PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Answer:
For the cuboidal matchbox, length l = 4 cm; breadth b = 2.5 cm and height h = 1.5 cm.
Volume of a cuboidal matchbox
= l × b × h
= 4 × 2.5 × 1.5 cm3
= 15 cm3
Then, volume of 12 matchboxes = 12 × 15 cm3 = 180 cm3
Thus, the volume of a packet containing 12 matchboxes is 180 cm3.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l)
Answer:
For the cuboidal water tank, length l = 6m; breadth b = 5 m and height h = 4.5 m.
Capacity of the cuboidal tank = l × b × h
= 6 × 5 × 4.5 m3
= 135 m3
1 m3 = 1000 litres
∴ 135 m3 = 135000 litres
Thus, the given cuboidal water tank can hold 1,35,000 litres of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer:
For the cuboidal vessel, length l = 10 m;
breadth b = 8 m and capacity = 380 m3.
Capacity of a cuboidal vessel = l × b × h
∴ 380 m3 = 10 m × 8 m × h m
∴ h = \(\frac{380}{10 \times 8}\) m
∴ h = 4.75 m
The height of the cuboidal vessel must be made 4.75 m.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹ 30 per m3.
Answer:
For the cuboidal pit, length l = 8m; breadth b = 6 m and height (depth) h = 3 m.
Volume of the earth to be dugout to make the cuboidal pit = Volume of a cuboid
= l × b × h
= 8 × 6 × 3 m3
= 144 m3
Cost of digging out 1 m3 of earth = ₹ 30
∴ Cost of digging out 144 m3 of earth
= ₹ (30 × 144)
= ₹ 4320
Thus, the cost of digging the cuboidal pit is ₹ 4320.

Question 5.
The capacity of a cuboidal tank is 50,000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Answer:
For the cuboidal tank, length l = 2.5 m;
height (depth) h = 10 m and
capacity = 50,000 litres.
1000 litres = 1 m3
∴ 50,000 litres = \(\frac{50,000}{1000}\) m3 = 50 m3
Capacity of cuboidal tank = l × b × h
∴ 50 m3 = 2.5 m × b m × 10 m
∴ b = \(\frac{50}{2.5 \times 10}\) m
∴ b = 2 m
Thus, the breadth of the cuboidal tank is 2 m.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Answer:
Total requirement of water per day
= No. of people × daily requirement per person
= 4000 × 150 litres
= 6,00,000 litres
= \(\frac{6,00,000}{1000}\) m3
= 600 m3
For the cuboidal tank, length l = 20 m;
breadth b = 15 m and height h = 6 m
Capacity of the cuboidal tank = l × b × h
= 20 × 15 × 6 m3
= 1800 m3
600 m3 of water can last for 1 day in the village.
∴ 1800 m3 of water can last for \(\frac{1800}{600}\) = 3 days in the village.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Answer:
For the cuboidal godown, length l = 40 m;
breadth b = 25 m and height h = 15 m.
Capacity of cuboidal godown = l × b × h
= 40 × 25 × 15 m3
For the wooden cuboidal crate, length l = 1.5 m; breadth b = 1.25 m and height h = 0,5 m.
Volume of 1 cuboidal crate
= l × b × h
= 1.5 × 1.25 × 0.5 m3
∴The no. of crates that can be stored in the godown = \(\)
= \(\left(\frac{40}{1.25}\right) \times\left(\frac{25}{0.5}\right) \times\left(\frac{15}{1.5}\right)\)
= 32 × 50 × 10
= 16,000

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube ? Also, find the ratio between their surface areas.
Answer:
For the original cube, edge a =12 cm.
Volume of original cube = a3 = 123 cm3
= 1728 cm3
8 cubes of equal volume are made from the original cube.
∴ Volume of each new cube = \(\frac{1728}{8}\) cm3
= 216 cm3
Let the edge of new cube be A cm.
Volume of new cube = A3
∴ 216 cm3 = A3
∴ A = \(\sqrt[3]{216}\) cm = 6 cm
Thus, the side of each new cube is 6 cm.
Total surface area of original cube
= 6a2
= 6 (12)2 cm2
Total surface area of a new cube = 6A2
= 6 (6)2 cm2
\(\frac{\text { Total surface area of original cube }}{\text { Total surface area of a new cube }}\) = \(\frac{6(12)^{2} \mathrm{~cm}^{2}}{6(6)^{2} \mathrm{~cm}^{2}}\)
= \(\left(\frac{12}{6}\right)^{2}\)
= 4
= 4:1
Thus, the required ratio of the total surface area of the original cube and the total surface area of a new cube is 4:1.
Note: If the ratio of TSA of the original ‘ cube and TSA of all the new cubes is required, then it will be 1 : 2.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer:
2 km = 2000 m and 1 hour = 60 minutes
Rate of flow of water in the river
= 2 km/hour
= \(\frac{2000}{60}\) m/min
Thus, during 1 minute, water of length will flow in the sea.
Then, the water falling in sea per minute takes cuboidal shape with length l = \(\frac{2000}{60}\) m,
breadth b = 40 m and height (depth) h = 3 m.
Volume of water falling in sea per minute
= l × b × h
= \(\frac{2000}{60}\) × 40 × 3 m3
= 4000 m3
Thus, 4000 m3 of water will fall into the sea in a minute.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

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

Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm
Answer:
For the given sphere,
radius r = 10.5 cm = \(\frac{21}{2}\) cm.
Surface area of a sphere
= 4πr2
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm2 = 1386 cm2

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

(ii) 5.6 cm
Answer:
For the given sphere, radius r = 5.6cm.
Surface area of a sphere
= 4πr2
= 4 × \(\frac{22}{7}\) × 5.6 × 5.6 cm2
= 394.24 cm2

(iii) 14 cm
Answer:
For the given sphere, radius r= 14 cm.
Surface area of a sphere
= 4πr2
= 4 × \(\frac{22}{7}\) × 14 × 14 cm2
= 2464 cm2

Question 2.
Find the surface area of a sphere of diameter:
(i) 14cm
Answer:
For the given sphere, diameter d = 14 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{14}{2}\) cm = 7 cm
Surface area of a sphere
= 4πr2
= 4 × \(\frac{22}{7}\) × 7 × 7 cm2
= 616 cm2

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

(ii) 21cm
Answer:
For the given sphere, diameter d = 21 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{21}{2}\) cm
Surface area of a sphere
= 4πr2
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm2
= 1386 cm2

(iii) 3.5 m
Answer:
For the given sphere, diameter d = 3.5 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) m
= \(\frac{35}{20}\) m
Surface area of a sphere
= 4πr2
= 4 × \(\frac{22}{7}\) × \(\frac{35}{20}\) × \(\frac{35}{20}\) m2
= 38.5 m2

Note: We can also use the formula “Surface area of a sphere = πd2” as
4πr2 = π × 4r2 = π × (2r)2 = πd2, where r and d are radius and diameter of the sphere respectively.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π =3.14)
Answer:
For the given hemisphere, radius r = 10 cm.
Total surface area of a hemisphere
= 3πr2
= 3 × 3.14 × 10 × 10 cm2
= 942 cm2

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Answer:
For the first case, radius r1 of the spherical balloon is 7 cm.
Surface area of the spherical balloon in the
first case = 4πr12
= 4 × \(\frac{22}{7}\) × 7 × 7cm2
For the second case, radius r2 of the spherical balloon is 14 cm.
Surface area of the spherical balloon in the second case = 4πr22
Then, the required ratio of surface areas in two cases
= \(\frac{4 \times \frac{22}{7} \times 7 \times 7}{4 \times \frac{22}{7} \times 14 \times 14}\)
= \(\frac{1}{4}\) = 1 : 4
Thus, the required ratio is 1 : 4.
Note: Here, the ratio of radii = 7 : 14 = 1 : 2
Hence, the ratio of surface areas = \(\left(\frac{1}{2}\right)^{2}\) = \(\frac{1}{4}\) = 1 : 4, because in the formula of surface area of sphere, the degree of r is 2.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm2.
For the given hemispherical bowl, diameter = 10.5 cm.
Then, the radius r of the bowl = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= \(\frac{\frac{21}{2}}{2}\) cm
= \(\frac{21}{4}\) cm
Inner curved surface area of the hemispherical bowl
= 2 πr2
= 2 × \(\frac{22}{7}\) × \(\frac{21}{4}\) × \(\frac{21}{4}\) cm2
= \(\) cm2
= 173.25 cm2
Cost of tin-plating 100 cm2 region = ₹ 16
∴ Cost of tin-plating 173.25 cm2 region
= ₹ \(\left(\frac{16 \times 173.25}{100}\right)\)
= ₹ 27.72
Thus, the cost of tin-plating on the inner surface of the bowl is ₹ 27.72.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 6.
Find the radius of a sphere whose surface area is 154 cm2.
Answer:
For the given sphere, surface area = 154 cm2.
Surface area of a sphere = 4 πr2
∴ 154 cm2 = 4 × \(\frac{22}{7}\) × r2 cm2
∴ r2 = \(\frac{154 \times 7}{4 \times 22}\) cm2
∴ r2 = \(\frac{49}{4}\) cm2
∴ r = \(\frac{7}{2}\) cm
∴ r = 3.5 cm
Thus, the radius of the given sphere is 3.5 cm.

Question 7.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Answer:
Suppose, the diameter of the moon = d1
The diameter of the earth = 4 × d1 = 4d1
Then, the radius of the moon r1 = \(\frac{d_{1}}{2}\) and
the radius of the earth r2 = \(\frac{4 d_{1}}{2}\) = 2d1.
Now, \(\frac{\text { The surface area of the moon }}{\text { The surface area of the earth }}\) = \(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\)
= \(\frac{r_{1}^{2}}{r_{2}^{2}}\)
= \(\frac{\left(\frac{d_{1}}{2}\right)^{2}}{\left(2 d_{1}\right)^{2}}\)
= \(\frac{d_{1}^{2}}{4} \times \frac{1}{4 d_{1}^{2}}\)
= \(\frac{1}{16}\)
= 1 : 16
Thus, the ratio of the surface area of the moon and the surface area of the earth is 1 : 16.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Answer:
For the given hemispherical bowl, the inner radius is 5 cm and the thickness of steel is 0.25 cm.
∴ Outer radius r of the given hemispherical bowl = 5 + 0.25 cm = 5.25 cm.
Curved surface area of a hemisphere
= 2πr2
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25 cm2
= 2 × \(\frac{22}{7}\) × \(\frac{525}{100}\) × \(\frac{525}{100}\) cm2
= \(\frac{693}{4}\) cm2
= 173.25 cm2
Thus, the outer curved surface area of the given hemispherical bowl is 173.25 cm2.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 9.
A right circular cylinder just encloses a sphere of radius r (see the given figure). Find :
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 1
Answer:
Here,
radius of the cylinder = radius of the sphere = r and height of the cylinder h
= 2 × radius of the sphere = 2r
(i) Surface area of the sphere = 4πr2

(ii) Curved surface area of the cylinder
= 2 πrh
= 2 × 1 × r × 2r
= 4 πr2

(iii) Ratio of areas obtained in ( i ) and (ii)
= \(\frac{4 \pi r^{2}}{4 \pi r^{2}}\)
= \(\frac{1}{1}\)
= 1 : 1

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

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

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:
For the given cone, diameter d = 10.5 cm.
Then, radius r = \(\frac{10.5}{2}\) cm and slant height
l = 10 cm.
Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × \(\frac{10.5}{2}\) × 10 cm2
= \(\frac{22}{7}\) × \(\frac{105}{2}\) × cm2
= 11 × 15 cm2
= 165 cm2
Thus, the curved surface area of the given cone is 165 cm2.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:
For the given cone, diameter d = 24 m.
Then, radius r = \(\frac{24}{2}\) = 12 m and slant height l = 21 m.
Total surface area of a cone
= πr (l + r)
= \(\frac{22}{7}\) × 12(21 + 12) m2
= \(\frac{22 \times 12 \times 33}{7}\) m2
= \(\frac{8712}{7}\) m2
= 1244.57 m2
Thus, the total surface area of the given cone is 1244.57 m2.

Question 3.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find,
(i) radius of the base and
(ii) total surface area of the cone.
Answer:
For the given cone, slant height l = 14 cm and curved surface area = 308 cm2
(i) Curved surface area of a cone = πrl
∴ 308 cm2 = \(\frac{22}{7}\) × r × 14 cm
∴ \(\frac{308 \times 7}{22 \times 14}\) cm = r
∴ r = 7 cm
Thus, the radius of the base of the cone is 7 cm.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

(ii) Total surface area of a cone
= πrl + πr2
= 308 + \(\frac{22}{7}\) × 7 × 7 cm2
= 308 + 154 cm2
= 462 cm2
Thus, the total surface area of the cone is 462 cm2.

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹ 70.
Answer:
For the conical tent,
radius r = 24 m and height h = 10 m.

(i) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{10^{2}+24^{2}}\)
= \(\sqrt{100+576}\)
= \(\sqrt{676}\)
∴ l = 26 m
Thus, the slant height of the tent is 26 m.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

(ii) Area of the canvas used to make tent
= Curved surface area of the conical tent
= πrl
= \(\frac{22}{7}\) × 24 × 26 m2
= \(\frac{13728}{7}\) m2
Cost of 1 m2 canvas = ₹ 70
∴ Cost of \(\frac{13728}{7}\) m2 canvas
= ₹ \(\left(70 \times \frac{13728}{7}\right)\)
= ₹ 1,37,280
Thus, the cost of canvas required is ₹ 1,37,280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.
(Use π = 3.14)
Answer:
For the conical tent to be made, radius r = 6 m and height h = 8 m.
l2 = h2 + r2 = 82 + 62 = 64 + 36 = 100
∴ l = √100 = 10 m
Area of the tarpaulin used in making tent
= Curved surface area of conical tent
= πrl
= 3.14 × 6 × 10 m2
= 188.4 m2
Now, the width of the tarpaulin is 3 m.
∴ Length of tarpaulin required = \(\frac{188.4}{3}\) m
= 62.8 m
But, 20 cm, i.e., 0.2 m of tarpaulin is required more for margins and wastage.
∴ Total length of the tarpaulin required = 62.8 + 0.2 m = 63 m
Thus, total length of tarpaulin required is 63 m.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 210 per 100 m2.
Answer:
For the given conical tomb,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{14}{2}\) = 7 m
and slant height l = 25 m. .
Area of the region to be whitewashed
= Curved surface area of the conical tomb
= πrl
= \(\frac{22}{7}\) × 7 × 25 m2
= 550 m2
Cost of whitewashing 100 m2 region = ₹ 210
∴ Cost of whitewashing 550 m2 region
= ₹ \(\left(\frac{210 \times 550}{100}\right)\)
= ₹ 1155
Thus, the cost of whitewashing the curved surface of the tomb is ₹ 1155.

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:
For the conical cap, radius r = 7 cm and height h = 24 cm.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{24^{2}+7^{2}}\)
= \(\sqrt{576+49}\)
= \(\sqrt{625}\) = 25 cm
Area of the sheet required to make 1 conical cap
= Curved surface area of the conical cap
= πrl
= \(\frac{22}{7}\) × 7 × 25 cm2
= 550 cm2
Area of sheet required to make 1 cap = 550 cm2
∴ Area of sheet required to make 10 caps
= 550 × 10 cm2
= 5500 cm2
Thus, the area of the sheet required to make 10 caps is 5500 cm2

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled carboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}\) = 1.02)
Answer:
For the given cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{40}{2}\) = 20 cm = 0.2 m and
height h = 1 m.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{1^{2}+0.2^{2}}\)
= \(\sqrt{1.04}\)
= 1.02 m
Curved surface area of a cone
= πrl
= 3.14 × 0.2 × 1.02 m2
∴ Curved surface area of 50 cones
= 50 × 3.14 × 0.2 × 1.02 m2
= 32.028 m2
Cost of painting 1 m2 region = ₹ 12
∴ Cost of painting 32.028 m2 region
= ₹ (12 × 32.028)
= ₹ 384.34 (approx.)
Thus, the cost of painting all the 50 cones is ₹ 384.34 (approx.)

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

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.

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

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
PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 1
(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.

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

(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

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

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.

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

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

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

(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.

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

(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.
PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 2
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.

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

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.

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

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

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

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

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

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

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

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

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

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).

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

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

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

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

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

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

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

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

PSEB 9th Class Maths MCQ Chapter 12 Heron's Formula

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

PSEB 9th Class Maths MCQ Chapter 12 Heron's Formula

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

PSEB 9th Class Maths MCQ Chapter 12 Heron's Formula

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

PSEB 9th Class Maths MCQ Chapter 12 Heron's Formula

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

PSEB 9th Class Maths MCQ Chapter 12 Heron's Formula

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

PSEB 9th Class Maths Solutions Chapter 12 Heron’s Formula Ex 12.2

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2

Question 1.
A park, in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Answer:
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 1

In ∆ BCD, ∠C = 90°
∴ BD2 = BC2 + CD2
= (12)2 + (5)2
= 144 + 25
= 169
= (13)2
∴ BD = 13 m

In ∆ BCD, a = 5 m, b = 12 m and c = 13 m
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{5+12+13}{2}\) = \(\frac{30}{2}\) = 15 m
Then, s – a = 15 – 5 = 10m,
s – b = 15 – 12 = 3m and
s – c = 15 – 13 = 2 m.

Area of ∆ BCD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 10 \times 3 \times 2}\) m2
= \(\sqrt{900}\) m2
= 30 m2

Note: ∆ BCD is a right triangle.
∴ Area of ∆ BCD = \(\frac{1}{2}\) × BC × CD
= \(\frac{1}{2}\) × 12 × 5 = 30 m2

Now, in ∆ ABD, a = 9 m, b = 13 m arid c = 8 m
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{9+13+8}{2}\) = \(\frac{30}{2}\) = 15 m
Then,
s – a = 15 – 9 = 6m,
s – b = 15 – 13 = 2m and
s – c = 15 – 8 = 7 m.

Area of ∆ ABD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 6 \times 2 \times 7}\) m2
= \(\sqrt{5 \times 3 \times 3 \times 2 \times 2 \times 7}\) m2
= 6 √35 m2
= 35.5 m2 (approx.)
Then, the area of park in the shape of quadrilateral ABCD
= Area of ∆ BCD + Area of ∆ ABD
= (30 + 35.5) m2 (approx.)
= 65.5 m2 (approx.)
Thus, the area of the park is 65.5 m2 (approx.)

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Answer:
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 2
In ∆ ABC, a = 3 cm; b = 4 cm and c = 5 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{3+4+5}{2}\)
= \(\frac{12}{2}\) = 6 cm
Then,
s – a = 6 – 3 = 3 cm,
s – b = 6 – 4 = 2 cm and,
s – c = 6 – 5 = 1 cm.
Area of ∆ ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{6 \times 3 \times 2 \times 1}\) cm2
= 6 cm2
Note: Proving that ∆ ABC is a right triangle, Area of ∆ ABC = \(\frac{1}{2}\) × 3 × 4 = 6 cm2 can be obtained easily.
In ∆ ACD, a = 4 cm; b = 5 cm and c = 5 cm
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 3
Area of quadrilateral ABCD
= Area of ∆ ABC + Area of ∆ ACD
= (6 + 9.2) cm2 (approx.)
= 15.2 cm2 (approx.)

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in the given figure, s Find the total area of the paper used. ;
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 4
Answer:
The sides of the triangle in part 1 measure 5 cm, 5 cm and 1 cm.
∴ a = 5 cm, b = 5 cm and c = 1 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{5+5+1}{2}\) = \(\frac{11}{2}\) cm
Area of part 1
= Area of triangle
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 5
The length and breadth of rectangle in part II are 6.5 cm and 1 cm respectively.
Area of part II = Area of rectangle
= length × breadth
= (6.5 × 1) cm2
= 6.5 cm2
For the trapezium in part III, the parallel sides measure 1 cm and 2 cm, while both the non-parallel sides measure 1 cm each.
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 6
Drawing DM ⊥ AB and CN ⊥ AB. we get
AM = BM = \(\frac{2-1}{2}\) = \(\frac{1}{2}\) cm.
In ∆ DMA, ∠M = 90°
Area of trapezium ABCD
= \(\frac{1}{2}\) × Sum of parallel sides X Distance between parallel sides
= \(\frac{1}{2}\) × (AB + CD) × DM
= \(\frac{1}{2}\) × (2 + 1) × \(\frac{\sqrt{3}}{2}\)cm2
= \(\frac{1}{2}\) × 3 × \(\frac{\sqrt{3}}{2}\) cm2
= 1.3 cm2 (approx.)
For the right triangle in part IV the sides forming the right angle measure 6 cm and 1.5 cm.
Area of right triangle in part IV.
= \(\frac{1}{2}\) × Product of sides forming the right angle
= \(\frac{1}{2}\) × 6 × 1.5 cm2
= 4.5 cm2
The right triangle in part V is congruent to the right triangle in part IV.
∴ Area of right triangle in part V = 4.5 cm2
Now, total area of the paper used
= Areas of figures in part I to part V
= (2.5 + 6.5 + 1.3 + 4.5 + 4.5) cm2
= 19.3 cm2

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer:
In the given triangle, a = 26 cm, b = 28 cm and c = 30 cm
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{26+28+30}{2}\) = \(\frac{84}{2}\) = 42 cm
Area of triangle
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 7
= 7 × 6 × 4 × 2 cm2
= 336 cm2
The area of the triangle and the area of the parallelogram are equal.
∴ Area of the parallelogram = 336 cm2
∴ Base × Corresponding altitude = 336 cm2
∴ 28 cm × Corresponding altitude = 336 cm2
∴ Corresponding altitude = \(\frac{336}{28}\) cm
∴ Corresponding altitude = 12 cm
Thus, the height of the parallelogram is 12 cm.

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Answer:
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 8
Rhombus ABCD in the given figure represents the field.
A diagonal of a rhombus divides it into two congruent triangles.
∴ Area of rhombus ABCD = 2 × Area of ∆ ABC
In ∆ ABC, a = 30 m; b = 30 m; and c = 48 m.
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{30+30+48}{2}\) = \(\frac{108}{2}\) = 54 cm
Area of ∆ ABC
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 9
= 3 × 6 × 24 m2
= 432 m2
Now, area of the field
= area of rhombus ABCD
= 2 × area of ∆ ABC
= 2 × 432 m2
= 864 m2
Now, area of grass field available for 18 cows to graze = 864 m2
∴ Area of grass field available for 1 cow to graze = \(\frac{864}{18}\) m2 = 48 m2
Thus, each cow gets 48 m2 of grass field to graze.

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 10
Answer:
Out of 10 triangular pieces, 5 are dark coloured and 5 are light coloured.
For each triangle, a = 20 cm, b = 50 cm and c = 50 cm
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 11
Hence, the total area of 5 dark coloured cloth pieces = 5 × 200 √6 cm2 = 1000 √6 cm2
Similarly, the total area of 5 light coloured cloth pieces = 5 × 200 √6 cm2 = 1000 √6 cm2

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 12
Answer:
Let us name the square part as ABCD and the triangular part as CMN.
Suppose the length of square ABCD is xcm.
∴ In ∆ ABD, AB = AD = x cm and ∠A = 90°
The length of hypotenuse BD is given to be 32 cm.
AB2 + AD2 = BD2 (Pythagoras’ theorem)
∴ (x)2 + (x)2 = (32)2
∴ 2x2 = 1024
∴ x2 = 512
∴ x = √512
∴ x = \(\sqrt{256 \times 2}\)
∴ x = 16√2
Thus, the length of each side of square ABCD is 16 √2 cm.
Area of part I = Area of ∆ ABD
= \(\frac{1}{2}\) × AB × AD (∠A is a right angle.)
= \(\frac{1}{2}\) × 16 √2 × 16 √2 cm2
= 256 cm2
Area of part II = Area of A BCD
= \(\frac{1}{2}\) × BD × CD (∠A is a right angle.)
= \(\frac{1}{2}\) × 16 √2 × 16 √2 cm2
= 256 cm2
Note: Here, area of square ABCD can easily be found as below:
Area of square ABCD = \(\frac{(\text { Hypotenuse })^{2}}{2}\)
= \(\frac{(32)^{2}}{2}\)
= \(\frac{1024}{2}\)
= 512 cm2
To find the area of part III, we find the area of ∆ CMN.
In ∆ CMN, a = 6 cm, b = 8 cm and c = 6 cm.
∴ Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{6+8+6}{2}\) = \(\frac{20}{2}\) = 10 cm

Area of part III
= Area of ∆ CMN
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 13
= 8 × 2.24 cm2 (approx.)
= 17.92 cm2 (approx.)

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50 p per cm2.
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 14
Answer:
For each of 16 triangular tiles,
a = 9 cm; b = 28 cm and c = 35 cm
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 15
= 88.2 cm2 (approx.)
∴ Area of 16 tiles = 16 × 88.2 cm2
= 1411.2 cm2
50 paise = ₹ 0.50
Cost of polishing 1 cm2 region = ₹ 0.50
∴ Cost of polishing 1411.2 cm2 region
= ₹ (1411.2 × 0.50)
= ₹ 705.60

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Answer:
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 16
In the given figure, trapezium ABCD represents the field in which AB || CD,
AB = 25 m, BC = 14 m, CD = 10 m and DA = 13 m.
Through C, draw a line parallel to DA to intersect AB at E.
In quadrilateral AECD, AE || CD and DA || CE
∴ AECD is a parallelogram.
∴ CE = DA = 13 m and AE = CD = 10 m
Now, BE = AB – AE = 25 – 10 = 15 m
In ∆ CEB, a = 13 m; b = 15 m and c = 14 m
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.2 17
In ∆ CEB, draw CM ⊥ BE.
Area of ∆ CEB = \(\frac{1}{2}\) × BE × CM
∴ 84 m2 = \(\frac{1}{2}\) × 15 m × CM
∴ CM = \(\frac{84 \times 2}{15}\) m
∴ CM = 11.2 m
Area of parallelogram AECD
= Base × Corresponding altitude
= AE × CM
= 10 × 11.2 m2
= 112 m2
Hence, area of the field
= Area of trapezium ABCD
= Area of ∆ CEB + Area of parallelogram AECD
= 84 m2 + 112 m2
= 196 m2
Note: After finding CM = 11.2m, the area of . the field can also be found as below:
Area of the field
= Area of trapezium ABCD
= \(\frac{1}{2}\) × sum of parallel sides × distance between parallel sides
= \(\frac{1}{2}\) × (AB + CD) × CM
= \(\frac{1}{2}\) × (25 + 10) × 11.2 m2
= \(\frac{1}{2}\) × 35 × 11.2 m2
= 196 m2

PSEB 9th Class English Paragraph Writing

Punjab State Board PSEB 9th Class English Book Solutions English Paragraph Writing Exercise Questions and Answers, Notes.

PSEB 9th Class English Paragraph Writing

1. My Father

My father is an able person. He is a farmer. He is not well-educated. But he knows his work well. He is very hard-working. He is truthful and honest. So people respect him. They greet him respectfully. My father has an open mind. He tries to settle the quarrels among the people of the village. He is the wisest man of the village. He leads a simple and clean life. He does not lose temper with anybody. He is respected by one and all. I am proud of my father.

2. My Mother

Shrimati Asha is my mother. She is 40 years old. She is a kind and noble lady. She is active and smart. She is an M.A. She is a very simple lady. She has good habits. She gets up early in the morning. Then she cleans the house. She takes a bath and prays to God. She goes to temple daily. She prepares food for us. She looks after us all. She helps me in my studies. At night, she tells us stories. She loves me very much. I am proud of my mother. May she live long !

PSEB 9th Class English Paragraph Writing

3. My Younger Brother

Surinder is my younger brother. He is twelve years old. He is strong and healthy. He is tall for his age. He is very intelligent. He is honest. He never tells a lie. He is hard-working and obedient. He reads in the 6th class. He is the monitor of his class. He stands first in his class. He does his homework daily. He does not mix with bad boys. All the teachers love him. He is very fond of cricket. He is a member of the school cricket team. He is a good singer also. He is fond of reading storybooks. We are proud of him.

4. The Person I Dislike Most

Mr. Chaudhry, our next-door neighbour, is the person I dislike most. He has made our life miserable. Whenever he sees that we are studying, he switches on his TV at full volume. He has a big dog. He keeps it unchained. Many a time the dog bites people going through the street. Mr. Chaudhry’s wife is a very quarrelsome lady. She quarrels over trifles. She has six children. They make mischiefs all the day. Their mother never scolds them. These children are very rude. They know no manners. They write dirty words on the walls. But nobody dares complain against them to their parents. It is really a curse to have such a neighbour.

5. My School

I read in Arya High School, Ludhiana. It is a very big school. It has one huge gate. It has two storeys. There are fifty rooms. The rooms are airy. Each room has two electric fans. The hall of our school is very big. The school has two big playgrounds. It has a beautiful garden also. There are ten classes in our school.

PSEB 9th Class English Paragraph Writing

Each class has four sections. Each section has about sixty students. Our Headmaster is very able. He is very kind to the students. He is very hardworking. The teachers of our school are also able and hard-working. They love the students and the students respect them. Our school shows very good results every year. I love my school. I am proud of it. May it propser day and night!

6. Our Headmaster

Sh. Sohan Lal is the headmaster of our school. He is forty years old. He is tall and strong. He is active and smart. He is an M.A., B.Ed. Our headmaster is true to his duty. He is very punctual. He comes to school in time. He sits in his office. He works very hard. He plans his work well. He is very intelligent. He watches the working of the school. All the teachers and students respect him. He is a good teacher. He is a good speaker. He is a good writer also. He has written many books. He is a good player. He plays games in the evening. He is all in all in our school. We are proud of him. May he live long !

7. The Prize Distribution Function

The prize distribution function of our school was held on the seventh of March this year. The Education Minister presided over the function. The Minister took his seat and the function began. The Headmaster spoke a few words to welcome the guests and the Minister. Then the Headmaster requested the Minister to give away the prizes. The Minister shook hands with the prize-winners. All the prize-winners were loudly cheered. After giving away the prizes, the Minister made a short speech. He congratulated the prize-winners. He congratulated the Headmaster and the staff on their excellent work. In the end, the Headmaster thanked the Minister and the function was over. Tea was served to the guests and the prize-winners.

8. Our School Peon

Ramu is our school peon. He is twenty-five years old. He is tall and strong. He wears a khaki uniform. He is obedient and honest. He is true to his duty. He respects the teachers. He always speaks the truth. He knows his job well. Ramu lives in the school. He gets up early in the morning. He sweeps the school. He dusts the office. He rings the bell at the right time. He is busy the whole day. He is loyal to the school. He looks after the school property. His duty is hard, but his pay is small. I pity his lot.

9. A Postman

The postman is a very useful public servant. His duty is very hard. He has to do his duty in sun and rain. He goes to the Head Post Office in the morning. There he gets the dak. He arranges the letters. He puts them in a bag. Then he goes on his beat. He goes from door to door. He is eagerly waited for. He brings good as well as bad news. He helps to bring the world closer. A postman has to work hard. But his pay is small. He can hardly make both ends meet. I pity his lot.

PSEB 9th Class English Paragraph Writing

10. A Rickshaw-Puller 

The life of a rickshaw-puller is very hard. He lives by the sweat of his brow. He pulls heavy loads. He pulls men, women and children. It is very painful to look at him. A rickshaw-puller hardly gets as much as he deserves. People try to give him as little as possible. He has to work in sun and rain from morning till evening. Even then, he gets very little to eat. He is in rags. It is unlucky that even in this age of science men have to work like beasts of burden to earn their bread.

11. The Diwali Festival

Diwali is an important Indian festival. It falls in the month of October or November. It comes twenty days after Dussehra. Shri Ram came back to Ayodhya on this day. Shri Guru Hargobind was set free by the Mughal Emperor on this day. This festival is celebrated in every village and town. Houses and shops are painted in new colours. People light their homes with candles and electric lights. They buy sweets and toys. They distribute gifts among friends and relatives. Children enjoy fireworks at night. On this day, people worship goddess Lakshmi Some people gamble on this day. It is evil. It should be ended.

12. The Dussehra Festival

Dussehra is an important Hindu festival. It comes off in October. Rama defeated Ravar on this day. It marks the victory of good over evil. The festival lasts for ten days. Ram Lila staged at night. Many people come to see this Lila. On the last day, a fair is held. Many people come to see the fair. Everyone looks happy. Effigies of Ravana, Meghnada and Kumbhakarn are set up. Rama shoots arrows at the effigies. At about sunset, Hanumana sets them on fire After this people come back to their homes. They feel happy.

13. The Independence Day

India became a free country on August 15, 1947. So, August 15 is called the Independence Day of India. The British rule came to an end on this day. It is a red-letter day in the history of the country. It is celebrated all over the country with great enthusiasm. On this day, all schools, colleges and offices remain closed. It is a national holiday. Public meetings are held in all towns and cities. A big function is held in Delhi. The Prime Minister unfurls the national flag at the Red Fort. Our freedom is a hard-won freedom. We should protect it.

14. The Republic Day

India became a Republic on January 26, 1950. The Constitution of the country came into force on this day. India became a secular democratic country. The power of government passed into the hands of the common people. All castes, creeds and religions are to be equal in the eyes of the law. It is a red-letter day in the history of the country. It is celebrated all over the country with great enthusiasm. The national flag is unfurled at all the public buildings. A big function is held in Delhi. The President of the country presides over this function. It is worth seeing. This day is a national holiday.

15. Mahatma Gandhi

Mahatma Gandhi was born on October 2, 1869. He was unlike other boys. He was very gentle. He loved truth. He respected his teachers. After doing law he started practice in India. He did not take up false cases. He went to Africa to fight a case. There he saw the poor Indians. The English treated them badly. Gandhiji fought for their rights for ten years. Then he came back to India. He fought for the freedom of India. He gave us a new way of fighting. It was ‘ahimsa’. It was more powerful than violence. He was able to free India in 1947. He was a real Mahatma. He led a very simple life. He is called the Father of our Nation. A mad person shot him dead on January 30, 1948. Gandhiji’s name will always be remembered.

16. An Ideal Student

An ideal student is a knowledge-seeker in the real sense. He obeys his teachers. He has full confidence in them. He is regular and punctual. He works hard at studies. But he takes part in games also. He does not read cheap and dirty literature. He reads only good and useful books. An ideal student believes in simple living and high thinking. He knows the value of discipline. He does not waste the hard-earned money of his parents. An ideal student is a true patriot. In short, he has all the qualities of head and heart.

PSEB 9th Class English Paragraph Writing

17. The Recess Period

The recess is the period of enjoyment. In this period, the students feel happy. They enjoy freedom for some time. As soon as the recess period begins, students rush out of their classrooms. Some of them run to the vendors. They buy things to eat. Others go to the taps to drink water. There is great rush in the playground. Some love to play there while others like to sit under the shady trees. They talk about their friends and teachers. Soon the bell goes. Students run back to their classes. The students feel fresh and start their studies once again.

18. A One-Day Cricket Match

Last Sunday, a one-day cricket match was played between our school and Arya High School. Each team played 40 overs. The match started at 10 a.m. We won the toss. We decided to bat first. Mohan and Gopal were our openers. Mohan made 30 runs and was out. Now Raja came in to bat. He did not play well. He was out for a duck. The next four players made 60 runs. Our team was out at 120 runs. Now it was the turn of Arya High School. They had good openers. They made 60 runs. Their third batsman was a hitter. He made 30 runs. But the other players were soon out. Their team could make only one hundred runs. We won the match by 20 runs. It was really a very interesting match.

19. A Football Match

Last Monday, a football match was played between our school and Khalsa School. It was played on our school grounds. Sh. Jaswant Singh was the referee. He blew the whistle. There was a toss. We won the toss. We chose our side. The match began. At first, the game was slow, but soon it became brisk. All the players played well. Our defence was very strong. There was no goal. The referee blew the whistle for interval. In the second half, Vinod passed the ball on to me. I ran with it into the Then I kicked it hard. It went through the poles. It was a goal. There were loud cheers. The referee blew the whistle. The game was over. We won the march by one goal.

20. A Kabaddi Match

I saw a kabaddi match last Sunday. It was played between our school and New High School. Sh. Mohan Lal was the referee. Many people came to see the match. There was a toss. We won the toss. We chose our side. Then the match began. First of all, our captain went running to the other side. He shouted, “Kabaddi, Kabaddi.” He came back. There was no point. Now it was the turn of New High School.

PSEB 9th Class English Paragraph Writing

One of their players came to our side. He was caught. He could not go back. We scored a point. There were loud cheers. We scored more points. New High School team got only 8 points. We had gained 20 points. The referee blew a long whistle. The match came to an end. We won the match by 12 points. It was an interesting match.

21. Morning Walk

Morning walk is the best form of exercise. It costs nothing. It is very useful for our health. It refreshes our mind. It strengthens our body. It saves us from many diseases. Morning walk keeps us fresh for the whole day. It develops in us the habit of rising early. It brings pure thoughts in our mind. The dew drops, the fresh flowers, the chirping birds and the rustling leaves charm our mind. We start loving these objects of natural beauty. Thus, morning walk is useful not only for our body but for our mind also.

22. A Journey by Bus

Last Sunday, I went to Delhi by bus. I went to the bus stand and bought a ticket. A bus bound for Delhi was standing there. I got in and took the front seat. The conductor gave a whistle and the bus started. ‘We were soon out of the city. The driver drove very fast. But he was very good at his job. We felt quite safe. He left many buses behind. I saw farmers working in their fields. Here and there, I saw carts going on the road. The conductor was a jolly fellow. He made the journey pleasant by his witty talk. The bus reached Delhi at 6 p.m. It was a very pleasant journey.

23. A Journey by Train

Last year, I went to Delhi by train. I packed my luggage. I hired a rickshaw. I reached the station. I bought a ticket. I went to the platform. Soon the train arrived. I got into it. There was a great rush. But I was lucky. I got a seat near the window. The train started. I saw many things on the way. Farmers were ploughing the fields. Children were playing. A ticket-checker came. He checked our tickets. A young man was without ticket. He was fined. The train stopped at many stations. I bought a newspaper, I read it. It was 10 a.m. The train reached Delhi. It was a happy journey.

24. A Visit to a Zoo

There is a zoo in our city. I visited it last Sunday. I went with my parents. We bought tickets and went in. First of all, we saw birds. There were many beautiful and rare kinds of birds. We saw parrots, canaries, swallows, peacocks, ducks, cranes, herons, gulls and geese. Then we saw some wild beasts. A lioness and her cub were basking in the sun. They roared now and then. We also saw wolves, tigers, elephants and rhinos. When we were coming back, we saw a muddy pond. There were many big snakes in it. It was fearful to look at them. We stayed in the zoo for about three hours. Then we came back home.

25. A Visit to a Fair 

I went to see a fair last Tuesday. This fair is held every year in our town. It is held in the memory of a pious faqir. Many people go to see this fair. They include men of all religions and faiths. This year I went to see the fair with my parents. We offered flowers at the faqir’s tomb.

PSEB 9th Class English Paragraph Writing

Then we went round the fair. There was a temporary bazaar. Stalls were arranged on either side. There was a great hustle and bustle. Sweets were in great demand. Children were enjoying rides in merry-go-rounds. A big shamiana was set up on one side of the fair. Qawalis were being sung there. We sat there for some time. Then we came back home.

26. A Visit to a Circus

A circus came to our town last month. I went to see it with my parents. We bought tickets and went in. We took our seats in the front row. First of all, a young girl came in. She had an umbrella in her hand. She walked on a rope. Then some more girls joined her. They showed various feats in gymnastics. They looked like rubber dolls. One of the girls jumped through a fire ring. Then there were animal shows. An elephant drank water from a bottle. A lion and a goat played with each other. A monkey drove a mini-cycle. The show came to an end at 7 p.m. I liked it very much.

27. A Visit to a Historical Place

During the last spring holidays, I went to Agra. There I visited the Taj. It is built outside the city on the bank of the Yamuna. A large gateway of red stone provides the entrance. The Taj is a large and beautiful building. It stands on a raised platform. In the middle of the platform, there is a splendid white dome. At its four corners, there are four stately towers. Underneath the white dome are the marble tombs of Mumtaz Mahal and Shah Jahan. The whole building is surrounded by a garden on three sides. On the fourth side, the river Yamuna grazes it. No words can describe the beauty of the Taj.

28. A Scene at the Railway Station

Last Sunday, my father went to Delhi. I went to the station to see him off. I bought a ticket and a platform ticket. We went to the platform. There was great hustle and bustle. Some men were buying books at the bookstall. The hawkers were going up and down the platform. The coolies were busy. People were waiting for the train. Soon the train arrived. There was a great rush in it. Some passengers got down. Others got in and took their seats. I got a seat for my father. The engine gave a whistle. The guard waved a green flag. The train again whistled and steamed off. Now there was all quiet on the platform. I came back home.

29. A Scene at the Bus Stand

Last Monday, I went to the bus stand to see off my uncle. The bus stand was humming with life. There were separate parking stands for different routes. A bus was parked at each stand. Men behind the counter were issuing tickets. The conductors were shouting to attract passengers for their respective buses. As soon as a bus was full, the conductor blew his whistle and the bus moved out of the stand. Another one immediately took its place. This activity was going on endlessly. I bought a ticket for my uncle, got him a good seat and then came back home.

30. A House on Fire

It was Sunday. I was sitting in my room with my friend, Atul. Suddenly, we saw clouds of smoke rising in the sky. There was a big fire in the next street. People were running to the site of fire. Children were shouting for help. People brought buckets of water. We also joined them. We threw sand and water on the flames. The fire was put out after half an hour. It was the house of a carpenter. The poor man suffered a big loss. All his wood, grain and money were gone. The house was reduced to ashes. He was very sad at his loss. People felt sorry for him. They gave him food, clothes and some money. The poor carpenter thanked them with folded hands.

PSEB 9th Class English Paragraph Writing

31. A Bus Accident

Last Monday, I was travelling from Panipat to Delhi by bus. We had hardly gone twenty kilometres when a dreadful accident took place. All of a sudden a scooterist, coming from a side-road, came in front of the bus. The driver at once applied the brakes, and also turned the bus to one side. All the passengers were thrown off their seats. In no time, the bus went off the road and fell into a ditch. There were loud cries. Many passengers were badly wounded. I, too, got a deep cut on my forehead. Many people gathered there. They helped us to get out of the bus. Luckily there was no death. The scooterist had sped away. I reached home with a bandaged head.

32. A Street Quarrel

Last evening, I was sitting near the window of my room. I saw two children playing in the street. Suddenly, they fell out. Other boys of the street gathered there. None tried to separate them. They kept looking on. Soon, the mothers of both the children reached there. They started abusing each other. They used very dirty words for each other. From hot words, they came to blows. They pulled each other’s hair. Luckily, an elderly woman came there. She separated the fighting ladies. She spoke to them very wisely. The two women realised their mistake. They went back to their homes. Both the children started playing together once again.

33. A Rainy Day

It was the month of July last year. One day, it was very hot. Men and animals were panting. All were perspiring. We longed for a shower of rain. In the afternoon, some clouds appeared in the east. Soon the whole sky was overcast with dark clouds. It started raining heavily. Streets and bazaars were flooded with water. Little children came out and played in the rain. They splashed water over one another. The rain stopped after two hours. It became very cool and pleasant. Streets and bazaars were washed clean. The city gave a fresh look.

34. Life in a Village

The three words that can amply describe the life in a village are — Simple, Pure and Fresh. The villagers are very simple-hearted people. They know no cunning. They are pure in their thoughts and actions. They are very hospitable. They live simply and happily. They have no anxiety. Life in a village is very calm and peaceful. It is free from the noise and din of cities. The air is fresh and health-giving. Says Leo Tolstoy in one of his stories, “A villager’s life is not a fat one, but it is a long one.” He may never grow rich, but he has always enough to eat. In short, we can say that life in a village is worth living.

35. How I Celebrated My Birthday

I gave a party on my birthday. I invited all my friends. The party was held at my house. The party began at 6 p.m. A big cake was placed on a table. All my friends stood round the table. I cut the cake with a knife. My friends and parents chanted three times : ‘Happy Birthday To You.’ Then everybody set to eating. The cake was served to all. It was very tasty. There were many things to eat. Everybody ate to their heart’s content. There was singing and dancing also. Everyone enjoyed the party. It was over by 8 p.m. My friends congratulated me once again and went back to their homes.

36. A Drowning Tragedy

One day, I was picnicking with some of my friends on the riverbank. A boy named Kamal fell into the river. He didn’t know how to swim. I saw him struggling with water. It was a painful sight. I at once jumped into the river. I swam to him and brought him out with great difficulty. He had swallowed a lot of water. He was unconscious. We, at once called a doctor. Someone ran to inform Kamal’s parents. The doctor pressed out the water from Kamal’s belly. Kamal opened his eyes. We felt great relief. After some time Kamal’s parents reached the place. They thanked me and the doctor again and again.

PSEB 9th Class English Paragraph Writing

37. The Golden Temple

Amritsar is also called Guru-ki-nagri. It is famous for the Golden Temple. The Temple is situated in the city. It is surrounded by many narrow lanes. The golden shrine built in the middle of the sarover shines at sunrise and sunset. It was built by Guru Arjun Dev Ji. It is a unique experience when Granth Sahib is brought out from the Akal Takhat Sahib amidst chanting of hymns and blowing of bugles.

The Akal Takhat Sahib, facing the Harmandir Sahib, was built by Guru Hargobind Ji. It was used for holding courts. The complex has a museum of rare paintings, books, shashtras, etc. The lives of the Gurus are described through them. There is a big bazaar near Darshani Deori. Gutakas, karas and other articles related to the Sikh religion are sold there. Home-made papad-varian, chura-bangles, dry fruit are also sold in many shops. There are number of hotels and guest houses near the Temple for tourists to stay. There is a sarai also for pilgrims in the Temple. The Golden Temple is indeed a worth-visiting place.

38. Canada

Canada is one of the largest country of the world. Its area is 9,976,139 sq. km. and population is about 32 million. The capital city of Canada is Ottawa. The currency of the country is Canadian dollar. English and French are the official languages of Canada. In winter, the climate of Canada is bitterly cold. In some regions, the mercury may dip to -65°C. The average temperature in Ottawa is from -15°C to -6°C in January.

In July, the average temperature is 15°C to 26°C. The main products of Canada are fruit, vegetables, livestock, tobacco, copper, zinc, iron, salt, oil and natural gas. And major industries of the country are agriculture, forestry, food processing, transport, chemicals, oil and gas refining and cement. Vehicles, machinery, food stuffs, natural gas, meat, coal and timber are exported to other countries. Canada is one of the most developed nations of the world.

39. Aruna Asif Ali

Aruna Asif Ali is known as the Grand Old Lady of India. She took active part in the Independence movement. She was born in an orthodox Hindu Bengali family in 1909 at Kalka. She married a Muslim, Mr. Asif Ali, thus breaking all conventions regarding marriage. Her husband, Mr. Asif Ali, was also involved in the freedom struggle. Aruna Asif Ali took part in Salt Satyagrah under the leadership of Gandhiji.

She addressed many public meetings and led processions for the cause of India’s independence from the British rule. As a result, she was sentenced to one-year imprisonment. But she didn’t give up the cause for which she was fighting against the British rule. She was again sentenced to jail.

She became the editor of the newspaper ‘Inquilab’. After Independence, she became a social worker. She fought for the rights of women. In 1992 she received Nehru Award for International Understanding. She passed away in July 1996. She was honoured with Bharat Ratna posthumously.

40. The Tribals of Odisha 

There are many tribal groups in Odisha. They live in remote places. One such group lives in the forests of Kalahandi. These people are one of most backward tribes in the world. They have dark skin and black hair. The women wear bright-coloured saris while the men wear nothing but loincloth.

They still believe that India is ruled by kings. These people are illiterate as they do not have any facility of schooling, means of transportation and proper motorable roads. As a result, they are cut off from the rest of the world. They do not have any idea of currency notes. They still use barter system.

PSEB 9th Class English Paragraph Writing

They usually live in groups and each group has common property. They cure diseases with herbs and set bones by rubbing oils. The government should launch schemes to educate them and bring them to the mainstream of the nation.

41. An Incident of Burglary

Mr. Ramanathan is an affluent businessman of our town. One day, he with his family went out of the town to attend a wedding. There was nobody at home and the house was locked from outside. A thief broke into the house at night. He decamped with the jewellery, valuables and money.

But the neighbours had seen the lights on and they informed the police about it. The police came along with a dog. They found the thief’s glove. The dog sniffed the scent of the thief. It took the policemen to the thief’s place. Thus the thief was arrested and the case was solved. The policemen were rewarded by the department for their efficiency.

42. Floods in Mumbai

On July 26, 2005, I was busy shopping in a famous crowded market, although it was raining. Gradually, it started raining heavily. Now, it was impossible to go from there. Therefore, I took shelter in a shop. Soon, the place got flooded and water started entering the shops. The articles in the shops started floating in the water and the shopkeeper tried to retrieve valuable articles. The entire area was submerged in the flood water.

Many vehicles couldn’t move in the flood water. The people had to stay in them. Some other people took shelter in shops and houses. Suddenly, it started raining like hell. Now water in the shops and houses rose up to 6-7 feet. The people had to move to first floor. In no time the army swung into action. The volunteers of many NGO’s started helping the affected people with food and water. All this went on for more than 24 hours. It was really a horrifying experience which I can never forget.

43. The Lohri Festival

Lohri is a festival of fun and frolic. It is generally celebrated on 13th of January every year. This festival is related to the folklore of Dulla Bhatti. At sunset, people light up bonfires in the open in front of their houses. Lohri is celebrated with more enthusiasm in the families where there is a newborn son or a newly married person.

Giddha or Bhangra is performed to the beats of the drums. On the day of Lohri, children go singing from house to house asking for money and sweets. Lohri is a busy festival. People meet their friends and relatives and exchange greetings and gifts.

44. How to Make Papier-mache -Toys

In order to make toys with papier-mache, old newspaper sheets are taken. They are torn into small pieces. These pieces are soaked in water overnight. Next day, the mixture is boiled for half an hour. After that, the mixture is whipped till it becomes soft and pulpy.

PSEB 9th Class English Paragraph Writing

The water is squeezed out from the mixture and two tablespoons of white gum are added into it. The mixture is stirred well and then the toys are made from it. These toys are left to dry overnight or more. Then they are painted with water-based colour. To make these toys waterproof, two or three coats of lacquer are given on them. Masks can also be made in the same manner.

45. How to Make Gajrela

It is very easy to make gajrela at home. Take three kilograms of large carrots and wash them properly. Then grate the carrots. Mix 242 litres milk with the carrots. After that put the mixture in a pan and boil it till the mixture becomes very thick. Add 3/4 cup of sugar and 250 gm of khoya in the mixture.

Stir the mixture till it becomes thick. Stir it continuously as the mixture should not stick to the pan. Now remove the pan from the fire. Add nuts to it. Your gajrela is ready. Let it cool before serving. It can also be served hot.

PSEB 9th Class Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1

Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Answer:
In equilateral ∆ ABC, the length of each side is a.
∴ a = a, b = a, c = a
and semiperimeter s = \(\frac{a+b+c}{2}\) = \(\frac{a+a+a}{2}\) = \(\frac{3}{2}\)a
Now,
s – a = \(\frac{3}{2}\)a – a = \(\frac{1}{2}\)a,
s – b = \(\frac{3}{2}\)a – a = \(\frac{1}{2}\)a, and
s – c = \(\frac{3}{2}\)a – a = \(\frac{1}{2}\)a,
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1 1
Now, the perimeter of equilateral ∆ ABC is 180 cm.
∴ Length of each side = \(\frac{180}{3}\) = 60 cm and
semiperimeter s = \(\frac{180}{2}\) = 90 cm.
Here, a = b = c = 60 cm and s = 90 cm
∴ s – a = 90 – 60 = 30 cm,
s – b = 90 – 60 = 30 cm and
s – c = 90 – 60 = 30 cm.
Area of ∆ ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{90 \times 30 \times 30 \times 30}\) cm2
= \(\sqrt{3 \times 900 \times 900}\) cm2
= 30 × 30 × √3 cm2
= 900 √3 cm2

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see the given figure). The advertisements yield an earning of ₹ 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1 2
Answer:
For triangular side wall of the flyover,
a = 122 m, b = 120 m and c = 22 m.
Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{122+120+22}{2}\)
= \(\frac{264}{2}\)
= 132 m
Area of triangular side wall
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{132(132-122)(132-120)(132-22)}\) m2
= \(\sqrt{132 \times 10 \times 12 \times 110}\) m2
= \(\sqrt{12 \times 11 \times 10 \times 12 \times 11 \times 10}\) m2
= 12 × 11 × 10 m2
= 1320 m2
∴ Annual rent of one wall = ₹ (1320 × 5000)
∴ Rent of one wall for 3 months
= ₹ (1320 × 5000 × \(\frac{3}{12}\))
= ₹ 16,50,000

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see the given figure). If the sides of the wall are 15 m, 11m and 6 m, find the area painted in colour.
PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1 3
Answer:
The lengths of the triangular side wall are
15 m, 11m and 6 m.
∴ a = 15 m, b = 11m, c = 6m and
semiperimeter s = \(\frac{a+b+c}{2}\) = \(\frac{15+11+6}{2}\) = \(\frac{32}{2}\) = 16 cm
Then, s – a = 16 – 15 = 1 m,
s – b = 16 – 11 = 5 m, and
s – c = 16 – 6 = 10m.
Area of the triangular region painted in colour
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{16 \times 1 \times 5 \times 10}\) m2
= \(\sqrt{16 \times 5 \times 5 \times 2}\) m2
= 4 × 5 × √2 m2
= 20√2 m2

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Answer:
Here, perimeter of the triangle = 42 cm
∴ Semiperimeter s = \(\frac{42}{2}\) = 21 cm.
Now, a = 18 cm and b = 10 cm.
s = \(\frac{a+b+c}{2}\)
∴ 21 = \(\frac{18+10+c}{2}\)
∴ 42 = 28 + c
∴ c = 14 cm
Now,
s – a = 21 – 18 = 3 cm,
s – b = 21 – 10 = 11 cm and
s – c = 21 – 14 = 7 cm.
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{21 \times 3 \times 11 \times 7}\) cm2
= \(\sqrt{21 \times 21 \times 11}\) cm2
= 21√11 cm2

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1

Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Answer:
Suppose the sides of the triangle measure 12x cm, 17x cm and 25x cm.
Perimeter of a triangle = Stun of three sides
∴ 540 = 12x + 17x + 25x
∴ 540 = 54x
∴ x = 10
Then, the measures of the sides of the triangle are,
a = 12 × 10 = 120 cm,
b = 17 × 10 = 170 cm and
c = 25 × 10 = 250 cm.
Now, s – a = 270 – 120 = 150 cm,
s – b = 270 – 170 = 100 cm and
s – c = 270 – 250 = 20 cm.
Area of a triangle
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= 7\(\sqrt{270(150)(100)(20)}\) cm2
= \(\sqrt{270 \times 30 \times 5 \times 100 \times 5 \times 4}\) cm2
= \(\sqrt{8100 \times 25 \times 400}\) cm2
= \(\sqrt{(90)^{2} \times(5)^{2} \times(20)^{2}}\) cm2
= 90 × 5 × 20 cm2
= 9000 cm2

PSEB 9th Class Maths Solutions Chapter 12 Heron's Formula Ex 12.1

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Answer:
Let, the sides of the isosceles triangle be a = 12 cm, b = 12 cm and c cm.
Perimeter of triangle = Sum of three sides
∴ 30 = 12 + 12 + c
∴ 30 = 24 + c
∴ c = 6 cm
Now, semiperimeter s = \(\frac{\text { Perimeter }}{2}\) = \(\frac{30}{2}\) = 15 cm
Then, s – a = 15 – 12 = 3 cm,
s – b = 15 – 12 = 3 cm and
s – c = 15 – 6 = 9 cm.
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 3 \times 3 \times 9}\) cm2
= \(\sqrt{15 \times 9 \times 9}\) cm2
= 9 √15 cm2

PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

Question 1.
Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13cm.
Answer:
PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2 1

Steps of construction:

  1. Draw any ray BX. With centre B and radius 7 cm draw an arc to intersect BX at C.
  2. At B, construct ∠YBC with measure 75°.
  3. With centre B and radius 13 cm, draw an arc to intersect BY at M.
  4. Draw line segment MC. Draw the perpendicular bisector of MC to intersect BM at A.
  5. Draw line segment AC.
    Then, ∆ ABC is the required triangle.

PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Answer:
PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2 2

Steps of construction:

  1. Draw any ray BX and from that obtain the line segment BC of length 8 cm.
  2. At B, draw ray BY such that ∠YBC = 45°.
  3. With centre B and radius 3.5 cm, draw an arc to intersect ray BY at D.
  4. Draw line segment DC. Draw the perpendicular bisector of DC to intersect ray BY at A.
  5. Draw line segment AC.
    Then, ∆ ABC is the required triangle.

PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2

Question 3.
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Answer:
PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2 3

Steps of construction:

  1. Draw any ray QX and from that obtain the line segment QR of length 6 cm.
  2. At Q, construct ray QY such that Z YQR = 60°.
  3. Produce ray QY on the side of Q to obtain ray QZ. Obtain point S on ray QZ such that QS = 2 cm.
  4. Draw line segment RS. Draw the perpendicular bisector of RS to intersect QY at E
  5. Draw line segment PR.
    Then, ∆ PQR is the required triangle.

PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer:
PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2 4

Steps of construction:

  1. Draw any ray AP and from that obtain the line segment AB of length 11 cm.
  2. Construct ray AL such that ∠LAB = 30°.
  3. Construct ray BM such that ∠MBA = 90°.
  4. Draw the bisectors of ∠LAB and ∠MBA to intersect each other at X.
  5. Draw line segment XB. Draw the perpendicular bisector of XB to intersect AB at Z.
  6. Draw line segment XA. Draw the perpendicular bisector of XA to intersect AB at Y.
  7. Draw line segments XY and XZ.
    Then, ∆ XYZ is the required triangle.

PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2

Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:
PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2 5

Steps of construction:

  1. Draw any ray BX and from that obtain the line segment BC of length 12 cm.
  2. Construct ray BY such that ∠YBC = 90°.
  3. Taking B as centre and radius 18 cm, draw an arc to intersect BY at M.
  4. Draw line segment CM. Draw the perpendicular bisector of CM to intersect BM at A.
  5. Draw line segment AC.
    Then, ∆ ABC is the require triangle in which ∠B is a right angle, BC = 12 cm and AB + AC = 18 cm.