NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.3#
Cone of radius $r$, height $h$ and slant height $l$:
$$l^2 = h^2 + r^2$$$$\text{CSA} = \pi rl, \qquad \text{TSA} = \pi r(l + r), \qquad V = \dfrac{1}{3}\pi r^2 h$$
For $\pi$, use $\dfrac{22}{7}$ or $3.14$.
1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. #
Solution
$r = 12$ m and $l = 21$ m.
$$\text{TSA} = \pi r(l + r)$$$$\text{TSA} = \dfrac{22}{7} \times 12 \times 33$$
$$\text{TSA} = \dfrac{8712}{7} \approx 1244.57 \text{ m}^2$$
(With $\pi \approx 3.14$: $3.14 \times 12 \times 33 = 1243.44$ m².)
2. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm. #
Solution
$$\text{CSA} = \dfrac{22}{7} \times 7 \times 10$$
$$\text{CSA} = 220 \text{ cm}^2$$
3. The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. #
Solution
$$l^2 = 400$$
$$\Rightarrow l = 20 \text{ cm}$$$$\text{CSA} = 3.14 \times 12 \times 20 = 753.6 \text{ cm}^2$$
$$\text{TSA} = 3.14 \times 12 \times (20 + 12) = 1205.76 \text{ cm}^2$$
4. A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base. #
Solution
$$\Rightarrow 15.7r^2 = 1570$$
$$\Rightarrow r^2 = 100$$
$$\Rightarrow r = 10$$
∴ The radius is 10 cm.
5. The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone. #
Solution
(i)
$$\pi rl = 308$$$$\Rightarrow \dfrac{22}{7} \times r \times 14 = 308$$
$$\Rightarrow 44r = 308$$
$$\Rightarrow r = 7 \text{ cm}$$
(ii)
$$\text{TSA} = \text{CSA} + \pi r^2$$$$\text{TSA} = 308 + \dfrac{22}{7} \times 49$$
$$\text{TSA} = 308 + 154$$
$$\text{TSA} = 462 \text{ cm}^2$$
6. 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. #
Solution
$$l^2 = 625$$
$$\Rightarrow l = 25 \text{ cm}$$
A cap is the curved surface only:
$$\text{One cap} = \dfrac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$$$$\text{10 caps} = 5500 \text{ cm}^2$$
*7. What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm. #
Solution
$$l^2 = 100$$
$$\Rightarrow l = 10 \text{ m}$$$$\text{CSA} = 3.14 \times 6 \times 10 = 188.4 \text{ m}^2$$
The tarpaulin is 3 m wide, so its length must be:
$$\text{Length} = \dfrac{188.4}{3} = 62.8 \text{ m}$$Adding the 20 cm (0.2 m) for margins and wastage:
$$\text{Length} = 62.8 + 0.2 = 63 \text{ m}$$∴ 63 m of tarpaulin is needed.
8. A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed. #
Solution
$6^2 + 8^2 = 10^2$, so the triangle is right-angled with legs 6 cm and 8 cm. Rotating about the 8 cm leg gives a cone with $h = 8$ cm, $r = 6$ cm and $l = 10$ cm.
$$V = \dfrac{1}{3}\pi \times 6^2 \times 8 = 96\pi \approx 301.71 \text{ cm}^3$$$$\text{CSA} = \pi \times 6 \times 10 = 60\pi \approx 188.57 \text{ cm}^2$$
*9. Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water? #
Solution
The cup points downwards, so the water forms a smaller cone at the bottom, with the same shape. Filling to half the depth makes its height $\dfrac{h}{2}$, and (by similar triangles) its radius $\dfrac{r}{2}$:
$$V_{\text{water}} = \dfrac{1}{3}\pi \left(\dfrac{r}{2}\right)^2 \left(\dfrac{h}{2}\right)$$$$V_{\text{water}} = \dfrac{1}{8} \times \dfrac{1}{3}\pi r^2 h$$
∴ The water fills only $\dfrac{1}{8}$ of the cup — much less than half!