EXERCISE 13.3
Q 1, Ex. 13.3, Page No 221, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Page No 221:
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. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of the base of cone $=\left(\frac{10.5}{2}\right) \mathrm{cm}$= 5.25 cm
Slant height (l) of cone = 10 cm
CSA of cone = πrl
$=\left(\frac{22}{7} \times 5.25 \times 10\right) \mathrm{cm}^{2}$
$=(22 \times 0.75 \times 10) \mathrm{cm}^{2}=165 \mathrm{cm}^{2}$
Therefore, the curved surface area of the cone is 165 cm2.
Q 2, Ex. 13.3, Page No 221, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of the base of cone $=\left(\frac{24}{2}\right) \mathrm{m}$= 12 mSlant height (l) of cone = 21 m
Total surface area of cone = πr(r + l)
$=\left[\frac{22}{7} \times 12 \times(12+21)\right] \mathrm{m}^{2}$
$=\left(\frac{22}{7} \times 12 \times 33\right) \mathrm{m}^{2}$\
$=1244.57 \mathrm{m}^{2}$
Q 3, Ex. 13.3, Page No 221, Surface Areas & Volumes (NCERT Maths Class 9th)
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.
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Slant height (l) of cone = 14 cmLet the radius of the circular end of the cone be r.
We know, CSA of cone = πrl
$(308) \mathrm{cm}^{2}=\left(\frac{22}{7} \times r \times 14\right) \mathrm{cm}$
$\Rightarrow r=\left(\frac{308}{44}\right) \mathrm{cm}$
=7cm
Therefore, the radius of the circular end of the cone is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base
= πrl + πr2
$=\left[308+\frac{22}{7} \times(7)^{2}\right] \mathrm{cm}^{2}$
$=(308+154) \mathrm{cm}^{2}$
$=462 \mathrm{cm}^{2}$
Therefore, the total surface area of the cone is 462 cm2
Q 4, Ex 13.3, Page No 221, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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 Rs 70.
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Let ABC be a conical tent.
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l.
In ΔABO,
AB2 = AO2 + BO2
l2 = h2 + r2
= (10 m)2 + (24 m)2
= 676 m2
∴ l = 26 m
Therefore, the slant height of the tent is 26 m.
(ii) CSA of tent = πrl
$=\left(\frac{22}{7} \times 24 \times 26\right) \mathrm{m}^{2}$
$=\frac{13728}{7} \mathrm{m}^{2}$
Cost of 1 m2 canvas = Rs 70
Cost of $\frac{13728}{7} \mathrm{m}^{2}$ canvas $=\operatorname{Rs}\left(\frac{13728}{7} \times 70\right)$
= Rs 137280
Therefore, the cost of the canvas required to make such a tent is
Rs 137280.
Q 5, Ex. 13.3, Page No 221, Surface Areas & Volumes, NCERT Class 9th Maths Solutions
Question 5:
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? 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:
Height (h) of conical tent = 8 mRadius (r) of base of tent = 6 m
Slant height (l) of tent $=\sqrt{r^{2}+h^{2}}$
$=(\sqrt{6^{2}+8^{2}}) \mathrm{m}=(\sqrt{100}) \mathrm{m}$
=10m
CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4 m2
Let the length of tarpaulin sheet required be l.
As 20 cm will be wasted, therefore, the effective length will be (l − 0.2 m).
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(l − 0.2 m) × 3] m = 188.4 m2
l − 0.2 m = 62.8 m
l = 63 m
Therefore, the length of the required tarpaulin sheet will be 63 m.
Q 6, Ex. 13.3, Page No 221, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 6:
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m2. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Slant height (l) of conical tomb = 25 mBase radius (r) of tomb $=\frac{14}{2}$ =7 m
CSA of conical tomb = πrl
$=\left(\frac{22}{7} \times 7 \times 25\right) \mathrm{m}^{2}$
= 550 m2
Cost of white-washing 100 m2 area = Rs 210
Cost of white-washing 550 m2 area $=\operatorname{Rs}\left(\frac{210 \times 550}{100}\right)$
= Rs 1155
Therefore, it will cost Rs 1155 while white-washing such a conical tomb.
Q 7, Ex. 13.3, Page No 221, Surface Areas & Volumes - Class 9th Maths Solutions
Question 7:
A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of conical cap = 7 cmHeight (h) of conical cap = 24 cm
Slant height (l) of conical cap $=\sqrt{r^{2}+h^{2}}$
$=[\sqrt{(7)^{2}+(24)^{2}}]cm$
$=(\sqrt{625}) \mathrm{cm}$
=25cm
CSA of 1 conical cap = πrl
$=\left(\frac{22}{7} \times 7 \times 25\right) \mathrm{cm}^{2}=550 \mathrm{cm}^{2}$
CSA of 10 such conical caps = (10 × 550) cm2 = 5500 cm2
Therefore, 5500 cm2 sheet will be required.
Q 8, Ex. 13.3, Page No 221, Surface Areas & Volumes - Class 9th Maths Solutions
Question 8:
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. 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 Rs 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:
Radius (r) of cone $=\frac{40}{2}=20 \mathrm{cm}$ = 0.2 mHeight (h) of cone = 1 m
Slant height (l) of cone $=\sqrt{h^{2}+r^{2}}$
$=[\sqrt{(1)^{2}+(0.2)^{2}}]m$
$=(\sqrt{1.04}) \mathrm{m}$
=1.02m
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02) m2 = 0.64056 m2
CSA of 50 such cones = (50 × 0.64056) m2
= 32.028 m2
Cost of painting 1 m2 area = Rs 12
Cost of painting 32.028 m2 area = Rs (32.028 × 12)
= Rs 384.336
= Rs 384.34 (approximately)
Therefore, it will cost Rs 384.34 in painting 50 such hollow cones.
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