ML Aggarwal Solution Class 10 Chapter 17 Mensuration Exercise 17.2

 Exercise 17.2

Question 1

Write whether the following statements are true or false. Justify your answer.

(i) If the radius of a right circular cone is halved and its height is doubled, the volume will remain unchanged.

(ii) A cylinder and a right circular cone are having the same base radius and same height. The volume of the cylinder is three times the volume of the cone.

(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle.

Sol :

(i) If the radius of a right circular cone is halved and its height is doubled,

then the volume will remain unchanged

It is wrong as

$\frac{1}{3} \pi r^{2} h \neq \frac{1}{3} \pi\left(\frac{r}{2}\right)^{2} \times 2 h$

$\frac{1}{3} \pi r^{2} h \neq \frac{1}{3} \pi \frac{r^{2}}{4} \times 2 h$

$\Rightarrow \frac{1}{3} \pi r^{2} h \neq \frac{1}{3} \times \frac{1}{2} \pi r^{2} h$

(ii) A cylinder and a right circular cone are having the same base radius and same height the volume of the cylinder is three times the volume of cone – It is true as

Volume of cylinder $=\pi r^{2} h=3 \times \frac{1}{3} \pi r^{2} h=3($ volume of cone $)$

(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle

It is true as in a cone and in a right-angled triangle.

Hypotenuse (slant x height) $=r^{2}+h^{2}$

and cone is formed by revolving the right triangle about the perpendicular.

Question 2

Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.

Sol :

10 Slant height of a cone (l) = 10 cm

and radius of the base = 7 cm

Curved surface area = πrl

$=\frac{22}{7} \times 7 \times 10=220 \mathrm{~cm}^{2}$

Question 3

Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.

Sol :

The diameter of the base of a cone = 10.5cm

Its radius $(r)=\frac{10.5}{7}=5.25 \mathrm{~cm}$

and slant height (l) = 10 cm

Curved surface area = πrl

$=\frac{22}{7} \times 5.25 \times 10 \mathrm{~cm}^{2}$

$=165.0 \mathrm{~cm}^{2}$

Question 4

Curved surface area of a cone is 308 $\mathrm{cm}^{2}$ and its slant height is 14 cm. Find ,

(i)radius of the base

(ii)total surface area of the cone.

Sol :

Curved surface area of a cone = 308 $\mathrm{cm}^{2}$

Slant height = 14 cm

(i) Radius $(r)=\frac{\text { Curved surface area }}{\pi l}$

$=\frac{308 \times 7}{22 \times 14} \mathrm{~cm}=7 \mathrm{~cm}$

(ii) Total surface area=Curved surface area+base area

$=308+\pi r^{2}=308+\frac{22}{7} \times 7 \times 7 \mathrm{~cm}^{2}$

$=308+154=462 \mathrm{~cm}^{2}$

Question 5

Find the volume of the right circular cone with

(i) radius 6 cm and height 7 cm

(ii) radius 3.5 cm and height 12 cm.

Sol :

(i) Radius of cone (r) = 6 cm

and height (h) = 7 cm

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 7 \mathrm{~cm}^{3}=264 \mathrm{~cm}^{3}$

(ii) Radius of a cone $(r)=3.5 \mathrm{~cm}$ and height $(h)=12 \mathrm{~cm}$

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12 \mathrm{~cm}^{3}=154 \mathrm{~cm}^{3}$

Question 6

Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

Sol :

(i) Radius = 7 cm

and slant height (l) = 25 cm

$\therefore h=\sqrt{l^{2}-r^{2}}=\sqrt{25^{2}-7^{2}}$

$=\sqrt{625-49}=\sqrt{576}=24 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \mathrm{~cm}^{3}=1232 \mathrm{~cm}^{3}$

$\therefore$ Capacity $=\frac{1232}{1000} l=1.232 l$

(ii) Height $(h)=12 \mathrm{~cm}$

and slant height $=13 \mathrm{~cm}$

$\therefore$ Radius $=\sqrt{l^{2}-h^{2}}=\sqrt{13^{2}-12^{2}}$

$=\sqrt{169-144}=\sqrt{25}=5 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12=\frac{2200}{7} \mathrm{~cm}^{3}$

$=\frac{2200}{7 \times 1000}$ litre $=\frac{11}{35}$ litres

Question 7

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?

Sol :

Diameter of top of conical pit = 3.5 m

Radius $(r)=\frac{3.5}{2}=1.75 \mathrm{~m}$

and depth (h) = 12m

Capacity $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 12 \mathrm{~m}=38.5 \mathrm{~m}^{3}$

=38.5 kilolitres

Question 8

If the volume of a right circular cone of height 9 cm is 48π $\mathrm{cm}^{3}$, find the diameter of its base.

Sol :

Volume of a right circular cone = 48π $\mathrm{cm}^{3}$

Height (h) = 9 cm

$\therefore$ Radius $(r)=\sqrt{\frac{\text { Volume }}{\frac{1}{3} \pi h}}=\sqrt{\frac{48 \pi \times 3}{1 \times \pi \times 9}}$

$=\sqrt{16}=4 \mathrm{~cm}$

$\therefore$ Diameter $=2 \times r=2 \times 4=8 \mathrm{~cm}$

Question 9

The height of a cone is 15 cm. If its volume is 1570 $\mathrm{cm}^{3}$, find the radius of the base. (Use π = 3.14)

Sol :

Height of cone (h) = 15 cm

Volume = 1570 $\mathrm{cm}^{3}$

$\therefore$ Radius $=\sqrt{\frac{\text { Volume }}{\frac{1}{3} \pi h}}=\sqrt{\frac{1570 \times 3}{3.14 \times 15}}$

$=\sqrt{100}=10 \mathrm{~cm}$

Question 10

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 area at the rate of Rs 210 per 100 $m^{2}$.

Sol :

Slant height of conical tomb (l) = 25 m

and base diameter = 14 m

Radius $(r)=\frac{14}{2}=7 m$

$\therefore$ Curved surface area $=\pi r l$

$=\frac{22}{7} \times 7 \times 25=550 \mathrm{~m}^{2}$

Rate of white washing $=₹ 210$ per $100 \mathrm{~m}^{2}$

$\therefore$ Total $\operatorname{cost}=\frac{550 \times 210}{100} \doteq ₹ 1155$

Question 11

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 $m^{2}$ canvas is Rs 70.

Sol :

Height of a conical tent (h) = 10 m

and radius (r) = 24 m

(i) $\therefore$ Slant height $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{24^{2}+10^{2}}=\sqrt{576+100}$

$=\sqrt{676}=26 \mathrm{~m}$

(ii) Curved surface area $=\pi r l$

$=\frac{22}{7} \times 24 \times 26 \mathrm{~m}^{2}=\frac{13728}{7} \mathrm{~m}^{2}$

Cost of $1 \mathrm{~m}^{2}$ canvas used $=₹ 70$

$\therefore$ Total cost $=₹ \frac{13728}{7} \times 70=₹ 137280$

Question 12

A Jocker’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 cloth required to make 10 such caps.

Sol :

Base radius of a conical cap = 7 cm

and height (h) = 24 cm

$\therefore$ Slant height $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{7^{2}+24^{2}}=\sqrt{49+576}$

$=\sqrt{625 \mathrm{~cm}}=25 \mathrm{~cm}$

$\therefore$ Area of cloth used to make it $=\pi r l$

$=\frac{22}{7} \times 7 \times 25 \mathrm{~cm}^{2}=550 \mathrm{~cm}^{2}$

Cloth used to make 10 such caps

= 550 x 10

$=5500 \mathrm{~cm}^{2}$

Question 13

(a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.

(b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.

(c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find:

(i) the ratio of their volumes.

(ii) the ratio of their lateral surface areas.

Sol :

(i) The ratio in base radii of two right circular cones of the same height = 3 : 4

Let h be the height and radius of first cone = 3x and

Radius of second cone = 4x

$\therefore$ Volume of first cone $=\pi r^{2} h$

$=\pi \times 3 x \times 3 x \times h=9 \pi h x^{2}$

and volume of second cone

$=\pi \times 4 x \times 4 x \times h$

$=16 \pi h x^{2}$

$\therefore$ Ratio in their volumes $=9 \pi h x^{2}: 16 \pi h x^{2}$

=9 : 16

(b) The ratio of the heights of two cones =5 : 2

and ratio in their radii =2 : 5

Let height of first cone =5x and radius =2y

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi(2 y)^{2} \times 5 x$

$=\frac{1}{3} \pi 20 x y^{2}$

and height óf the second cone=2x and radius =5y

$\therefore$ Volume $=\frac{1}{3} \pi(5 y)^{2} \times 2 x=\frac{1}{3} \pi \times 50 x y$

$\therefore$ Ratio in their volumes $=\frac{1}{3} \pi 20 x y: \frac{1}{3} \pi 50 x y$

=2 : 5

(c) Let the height of bigger cone= h

and radius= r

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

and height of the smaller cone $=\frac{h}{2}$

and radius $=\frac{r}{2}$

$\therefore$ Volume $=\frac{1}{3} \pi\left(\frac{r}{2}\right)^{2}\left(\frac{h}{2}\right)=\frac{1}{3} \pi \times \frac{r^{2}}{4} \times \frac{h}{2}$

$=\frac{1}{24} \pi r^{2} h$

and ratio in their values: $\frac{1}{24} \mathrm{pr}^{2} h: \frac{1}{3} \mathrm{pr}^{2} h$

$=\frac{1}{8}: 1=1: 4$

Question 14

Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per metre.

Sol :

Diameter of the base of the conical tent = 20 m

Radius $(r)=\frac{20}{2}=10 \mathrm{~m}$

and slant height (h) = 42 m

$\therefore$ Area of canvas used $=\pi r l$

$=\frac{22}{7} \times 10 \times 42 \mathrm{~m}^{2}=1320 \mathrm{~m}^{2}$

Width of cloth $=2 \mathrm{~m}$

$\therefore$ Length of canvas cloth $=\frac{1320}{2}=660 \mathrm{~m}$

Cloth used in stiching and folding $=10 \%$ of $660 \mathrm{~m}$

$=\frac{660 \times 16}{100}=66 \mathrm{~m}$

$\therefore$ Total length of cloth $=660 \mathrm{~m}+66 \mathrm{~m}=726 \mathrm{~m}$

Rate of cloth =₹80 per metre

$\therefore$ Total cost $=₹ 726 \times 80=₹ 58080$

Question 15

The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.

Sol :

Perimeter of the base of a cone = 44 cm

$\therefore$ Radius $(r)=\frac{P}{2 \pi}=\frac{44 \times 7}{2 \times 22}=7 \mathrm{~cm}$

and slant height $(l)=25 \mathrm{~cm}$

and height $(h)=\sqrt{l^{2}-r^{2}}=\sqrt{25^{2}-7^{2}}$

$=\sqrt{625-49}=\sqrt{576}=24 \mathrm{~cm}$

(i) $\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \mathrm{~cm}^{2}=1232 \mathrm{~cm}^{3}$

(ii) and curved surface area $=\pi r l$

$=\frac{22}{7} \times 7 \times 25 \mathrm{~cm}^{2}=550 \mathrm{~cm}^{2}$

Question 16

The volume of a right circular cone is 9856 $\mathrm{cm}^{3}$ and the area of its base is 616 $\mathrm{cm}^{2}$. Find

(i) the slant height of the cone.

(ii) total surface area of the cone.

Sol :

Volume of a circular cone = 9856 $\mathrm{cm}^{3}$

Area of the base = 616 $\mathrm{cm}^{2}$

$\therefore$ Radius $(r)=\sqrt{\frac{\text { Area of base }}{\pi}}$

$=\sqrt{\frac{616 \times 7}{22}} \mathrm{~cm}=\sqrt{28 \times 7}, \mathrm{~cm}$

$=\sqrt{196}=\sqrt{(14)^{2}}=14 \mathrm{~cm}$

Volume $=\frac{1}{3} \pi r^{2} h=9856 \mathrm{~cm}^{3}$

$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 h=9856 \mathrm{~cm}^{3}$

$\Rightarrow h=\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}=48 \mathrm{~cm}$

$(i) \therefore$ Slant height $(l)=\sqrt{r^{2}+h^{2}}$

$=\sqrt{14^{2}+48^{2}}$

$=\sqrt{196+2304}=\sqrt{2500}=50 \mathrm{~cm}$

(ii) Total surface area $=\pi r l+\pi r^{2}$ $=\pi r(l+r)$

$=\frac{22}{7} \times 14(50+14) \mathrm{cm}^{2}$

$=44(64)=2816 \mathrm{~cm}^{2}$

Question 17

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)

Sol :

Sides of a right triangle are 6 cm and 8 cm

It is revolved around 8 cm side

Radius (r) = 6 cm

Height (h) = 8 cm

Slant height (l) = 10 cm

(i) Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3}(3.14) \times 6 \times 6 \times 8 \mathrm{~cm}^{3}=301.44 \mathrm{~cm}^{3}$

(ii) Curved surface area $=\pi r l$ $=3.14 \times 6 \times 10 \mathrm{~cm}^{2}=188.4 \mathrm{~cm}^{2}$

Question 18

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be $\frac{1}{27}$ of the volume of the given cone, at what height above the base is the section cut?

Sol :

Height of a cone (H) = 30 cm

A small cone is cut off from the top of the cone given

$\therefore$ Volume of small cone $=\frac{1}{27}$ of the Volume of the given cone

Volume of given cone $=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{H}$

and volume of $=\frac{1}{3} \pi r^{2} h$

$\therefore \frac{1}{3} \pi r^{2} h=\frac{1}{27} \times \frac{1}{3} \pi \mathrm{R}^{2} \mathrm{H}$

$\frac{r^{2}}{\mathrm{R}^{2}} h=\frac{1}{27} \times 30$

$\Rightarrow \frac{r^{2}}{\mathrm{R}^{2}} h=\frac{10}{9}$...(i)

But $\frac{r}{R}=\frac{h}{H}=\frac{h}{30}$

$\Rightarrow \frac{r}{R}=\frac{h}{30}$...(ii)

From (i) and (ii)

 $\left(\frac{h}{30}\right)^{2} h=\frac{10}{9}$

$\Rightarrow \frac{h^{2} \times h}{900}=\frac{10}{9}$

$h^{3}=\frac{10 \times 900}{9}=1000=(10)^{3}$

h = 10 cm

∴ A line parallel to base at a distance of 30 – 10 = 20 cm is drawn.

Question 19

A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find

(i) the radius of the cone.

(ii) the (lateral) surface area of the cone.

Sol :

Radius of a semi-circular lamina = 35 cm

By folding it a cone is formed whose slant height (l) = r = 35

and half circumference = circumference of the top of the cone

$=\pi r=\frac{22}{7} \times 35 \mathrm{~cm}=110 \mathrm{~cm}$

(i) ∴New radii of the cone $=\frac{110}{2 \pi}=\frac{110 \times 7}{2 \times 22}$

$=\frac{35}{2} \mathrm{~cm}=17.5 \mathrm{~cm}$

(ii) Lateral surface area $=\pi r l$

$=\frac{22}{7} \times \frac{35}{2} \times 35=1925 \mathrm{~cm}^{2}$