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

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.