Class 9

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
Answer:
For the given sphere, radius r = 7 cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7 cm³
= \(\frac{4312}{3}\) cm³
= 1437\(\frac{1}{3}\) cm³

(ii) 0.63 m.
Answer:
For the given sphere, radius r = 0.63 m.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.63 × 0.63 × 0.63 m³
= 1.05 m³ (approx.)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Answer:
Amount of water displaced by a solid spherical ball = Volume of spherical ball

(i) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 14 × 14 × 14 cm³
= \(\frac{34496}{3}\) cm³
= 11498\(\frac{2}{3}\) cm³
Thus, the amount of water displaced by the given solid spherical ball is = 11498\(\frac{2}{3}\) cm³

(ii) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{0.21}{2}\) m
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) m³
= 11 × 0.01 × 0.21 × 0.21 m³
= 0.004851 m³
Thus, the amount of water displaced by the given solid spherical ball is 0.004851 m³.

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9g per cm³?
Answer:
For the given spherical ball,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{4.2}{2}\) cm
= 2.1 cm
= \(\frac{21}{10}\) cm
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) cm³
= 38.808 cm³
Now, the density of the metal of the ball is 8.9 g per cm³.
∴ Mass of the ball = Volume × Density
= 38.808 cm³ × 8.9 g/cm³
= 345.39 g (approx.)
Thus, the mass of the metallic ball is 345.39 g (approx.).

Question 4.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon ?
Answer:
As the diameter of the moon is one-fourth of the diameter of the earth, the radius of the moon is also one-fourth of the radius of the earth. In other words, the radius of the earth is four times the radius of the moon. Let, the radius of the moon be r and the radius of the earth be R.
Then, R = 4r
Now, \(=\frac{\text { volume of the moon }}{\text { volume of the earth }}\) = \(\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}\)
= \(\left(\frac{r}{R}\right)^{3}\)
= \(\left(\frac{r}{4 r}\right)^{3}\)
= \(\left(\frac{1}{4}\right)^{3}\)
= \(\frac{1}{64}\)
∴ Volume of the moon
= \(\frac{1}{64}\) × Volume of the earth
Thus, the volume of the moon is \(\frac{1}{64}\) times the volume of the earth.

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer:
For the hemispherical bowl,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= 5.25 cm
= \(\frac{21}{4}\) cm
Capacity of the hemispherical bowl
= Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3} \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4}\) cm³
= 303.19 cm³ (approx.)
= \(\frac{303.19}{1000}\) liters (approx.)
= 0.303 liters (approx)
Thus, the given hemispherical bowl can hold 0. 303 litres (approx.) of milk.

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Answer:
For the hemispherical tank, inner radius r = 1 m and the thickness of the iron sheet = 1 cm = 0.01 m.
∴ For the hemispherical tank, outer radius
R = 1 + 0.01 m = 1.01 m.
Volume of the iron used in the tank
= Volume of outer hemisphere – Volume of Inner hemisphere
= \(\frac{2}{3}\) πR³ – \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) π (R³ – r³)
= \(\frac{2}{3}\) × \(\frac{22}{7}\) (1.01³ – 1³) m³
= \(\frac{44}{21}\) (1.030301 – 1) m³
Thus, the volume of the iron used to make the tank is 0.06349 m³ (approx.).

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Answer:
For the given sphere, surface area = 154 cm².
Surface area of a sphere = 4πr²
∴ 154 cm² = 4 × \(\frac{22}{7}\) × r²cm²
∴ r² = \(\frac{154 \times 7}{4 \times 22}\) cm²
∴ r² = \(\frac{49}{4}\) cm²
∴ r = \(\frac{7}{2}\)
Thus, the radius of the given sphere is \(\frac{7}{2}\) cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\) cm³
= \(\frac{539}{3}\) cm³
= 179\(\frac{2}{3}\) cm³
Thus, the volume of the given sphere is 179\(\frac{2}{3}\) cm³.

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹ 4989.60. If the s cost of whitewashing is ₹ 20 per square metre, find the
(i) inside surface area of s the dome.
Answer:
(i) Area of the region whitewashed at the
cost of ₹ 20 = 1 m²
∴ Area of the region whitewashed at the cost of ₹ 4989.60 = \(\frac{4989.60}{20}\) m² = 249.48 m²
Hence, the inner surface area of the dome is 249.48 m²

(ii) volume of the air inside the dome.
Answer:
Curved surface area of hemispherical dome = 2πr²
∴ 249.48 m² = 2 × \(\frac{22}{7}\) × r² m²
∴ r² = \(\frac{249.48 \times 7}{2 \times 22}\) m²
∴ r² = 39.69 m²
∴ r² = \(\sqrt{39.69}\) m
∴ r = 6.3 m
Thus. the radius of the hemispherical dome is 6.3m.
Volume of air inside the hemispherical dome = Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 6.3 × 6.3 × 6.3 m³
= 523.908 m³
= 5239 m³ (approx.)
Thus, the volume of the air inside the dome is 523.9 m³ (approx.).

Question 9.
Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’, Find the
(i) radius r’ of the new sphere,
Answer:
(i) 27 solid iron spheres of radius r are melted to form 1 iron sphere of radius r’.
∴ Volume of 1 sphere of radius r’
= Volume of 27 spheres of radius r
∴ \(\frac{4}{3}\) πr’³ = 27 × \(\frac{4}{3}\) πr’³
∴ r’³ = 27r³
∴ r’³ = (3r)³
∴ r’ = 3r

(ii) ratio of S and S’.
Answer:
The surface area of the sphere with radius r is S and the surface area of the sphere with radius r’ is S’.
Then,
\(\frac{\mathrm{s}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{4 \pi r^{\prime 2}}=\frac{r^{2}}{r^{\prime 2}}=\frac{r^{2}}{(3 r)^{2}}=\frac{r^{2}}{9 r^{2}}=\frac{1}{9}\) = 1 : 9
Thus, the ratio of S and S’ is 1 : 9.

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule ?
Answer:
For the spherical capsule, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) mm
= 1.75 mm
Capacity of the spherical capsule
= Volume of a sphere
= \(\frac{4}{3}\) πr’³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 1.75 × 1.75 × 1.75 mm³
= 22.46 mm³ (approx.)
Thus, 22.46 mm³ (approx.) medicine is needed to fill the given capsule.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
Answer:
For the given cone, radius r = 6 cm and height h = 7 cm.
Volume of a cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7 cm³
= 264 cm³

(ii) radius 3.5 cm, height 12 cm
Answer:
For the given cone, radius
r = 3.5 cm = \(\frac{7}{2}\) cm and height h = 12 cm.
Volume of a cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 12 cm³
= 154 cm³

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
Answer:
For the given conical vessel, radius r = 7 cm and slant height l = 25 cm.
h = \(\sqrt{l^{2}-r^{2}}\)
= \(\sqrt{25^{2}-7^{2}}\)
= \(\sqrt{625-49}\)
= √576
∴ h = 24 cm
Capacity of conical vessel
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 24 cm³
= 1232 cm³
= \(\frac{1232}{1000}\) liters
= 1.232 litres

(ii) height 12 cm, slant height 13 cm
Answer:
For the given conical vessel, height
h = 12 cm and slant height l = 13 cm.
r = \(\sqrt{l^{2}-h^{2}}\)
= \(\sqrt{13^{2}-12^{2}}\)
= \(\sqrt{169-144}\)
= \(\sqrt{25}\)
∴ r = 5 cm
Capacity of conical vessel
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 12 cm³
= \(\frac{6600}{21}\)
= \(\frac{6600}{21 \times 1000}\) liters
= \(\frac{11}{35}\) liters

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14.)
Answer:
For the given cone, height h = 15 cm and
volume = 1570 cm³.
Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 1570 cm³ = \(\frac{1}{3}\) × 3.14 × r² × 15 cm³
∴ 1570 cm³ = 15.7 × r² cm³
∴ r² = \(\frac{1570}{15.7}\) cm²
∴ r² = 100 cm²
∴ r = 10 cm

Question 4.
If the volume of a right circular cone of height 9 cm is 48 πcm³, find the diameter of its base.
Answer:
For the given right circular cone, height h = 9 cm and volume = 48 πcm³
Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 48π cm³ = \(\frac{1}{3}\) × π × r² × 9 cm³
∴ r² = \(\frac{48 \pi \times 3}{\pi \times 9}\) cm²
∴ r² = 16 cm²
∴ r² = 16 cm²
∴ r = 4 cm
Now, diameter = 2r = 2 × 4 cm = 8 cm
Thus, the diameter of the right circular cone is 8 cm.

Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?
Answer:
For the given conical pit,
radius r = \(\frac{\text { diameter }}{\mathbf{2}}\) = \(\frac{3.5}{2}\) m = \(\frac{35}{2}\) m
and height (depth) h = 12 m
Capacity of the conical pit
= Volume of the cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{35}{20}\) × \(\frac{35}{20}\) × 12 m³
= 38.5 m³
= 38.5 kilioliters

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone,
(ii) slant height of the cone and
(iii) curved surface area of the cone.
Answer:
For the given right circular cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm and
volume = 9856 cm³.

(i) Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 9856 cm³ = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 14 × 14 × h
∴ h = \(\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}\) cm
∴ h = 48 cm

(ii) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{48^{2}+14^{2}}\)
= \(\sqrt{2304+196}\)
= \(\sqrt{2500}\)
∴ l = 50 cm

(iii) Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × 14 × 50 cm²
= 2200 cm²

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.. Find the volume of the solid so obtained.
Answer:
A right circular cone is received when ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
For the cone so obtained, radius r = 5 cm, height h = 12 cm and slant height l = 13 cm.
Volume of the cone obtained
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 5 × 5 × 12 cm³
= 100π cm³

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two ‘solids obtained in Questions 7 and 8.
Answer:
Now, if ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, again a right circular cone is received.
For the cone so obtained, radiqs r = 12cm; height h = 5 cm and slant height l = 13 cm.
Volume of the cone obtained
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 12 × 12 × 5 cm³
= 240π cm³
Ratio of the volumes of two cones obtained in question 7 and 8 = \(\frac{100 \pi}{240 \pi}\) = \(\frac{5}{12}\) = 5 : 12.

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Answer:
For the conical heap of wheat,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{10.5}{2}\) m = \(\frac{105}{2}\) m and
height h = 3 m
Volume of the conical heap of wheat
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{105}{20}\) × \(\frac{105}{20}\) × 3 m³
= 86.625 m³
To cover the heap with canvas, the area of the canvas required will be equal to the curved surface area of the heap.
Now, l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{\left(\frac{105}{20}\right)^{2}+(3)^{2}}\)
= \(\sqrt{(5.25)^{2}+9}\)
= \(\sqrt{27.5625+9}\)
= \(\sqrt{36.5625}\)
= 6.05 m (approx.)
Curved surface area of the conical heap
= πrl
= \(\frac{22}{7}\) × \(\frac{105}{20}\) × 6.05 m²
= 99.825 m²
Thus, 99.825 m² canvas is required to cover the conical heap of wheat to protect it from rain.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.
The circumference of the base of a 7 cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1l)
Answer:
For the given cylindrical vessel, height h = 25 cm and circumference of the base = 132 cm.
Circumference of base = 2πr
∴ 132 cm = 2 × \(\frac{22}{7}\) × r cm
∴ r = \(\frac{132 \times 7}{2 \times 22}\) cm
∴ r = 21 cm
Hence, for the cylindrical vessel, radius r = 21 cm
Capacity of cylindrical vessel
= Volume of cylinder
= πr²h
= \(\frac{22}{7}\) × 21 × 21 × 25 cm³
= 34650 cm³
= \(\frac{34650}{1000}\) liters
= 34.65 litres
Thus, the cylindrical vessel can hold 34.65 litres of water.

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer:
For the cylindrical wooden pipe,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm,
inner radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{24}{2}\) cm = 12 cm and
height (length) h = 35 cm.
Volume of the cylindrical wooden pipe
= Volume of outer cylinder – Volume of inner cylinder
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 35 × (14 + 12) (14 – 12) cm³
= 110 × 26 × 2 cm³
= 5720 cm³
Now, mass of 1 cm³ of wood = 0.6 g
∴ Mass of 5720 cm³ of wood = 5720 × 0.6 g
= 3432 g
= 3.432 kg
Thus, the mass of the given pipe is 3.432 kg.

Question 3.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much?
Answer:
(i) For the cuboidal container with rectangular base, length l = 5 cm; breadth b = 4 cm and height h = 15 cm.
Capacity of the cuboidal container
= Volume of cuboid
= l × b × h
= 5 × 4 × 15 cm³
= 300 cm³

(ii) For the cylindrical container,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and
height h = 10 cm.
Capacity of the cylindrical container = Volume of cylinder
= πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 10 cm³
= 385 cm³
Hence, the capacity of the cylindrical container is more than the cuboidal container by 385 – 300 = 85 cm³.

Question 4.
If the lateral surface area of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base and
(ii) its volume. (Use π = 3.14)
Answer:
(i) For the given cylinder, height h = 5 cm and lateral (curved) surface area = 94.2 cm².
Curved surface area of a cylinder = 2 πrh
∴ 94.2 cm² = 2 × 3.14 × r × 5 cm²
∴ r = \(\frac{94.2}{2 \times 3.14 \times 5}\) cm
∴ r = 3 cm
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of a cylinder
= πr²h
= 3.14 × 3 × 3 × 5 cm³
= 141.3 cm³
Thus, the volume of the cylinder is 141.3 cm³.

Question 5.
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base and
(iii) capacity of the vessel.
Answer:
(i) Area of the region painted at the cost of ₹ 20 = 1 m²
∴ Area of the region painted at the cost of ₹ 2200 = \(\frac{2200}{20}\) m² = 110m²
Thus, the inner curved surface area of the vessel is 110 m².

(ii) For the cylindrical vessel, height (depth) h- 10 m and curved surface area = 110 m².
Curved surface area of cylindrical vessel = 2πrh
∴ 110 m² = 2 × \(\frac{22}{7}\) × r × 10 m²
∴ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) m
∴ r = \(\frac{7}{4}\) m
∴ r = 1.75 m
Thus, the radius of the cylindrical vessel is 1.75 m.

(iii) Capacity of the cylindrical vessel
= Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × 1.75 × 1.75 × 10 m³
= 96.25 m³
= 96.25 kilolitres
Thus, the capacity of the cylindrical vessel is 96.25 kilolitres.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Answer:
For the closed cylindrical vessel, height h = 1 m and capacity = 15.4 litres
∴ Volume of the vessel = 15.4 litres
= \(\frac{15.4}{1000}\) m³
= 0.0154 m³
Volume of cylindrical vessel = πr²h
∴ 0.0154 m³ = \(\frac{22}{7}\) × r² × 1 m³
∴ r² = \(\frac{154}{10000} \times \frac{7}{22}\) m²
∴ r² = \(\frac{49}{10000}\) m²
∴ r² = \(\frac{7}{100}\) m
∴ r = 0.07 m
Thus, the radius of the cylindrical vessel is 0. 07 m.
Area of the metal sheet required to make closed cylindrical vessel
= Total surface area of a cylinder
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.07 (0.07 + 1) m²
= 0.44 × 1.07 m²
= 0.4708 m²
Thus, 0.4708 m² of metal sheet would be needed to make the closed cylindrical vessel.

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Answer:
For the solid cylinder of graphite,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{1}{2}\) mm = \(\frac{1}{20}\) cm and
height h = 14 cm.
Volume of cylinder of graphite
= πr²h
= \(\frac{22}{7}\) × \(\frac{1}{20}\) × \(\frac{1}{20}\) × 14 cm³ = 011 cm³
For the hollow cylinder of wood,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) mm = \(\frac{7}{20}\) cm,
inner radius r = \(\frac{1}{20}\) cm and height h = 14 cm.
Volume of hollow cylinder of wood
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 14 \(\left(\frac{7}{20}+\frac{1}{20}\right)\left(\frac{7}{20}-\frac{1}{20}\right)\) cm³
= 44 × \(\frac{8}{20}\) × \(\frac{8}{20}\) cm³
= \(\frac{528}{100}\) cm³
= 5.28 cm³
Thus, in the given pencil, the volume of wood is 5.28 cm³ and the volume of graphite is 0.11cm³.

Question 8.
A patient in a hospital is given soup daily in a cylindrical howl of diameter 7 cm. If the bowl is Oiled with soup to a height of 4 cm, how much soup the hospital has ‘ to prepare daily to serve 250 patients ?
Answer:
For the soup served in cylindrical bowl, radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and height h = 4 cm.
Volume of soup served to one patient = Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 4 cm³
= 154 cm³
Thus, the volume of soup served to 1 patient =154 cm³
∴ The volume of soup served to 250 patients
= 154 × 250 cm³
= 38500 cm³
= \(\frac{38500}{1000}\) litres
= 38.5 litres
Thus, the hospital has to prepare 38500 cm³, i.e., 38.5 litres of soup daily.

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.

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.

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.

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

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.

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.

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

(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

(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

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.

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.

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.

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.

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).

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

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.

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.

(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.

(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.

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.

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

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.)

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 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:

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.)

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:

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

Area of quadrilateral ABCD
= Area of ∆ ABC + Area of ∆ ACD
= (6 + 9.2) cm² (approx.)
= 15.2 cm² (approx.)

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. ;

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

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.

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

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

= 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:

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

= 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.

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?

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

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

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?

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

= 8 × 2.24 cm² (approx.)
= 17.92 cm² (approx.)

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.

Answer:
For each of 16 triangular tiles,
a = 9 cm; b = 28 cm and c = 35 cm

= 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

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:

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

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