PSR \text { or } \angle RSP\)

Question (ii)
\(\angle \mathrm{2}\)
Solution:
\(\angle \mathrm{RPQ} \text { or } \angle \mathrm{QPR}\)

Question (iii)
\(\angle \mathrm{3}\)
Solution:
\(\angle \mathrm{SRP} \text { or } \angle \mathrm{PRS}\)

Question (iv)
\(\angle \mathrm{a}\)
Solution:
\(\angle \mathrm{Q} \text { or } \angle \mathrm{RQP} \text { or } \angle \mathrm{PQR}\)

Question (v)
\(\angle \mathrm{b}\)
Solution:
\(\angle \mathrm{PRQ} \text { or } \angle \mathrm{QRP}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Answer:
Height of cylinder h = 14 cm.
Curved surface area of a cylinder = 2 πrh
∴ 88 cm² = 2 × r × 14cm
∴ \(\frac{88 \times 7}{2 \times 22 \times 14}\) cm = r
∴ r = 1 cm
Now, diameter of the cylinder = 2r = 2 × 1 cm
= 2 cm
Thus, the diameter of the base of the cylinder is 2 cm.

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same ?
Answer:
Height of cylindrical tank h = 1 m
Diameter of the cylinder =140 cm
∴ Radius of the cylinder r = \(\frac{\text { diameter }}{2}\)
= \(\frac{140}{2}\) cm
= 70 cm
= 0.7 m
Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.7 (0.7 + 1) m²
= 4.4 × 1.7 m²
= 7.48 m²
Thus, 7.48 m² sheet is required to make the closed cylindrical tank.

Question 3.
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see the given figure). Find its

(i) inner curved surface area,
Answer:
For inner cylinder, diameter = 4 cm
∴ For inner cylinder,
radius r = \(\frac{\text { diameter }}{2}\) = 2 cm
and height (length) h = 77 cm.
Inner curved surface area of the pipe
= 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 968 cm²
Thus, the inner curved surface area is 968 cm².

(ii) outer curved surface, area,
Answer:
For outer cylinder, diameter = 4.4 cm
∴ For outer cylinder,
radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{4.4}{2}\) = 2.2
and height h = 77 cm.
Outer curved surface area of the pipe
= 2πRh
= 2 × \(\frac{22}{7}\) × 2 × 77 cm²
= 1064.8 cm²
Thus, the outer curved surface area is
1064.8 cm².

(iii) total surface area.
Answer:
Total surface area includes the area of two circular rings at the ends together with the inner and outer curved surface areas.
For each circular ring, outer radius R = 2.2 cm and inner radius r = 2 cm
Area of one circular ring
= π(R² – r²)
= \(\frac{22}{7}\)(2.2² – 2²)cm²
= \(\frac{22}{7}\) (4.84 – 4) cm²
= \(\frac{22}{7}\) × 0.84 cm²
= 2.64 cm²
∴ Area of two circular rings.
= 2 × 2.64 cm²
= 5.28 cm²
Now, total surface area of the pipe = Inner curved surface area + outer curved surface area + area of two circular rings
= 968 + 1064.8 + 5.28 cm²
= 2038.08 cm²

Question 4.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Answer:
For the cylindrical roller, diameter d = 84 cm and height (length) h = 120 cm.
Curved surface area of the cylindrical roller
= πdh
= \(\frac{22}{7}\) × 84 × 120 cm²
= 31680 cm²
= \(\frac{31680}{10000}\) m²
= 3.168 m²
Thus, the area of playground levelled in 1 complete revolution of the roller = 3.168 m²
∴ The area of playground levelled in 500 complete revolutions of the roller
= 3.168 × 500 m² = 1584 m²
Thus, the area of the playground is 1584 m².

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹ 12.50 per m².
Answer:
For the cylindrical pillar, diameter d = 50 cm = 0.5 m and height h = 3.5 m.
Curved surface area of the cylindrical pillar
= πdh
= \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 5.5 m²
Cost of painting 1 m² area = ₹ 12.50
∴ Cost of painting 5.5 m2 area = ₹ (12.50 x 5.5)
= ₹ 68.75
Thus, the cost of painting the curved surface of the pillar is ₹ 68.75.

Question 6.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Answer:
For the given cylinder, radius r = 0.7 m and
curved surface area = 4.4 m².
Curved surface area of a cylinder = 2πrh
∴ 4.4 m² = 2 × \(\frac{22}{7}\) × 0.7m × h
∴ h = \(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\)m
∴ h = 1 m
Thus, the height of the cylinder is 1 m.

Question 7.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of ₹ 40 per m².
Answer:
A circular well means a cylindrical well. For the cylindrical well, diameter d = 3.5 m and height (depth) h = 10 m.
(i) Curved surface area of the well
= πdh
= \(\frac{22}{7}\) × 3.5 × 10 m²
= 110 m²

(ii) Cost of plastering 1 m² region = ₹ 40
∴ Cost of plastering 110 m2 region
= ₹ (40 × 110)
= ₹ 4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Answer:
For the cylindrical pipe, diameter d = 5 cm = 0.05 m and height (length) h = 28 m.
The radiation surface in the system is the •curved surface of the pipe.
Hence, we find the curved surface area of the cylindrical pipe.
Curved surface area of the cylindrical pipe
= πdh
= \(\frac{22}{7}\) × 0.05 × 28 m²
= 4.4 m²
Thus, the total radiating surface in the system is 4.4 m².

Question 9.
Find: (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \(\frac{1}{12}\) of the steel actually used was wasted in making the tank.
Answer:
For the closed cylindrical tank, diameter d = 4.2 m, hence radius
r = \(\frac{4.2}{2}\) = 2.1 m and height h = 4.5 m.

(i) Curved surface area of the cylindrical tank
= 2 πrh
= 2 × \(\frac{22}{7}\) × 2.1 × 4.5 m²
= 59.4 m²

(ii) Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 2.1 (2.1 + 4.5) m²
= 13.2 × 6.6 m²
= 87.12 m²
Suppose, x m² steel was used for making the tank. But during production, \(\frac{1}{12}\) of the steel was wasted.
∴ Actual quantity of steel used = \(\frac{11}{12}\)x m².
Hence, \(\frac{11}{12}\)x = 87.12
∴ x = \(\frac{8712}{100} \times \frac{12}{11}\)
∴ x = 95.04 m²
Thus, the quantity of steel actually used during the preparation of the tank is 95.04 m².

Question 10.
In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer:
The shape of the decorative cloth will be cylindrical.
For the cylinder of cloth, diameter d = 20 cm and height h = 30 cm + 2.5 cm + 2.5 cm = 35 cm.
Curved surface area of the cylinder of cloth
= πdh
= \(\frac{22}{7}\) × 20 × 35 cm²
Thus, 2200 cm² cloth is required for covering the lampshade.

Question 11.
The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboards. If there were 35 competitors, how much cardboard was required to be bought for the competition ?
Answer:
The cylindrical penholders to be made have base but open at the top. Thus, to prepare a penholder, the area of the cardboard required will be given by the curved surface area of the cylinder and the area of base.

For cylindrical penholder, radius r = 3 cm and height h = 10.5 cm.
Area of cardboard required for 1 penholder
= Curved surface area of cylinder + Area of base
= 2πrh + πr²
= πr (2h + r)
= \(\frac{22}{7}\) × 3(2 × 10.5 + 3) cm2
= \(\frac{66 \times 24}{7}\) cm²
∴ Area of the cardboard required for 35 penholders
= 35 × \(\frac{66 \times 24}{7}\) cm²
= 7920 cm²
Thus, 7920 cm² cardboard was required to be bought for the competition.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m² costs ₹ 20.
Answer:
The plastic box to be made is open at the top. Hence, the plastic sheet is required for the lateral surfaces and the base.
Here, for the box to be made,
length l = 1.5 m;
breadth b = 1.25 m and
height h = 65 cm = 0.65 m.
Area of the plastic sheet required for open box = Lateral surface area + Area of base
= 2 h(l + b) + l × b
= 2 × 0.65 (1.5 + 1.25) + 1.5 × 1.25 m²
= 1.3 × 2.75 + 1.875 m²
= 3.575 + 1.875 m²
= 5.45 m²
Cost of 1 m² sheet = ₹ 20
∴ Cost of 5.45 m² sheet = ₹ (5.45 × ₹ 20)
= ₹ 109

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².
Answer:
For the given room, length 1 = 5 m; breadth b = 4 m and height h = 3 m.
Area of the region to be white washed
= Area of four walls + Area of ceiling
= 2 h(l + b) + l × b
= 2 × 3 (5 + 4) + 5 × 4 m²
= 54 + 20 m²
= 74 m²
Cost of white washing 1 m² region = ₹ 7.5
∴ Cost of white washing 74 m² region
= ₹ (74 × 7.5)
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Answer:
Area painted at the cost of ₹ 10 = 1 m²
∴ Area painted at the cost of ₹ 15,000
= \(\frac{15000}{10}\)
= 1500 m²
∴ Area of the four walls = 1500m²
∴ Lateral surface area = 1500 m²
∴ Perimeter Of the floor × Height = 1500 m²
∴ 250 m × Height = 1500 m²
∴ Height = \(\frac{15000}{250}\)
∴ Height = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm× 7.5 cm can be painted out of this container?
Answer:
For each brick, length l = 22.5 cm; breadth b = 10 cm and height h = 7.5 cm.
Total surface area of one brick
= 2 (lb + bh + hl)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2 (225 + 75 + 168.75) cm²
= 2 (468.75) cm²
= 937.5 cm²
= \(\frac{937.5}{10000}\) m² = 0.09375 m²
No. of bricks that can be painted with paint sufficient to paint 0.09375 m² area = 1
∴ No. of bricks that can be painted with paint sufficient to paint 9.375 m² area 9.375
= \(\frac{9.375}{0.09375}\) = 100

Question 5.
A cubical box has each edge 10 cm and |> another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much ?
Answer:
For the cubical box, edge a = 10 cm and for the cuboidal box, length l = 12.5 cm; breadth b = 10 cm and height h = 8 cm
(i) Lateral surface area of cubical box
= 4a²
= 4 (10)² cm²
= 400 cm
Lateral surface area of cuboidal box
= 2h(l + b)
= 2 × 8(12.5 + 10) cm²
= 16 × 22.5 cm²
= 360 cm²
Thus, the lateral surface area of cubical box is greater by 40 cm² (400 – 360).

(ii) Total surface area of cubical box = 6a2
= 6 (10)² cm²
= 600 cm²
Total surface area of cuboidal box
= 2 (lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm²
= 2 (125 + 80 + 100) cm²
= 2 (305) cm²
= 610 cm²
Thus, the total surface area of cubical box is smaller by 10 cm² (610 – 600).

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass ?
(ii) How much of tape is needed for all the 12 edges ?
Answer:
(i) For the cuboidal greenhouse, length l = 30 cm; breadth fa = 25 cm and height h = 25 cm.
Area of glass used
= Total surface area of cuboid
= 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 25 + 25 × 30) cm²
= 2 (750 + 625 + 750) cm²
= 2 (2125) cm²
= 4250 cm²

(ii) 12 edges of the cuboidal greenhouse is made-up of 4 lengths, 4 breadths and 4 heights.
∴ Length of tape needed for 12 edges
= 4l + 4b + 4h
= 4 (l + b + h)
= 4 (30 + 25 + 25) cm
= 4 (80) cm
= 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
For bigger cuboidal boxes, length l = 25 cm;
breadth b = 20 cm and height h = 5 cm.
Total surface area of a bigger box
= 2 (lb + bh + hl)
= 2 (25 × 20 + 20 × 5 + 5 × 25) cm²
= 2 (500 + 100 + 125) cm²
= 1450 cm²
Area of cardboard required for overlap
= 5 % of 1450 cm²
= 72.5 cm²
Thus, the total area of cardboard required for 1 bigger box = 1450 + 72.5 cm²
= 1522.5 cm²
∴ The total area of cardboard required for 250 bigger boxes = (1522.5 × 250) cm²
For smaller cuboidal boxes, length l = 15 cm; breadth b = 12 cm and height h = 5 cm.
Total surface area of a smaller box
= 2 (lb + bh + hl)
= 2 (15 × 12 + 12 × 5 + 5 × 15) cm²
= 2(180 +60 + 75) cm²
= 2 (315) cm²
= 630 cm²
Area of cardboard required for overlap
= 5% of 630 cm²
= 31.5 cm²
Thus, the total area of cardboard required for 1 smaller box = 630 + 31.5 cm² = 661.5 cm²
∴ The total area of cardboard required for 250 smaller boxes = (661.5 × 250) cm²
Now, the total area of cardboard required for all the boxes
= (1522.5 × 250) + (661.5 × 250) cm²
= 250(1522.5 + 661.5) cm²
= 250 × 2184 cm²
Cost of 1000 cm² cardboard = ₹ 4
∴ Cost of 250 × 2184 cm² cardboard
= ₹ \(\left(\frac{4 \times 250 \times 2184}{1000}\right)\)
= ₹ 2184

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m ?
Answer:
For the box-like structure without base, length
l = 4m; breadth b = 3m and height h = 2.5m.
Area of tarpaulin required
= Area of lateral surfaces + Area of top
= 2 h(l + b) + l × b
= 2 × 2.5 (4 + 3) + 4 × 3 m²
= 35 + 12 m²
= 47 m²

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 12 Heron’s Formula MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The sides of a triangle measure 8cm, 12cm and 6 cm. Then, the semiperimeter of the triangle is ……………………… cm.
A. 26
B. 52
C. 13
D. 6.5
Answer:
C. 13

Question 2.
Each side of an equilateral triangle measures 8 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 4
B. 24
C. 12
D.36
Answer:
C. 12

Question 3.
In a right angled triangle, the length of the hypotenuse is 15 cm and one of the sides forming right angle is 9 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 36
B. 18
C. 12
D. 15
Answer:
B. 18

Question 4.
The ratio of the measures of the sides of a triangle is 3:4:5. If the semiperimeter of the < triangle is 36 cm, the measure of the longest side of the triangle is ……………………. cm.
A. 12
B. 15
C. 20
D. 30
Answer:
D. 30

Question 5.
The area of a triangle is 48 cm2 and one of its sides measures 12 cm. Then, the length of the altitude corresponding to this side is …………………. cm.
A. 4
B. 8
C. 16
D. 6
Answer:
B. 8

Question 6.
The sides of a triangle measure 12 cm, 17 cm and 25 cm. Then, the area of the triangle is ……………………….. cm².
A. 54
B. 90
C. 180
D. 135
Answer:
B. 90

Question 7.
Two sides of a triangle measure 9 cm and 10 cm. If the perimeter of the triangle is 36cm, then its area is …………………. cm².
A. 17
B. 36
C. 72
D. 18
Answer:
B. 36

Question 8.
The area of an equilateral triangle with each side measuring 10 cm is ………………….. cm².
A. \(\frac{5 \sqrt{3}}{2}\)
B. 25√3
C. 5√3
D. 3√5
Answer:
B. 25√3

Question 9.
∆ ABC is an isosceles triangle in which BC = 8 cm and AB = AC = 5 cm. Then, area of ∆ ABC = ……………………….. cm².
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12

Question 10.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar(ABCD) = …………………. cm².
A. 18
B. 9
C. 36
D. 27
Answer:
C. 36

Question 11.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar (ABCD) = …………………. cm².
A. 3.6
B. 7.2
C. 7.5
D. 6
Answer:
B. 7.2

Question 12.
In quadrilateral ABCD, AC = 10 cm. BM and DN are altitudes on AC from B and D respectively. If BM = 12cm and DN = 4 cm, then ar (ABCD) = …………………. cm².
A. 160
B. 80
C. 320
D. 480
Answer:
B. 80

Question 13.
The perimeter of rhombus ABCD is 40 cm and BD =16 cm. Then, ar (ABCD) = ……………………. cm².
A. 96
B. 48
C. 24
D. 72
Answer:
A. 96

Question 14.
The area of a rhombus is 72 cm² and one of its diagonals measures 16 cm. Then, the length of the other diagonal is ………………… cm.
A. 12
B. 9
C. 18
D. 15
Answer:
B. 9

Question 15.
PQRS is a square. If PQ = 10 cm, then PR = ……………………….. cm.
A. 10
B. 20
C. 10√2
D. 2√10
Answer:
C. 10√2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.1

1. Measure the line segments using a ruler and a divider:

Solution:
(i) PQ = 4.4 cm
(ii) CD = 3.6 cm
(iii) XY = 2.5 cm
(iv) AB = 5.8 cm
(v) LM = 5 cm.

2. Compare the line segments in the figure and fill in the blanks:

Question (i)
AB _ AB
Solution:
AB = AB

Question (ii)
CD _ AC
Solution:
CD < AC Question (iii) AC _ AD Solution: AC > AD

Question (iv)
BC _ AC
Solution:
BC < AC Question (v) BD _ CD. Solution: BD > CD.

3. Draw any line segment AB. Take any point C between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
Solution:
If A, B, C are any three points on a line such that AC + CB = AB, then we are sure that C lies between A and B.

On measuring the lengths of AB, BC and AC, we get
AB = 6 cm, AC = 4 cm, CB = 2 cm
Now, AC + CB = 4 cm + 2 cm = 6 cm
Hence, AB = AC + CB.

4. Draw a line segment AB = 5 cm and AC = 9 cm in such a way that points A, B, C are collinear. What is the length of BC?
Solution:

AB = 5 cm and AC = 9 cm
Since, A, B and C are collinear
∴ AB + BC = AC
⇒ 5 cm + BC = 9 cm
⇒ BC = 9 cm – 5 cm
= 4 cm.
Hence, Length of BC = 4 cm

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 14 Symmetry MCQ Questions

Multiple Choice Questions :

Question 1.
A polygon is said to be a regular polygon if its :
(a) All sides are equal
(b) All angles are equal
(c) Both (A) and (B)
(d) None of these.
Answer:
(c) Both (A) and (B)

Question 2.
The number of lines of symmetry for an equilater triangle is :
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(c) Three

Question 3.
The number of lines of symmetry for a square will be :
(a) Two
(b) Four
(c) Three
(d) One
Answer:
(b) Four

Question 4.
What other name can you give to the line of symmetry of an isoscele triangle ?
(a) Perpendicular name
(b) Height
(c) Median
(d) Altitude
Answer:
(c) Median

Question 5.
Which letter has only one line of symmetry ?
(a) Z
(b) H
(c) E
(d) N
Answer:
(c) E

Fill in the blanks :

Question 1.
The objects or figures that do not have any line of symmetry are called ……………… figures.
Answer:
Asymmetrical

Question 2.
Mirror reflection leads to ………………
Answer:
Symmetry

Question 3.
The angle by which the object rotates is called the angle of ………………
Answer:
rotation

Question 4.
The number of lines of symmetry for regular pentagon is ………………
Answer:
five

Question 5.
The number of lines of symmetry scalar for scalar triangle is ………………
Answer:
none

Write True/False :

Question 1.
A square has four lines of symmetry. (True/False)
Answer:
True

Question 2.
An isosceles triangle has a line of symmetry but not rotational symmetry. (True/False)
Answer:
True

Question 3.
A square has both line symmetry as well as rotational symmetry. (True/False)
Answer:
True

Question 4.
Some figures have only line symmetry. (True/False)
Answer:
True

Question 5.
The number of lines of symmetry for a quadrilateral is four. (True/False)
Answer:
False

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3

1. In the following figures, find the number of lines of symmetry and angle of rotation for rotational symmetry.

Solution:
(a) Line of symmetry 3, angle of rotation 120°
(b) Line of symmetry 4, angle of rotation 90°.

2. Name any two figures that have both line of symmetry and rotational symmetry.
Solution:
Equilateral triangle and circle

3. If a figure has two or more lines of symmetry should it have a rotational symmetry of order more than 1 ?
Solution:
Yes, Square has four lines of symmetry and rotational symmetry of order 4.

4. Following shapes have both, line symmetry and rotational symmetry. Find the number of lines of symmetry, centre of rotation and order of rotational symmetry.



Solution:
(a) 3, centroid, 3
(b) 2, Intersection of diagonals, 2
(c) 6, centre of hexagon, 6

5. Some of the english alphabets have fascinating symmetrical structures. Which capital letters have just one line of symmetry (Like E) ? Which capital letters have a rotational symmetry of order 2 (Like I) ? By attempting to think on such lines, you will be able to fill in the following table.

Solution:

6. Multiple Choice Questions :

Question (i).
If 60° is the smallest angle of rotational for a given figure what will be the angle of rotation for same figure.
(a) 150°
(b) 180°
(c) 90°
(d) 330°
Answer:
(b) 180°

Question (ii).
Which of these can not be a measure of an angle of rotation for any figure.
(a) 120°
(b) 180°
(c) 17°
(d) 90°
Answer:
(c) 17°

Question (iii).
Which of the following have both line symmetry and rotational symmetry ?
(a) An isosceles triangle
(b) A scalene triangle
(c) A square
(d) A parallelogram
Answer:
(c) A square

Question (iv).
Which of the alphabet has both multiple line and rotational symmetries ?
(a) S
(b) O
(c) H
(d) L
Answer:
(b) O

Question (v).
In the word ‘MATHS’ which of the following pairs of letters shows rotational symmetry ?
(a) M and T
(b) H and S
(c) A and S
(d) T and S
Answer:
(b) H and S

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 14 Symmetry Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2

1. Write the order of rotation for the following figures.

Solution:
(a) 2
(b) 2
(c) 5
(d) 6

2. Specify the centre of rotation, direction of rotation, angle of rotation and order of rotation for the following.

(i)

Solution:
Centre of rotation is O, direction of rotation is clockwise, Angle of rotation is 120° and order of rotation is 3.

(ii)

Solution:
Centre of rotation is P, direction of rotation is clockwise, Angle of rotation is 90° and order of rotation is 4.

(iii)

Solution:
Centre of rotation is O, direction of rotation is clockwise, Angle of rotation is 90° and order of rotation is 4.

3. Which of the following figures have rotational symmetry about the marked point (×) give the angle of rotation and order of the rotation of the figures.
(a)

Solution:
It has rotational symmetry, angle of rotation 180° and order of rotation 2.

(b)

Solution:
It has rotational symmetry, angle of rotation 90° and order of rotation 4.

(c)

Solution:
It has rotational symmetry, angle of rotation 72° and order of rotation 5.

(d)

Solution:
It has rotational symmetry, angle of rotation 60° and order of rotation 6.

(e)

Solution:
It has rotational symmetry, angle of rotation 90° and order of rotation 4.

4. Multiple choice questions :

Question (i).
The angle of rotation in an equilateral triangle is :
(a) 60°
(b) 70°
(c) 90°
(d) 120°
Answer:
(d) 120°

Question (ii).
A square has a rotational symmetry of order 4 about its centre what is the angle of rotation ?
(a) 45°
(b) 90°
(c) 180°
(d) 270°
Answer:
(b) 90°

Question (iii).
What is the order of rotational symmetry of the english alphabet Z ?
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question (iv).
Which of these letters has only rotational symmetry ?
(a) S
(b) E
(c) B
(d) P
Answer:
(a) S

Question (v).
If the smallest angle of rotation is 90° then order of symmetry is ?
(a) 1
(b) 3
(c) 4
(d) 2
Answer:
(c) 4