 # RS Aggarwal Class 9 Solutions Chapter 13 - Volume And Surface Area Ex 13B(13.1)

## RS Aggarwal Class 9 Chapter 13 - Volume And Surface Area Ex 13B(13.1) Solutions Free PDF

The RS Aggarwal Class 9 solution of Chapter 13 provided here are in accordance with the CBSE syllabus and are designed by our subject experts with easy and effective explanations of the solutions given in RS Aggarwal Class 10 maths textbook. It helps the students in solving all the exercise questions in a step-by-step manner and develop a strong foundation of the important mathematical concepts. It also improves their question solving-skills.

Moreover, these solutions provide easy explanations of each topic so that students don’t end up wasting their time referring to other books. These RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area act as a crucial preparation guide to help students score well in the exam. Practicing these solutions will improve the speed and accuracy of the students.

## Download PDF of RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area Ex 13B(13.2)

Q.1: Find the volume, the curved surface area, the total surface area of a cone having base radius 35 am and height 84 cm.

Solution:

Here, r = 35 cm and h = 84 cm

∴ Volume of cone = 13πr2h$\frac{1}{3} \pi r^{2} h$

= [13×227×35×35×84]cm3$[\frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 84] cm^{3}$ = 107800 cm3

∴ Curved surface area = [πrh2+r2]=h2+r2$[ \pi r \sqrt{h^{2} + r^{2}} ] = \sqrt{h^{2} + r^{2}}$

= πr842+352$\pi r \sqrt{84^{2} + 35^{2}}$ = πr8281$\pi r \sqrt{8281}$

= 227×35×91$\frac{22}{7} \times 35 \times 91$ = 10010 cm2

∴ Total surface area = 𝜋r(l + r)

Now,  l = h2+r2$\sqrt{h^{2} + r^{2}}$ = 842+352$\sqrt{84^{2} + 35^{2}}$

= 7056+1225=8281=91cm$\sqrt{7056 + 1225} = \sqrt{8281} = 91 cm$

∴ Total surface area = 227×35(91+35)$\frac{22}{7} \times 35(91 + 35)$

= (22×5×126)cm2=13860cm2$(22 \times 5 \times 126) cm^{2} = 13860 cm^{2}$

Q.2: Find the volume, the curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively. (take 𝜋 = 3.14)

Solution:

Here, height (h) = 6 cm and slant height (l) = 10 cm

∴ radius (r) = l2h2$\sqrt{l^{2} – h^{2}}$ = 10262=10036$\sqrt{10^{2} – 6^{2}} = \sqrt{100 – 36}$ = 64=8cm$\sqrt{64} = 8 cm$

∴ Volume of cone = 13πr2h$\frac{1}{3} \pi r^{2} h$

= [13×227×8×8×6]cm3$[\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 6] cm^{3}$ = 401.92 cm3

∴ Curved surface area = 𝜋rl = (3.14 × 8 × 10) cm2 = 251.2 cm2

∴ Total surface area = 𝜋r(l + r) = 𝜋r(10 + 8) = (3.14 × 8 × 18) cm2 = 452.16 cm2

Q.3: The volume of a right circular cone is (100𝜋) cm3 and its height is 12 cm. Find its slant height and its curved surface area.

Solution:

Here, Volume = (100𝜋) cm3, height (h) = 12 cm

Volume of the cone =   13πr2h$\frac{1}{3} \pi r^{2} h$

$\Rightarrow$ 100𝜋 = 13πr212$\frac{1}{3} \pi r^{2} 12$

r2=100π×3π×12$\Rightarrow r^{2} = \frac{100 \pi \times 3}{ \pi \times 12}$
r2=25$\Rightarrow r^{2} = 25$
r=25=5cm.$\Rightarrow r = \sqrt{25} = 5 cm.$

Slant height (l) = h2+r2$\sqrt{h^{2} + r^{2}}$ = 122+52$\sqrt{12^{2} + 5^{2}}$

l = 144+25=169=13cm.$\sqrt{144 + 25} = \sqrt{169} = 13 cm.$

∴ Slant height (l) = 13 cm

∴ Curved surface area = 𝜋rl = (5)(13) cm2 = 65𝜋 cm2

Q.4: A cone of slant height 25 cm has a curved surface area 550 cm2. Find the height and volume of the cone.

Solution:

Here, curved surface area = 550 cm2 and slant height (l) = 25 cm

∴ Curved surface area = 𝜋rl

550=227×r×25$\Rightarrow 550 = \frac{22}{7} \times r \times 25$
r=550×722×25cm=7cm$\Rightarrow r = \frac{550 \times 7}{22 \times 25} cm = 7 cm$

Now, height (h) = l2r2$\sqrt{l^{2} – r^{2}}$ = (25)2(7)2$\sqrt{(25)^{2} -(7)^{2}}$ = 62549$\sqrt{625 – 49}$ = 576=24cm$\sqrt{576} = 24 cm$

∴ Height of the cone = 24 cm.

Volume of the cone = 13πr2h$\frac{1}{3} \pi r^{2} h$ = [13×227×7×7×24]cm3$[ \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24] cm^{3}$ = 1232 cm3

∴ Volume of the cone = 1232 cm3

Q.5: Find the volume of a cone having radius of the base 35 cm and slant height 37 cm.

Solution:

Here, radius (r) = 35 cm and slant height (l) = 37 cm

h=l2r2$h = \sqrt{l^{2} – r^{2}}$ =  372352$\sqrt{37^{2} – 35^{2}}$

=13691225=144=12cm$= \sqrt{1369 – 1225} = \sqrt{144} = 12 cm$

∴ Height (h) = 12 cm.

Volume of the cone = 13πr2h$\frac{1}{3} \pi r^{2} h$

= [13×227×35×35×12]cm3$[ \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 12] cm^{3}$ = 15400 cm3

∴ Volume of the cone = 15400 cm3

Q.6: The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. Find its slant height.

Solution:

Here, curved surface area = 4070 cm2

Diameter = 70 cm $\Rightarrow$ radius = 35 cm

∴ Curved surface area = 𝜋rl

4070=227×35×l$\Rightarrow 4070 = \frac{22}{7} \times 35 \times l$
l=4070110=37cm$\Rightarrow l = \frac{4070}{110} = 37 cm$

∴ Slant height = 37 cm.

Q.7: How many meters of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 m?

Solution:

Here, radius = 7 m and height = 24 m

∴ slant height (l) = h2+r2$\sqrt{h^{2} +r^{2}}$ = 242+72$\sqrt{24^{2} +7^{2}}$

= 576+49=625=25m$\sqrt{576 +49} = \sqrt{625} = 25 m$

Now, area of the cloth = 𝜋rl

area=227×7×25=550m2$\Rightarrow area = \frac{22}{7} \times 7 \times 25 = 550 m^{2}$

∴ Length of cloth = Areaofclothwidthofcloth=5502.5=220m$\frac{Area \, of \, cloth}{width \, of \, cloth} = \frac{550}{2.5} = 220 m$

∴ Length of cloth required to make a conical tent = 220 m.

Q.8: A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and re-casted into a right circular cone having a base radius 1.2 cm. Find its height.

Solution:

Here, height of cone = 3.6 cm and radius = 1.6 cm

After melting, its radius = 1.2 cm

Volume of original cone = Volume of cone after melting

13π×1.6×1.6×3.6=13π×1.2×1.2×h$\frac{1}{3} \pi \times 1.6 \times 1.6 \times 3.6 = \frac{1}{3} \pi \times 1.2 \times 1.2 \times h$

h=13π×1.6×1.6×3.613π×1.2×1.2=6.4cm$\Rightarrow h = \frac{\frac{1}{3} \pi \times 1.6 \times 1.6 \times 3.6}{\frac{1}{3} \pi \times 1.2 \times 1.2} = 6.4 cm$

∴ Height of new cone = 6.4 cm

Q.9: Two cones have their heights in the ratio 1 : 3 and their radii of their bases in the ratio 3 :1. Show that their volumes are in the ratio 3 : 1.

Solution:

Let their heights be ‘h’ and ‘3h’

And, their radii be ‘3r’ and ‘r’

Then, V1=13π(3r)2×h$V_{1} = \frac{1}{3} \pi (3r)^{2} \times h$

And, V2=13π(r)2×3h$V_{2} = \frac{1}{3} \pi (r)^{2} \times 3h$

V1V2=13π(3r)2×h13π(r)2×3h=31$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{\frac{1}{3} \pi (3r)^{2} \times h}{\frac{1}{3} \pi (r)^{2} \times 3h} = \frac{3}{1}$

∴ V1:V2= 3 : 1

Q.10: A circus tent is in cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.

Solution:

Radius of the cylinder, R = 1052$\frac{105}{2}$ m and its height, H = 3m

Slant height (l) = 53 m

∴ Area of canvas = (2𝜋RH + 𝜋Rl)

= [(2×227×1052×3)+(227×1052×53)]m2$[( 2 \times \frac{22}{7} \times \frac{105}{2} \times 3 ) + ( \frac{22}{7} \times \frac{105}{2} \times 53 ) ] m^{2}$ = (990 + 8745) m2 = 9735 m2

∴ Length canvas = areaofcanvaswidthofcanvasm$\frac{area \, of \, canvas}{width \, of \, canvas} m$ = 97355=1947m$\frac{9735}{5} = 1947 m$

Q.11: A conical tent is to accommodate 11 persons. Each person must have 4 m2 of the space on the ground and 20 m3 of air to breathe. Find the height of the cone.

Solution:

Let the radius be ‘r’ meters and height be ‘h’ meters.

Area of the base = (11 × 4) m2 = 44 m2

Therefore, 𝜋r2 = 44

r2=(44×722)=14m$\Rightarrow r^{2} = (44 \times \frac{7}{22} ) = 14 m$
r2=14m$\Rightarrow r^{2} = 14 m$

Volume of the cone = 13πr2h$\frac{1}{3} \pi r^{2} h$

∴ Volume of the cone = (11 × 20) m3 = 220 m3

220=13×227×14×h$\Rightarrow 220 = \frac{1}{3} \times \frac{22}{7} \times 14 \times h$
h=220×322×2=15m$h = \frac{220 \times 3}{22 \times 2} = 15 m$

∴ The height of the cone = 15 m.

Q.12: A cylindrical bucket, 32 cm high and 18 cm of the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and the slant height of the heap.

Solution:

Here, the height of the cylindrical bucket = 32 m and radius = 18 cm.

Now, let the radius of the heap be R cm and its slant height be ‘l’ cm.

Then, π×(18)2×32=13×π×R2×24$\pi \times (18)^{2} \times 32 = \frac{1}{3} \times \pi \times R^{2} \times 24$

R2=π×18×18×32×3π×24=1296$\Rightarrow R^{2} = \frac{ \pi \times 18 \times 18 \times 32 \times 3}{ \pi \times 24} = 1296$
R=1296=36cm$\Rightarrow R = \sqrt{1296} = 36 cm$

∴ Radius of the heap = 36 cm

Slant height (l) = h2+R2$\sqrt{h^{2} + R^{2}}$ = 242+362$\sqrt{24^{2} + 36^{2}}$ = 576+1296$\sqrt{576 + 1296}$ = 1872=43.27cm$\sqrt{1872} = 43.27 cm$

∴ Slant height of the heap = 43.27 cm.

Q.13: A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius and height of each have the ratio 3 :4.

Solution:

Let the curved surface area of the cylinder and cone b 8x and 5x.

Then,  2𝜋rh = 8x . . . . . . . (i)

And, πrh2+r2=5x$\pi r \sqrt{h^{2} + r^{2}} = 5x$ . . . . . . . (ii)

Squaring both sides of equation (i), we have

(2πrh)2=(8x)2$(2 \pi r h)^{2} = (8x)^{2}$

4π2r2h2=64x2$4 \pi^{2} r^{2} h^{2} = 64x^{2}$ . . . . . . . . . (iii)

From (ii), we have,

πrh2+r2=5x$\pi r \sqrt{h^{2} + r^{2}} = 5x$

Squaring both sides,

π2r2(h2+r2)=25x2$\Rightarrow \pi^{2} r^{2} ( h^{2} + r^{2} ) = 25x^{2}$ . . . . . . . . . .(iv)

4π2r2h2π2r2(h2+r2=6425$\Rightarrow \frac{4 \pi^{2} r^{2} h^{2}}{\pi^{2} r^{2} (h^{2} + r^{2}} = \frac{64}{25}$  [Divide (iii) by (iv)]

h2(h2+r2)=1625$\Rightarrow \frac{h^{2}}{ ( h^{2} + r^{2} )} = \frac{16}{25}$
9h2=16r2$\Rightarrow 9h^{2} = 16r^{2}$
r2h2=916$\Rightarrow \frac{r^{2}}{h^{2}} = \frac{9}{16}$
rh=34$\Rightarrow \frac{r}{h} = \frac{3}{4}$

∴ The ratio of radius and height = 3:4

Q.14: An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surrounding it. Find the weight of the pillar, given that 1 cm3 of iron weighs 7.5 g.

Solution:

Here, height (h) of cylinder = 2.8 m = 280 cm and diameter = 20 cm

$\Rightarrow$ radius = 10 cm

Height of the cone (H) = 42 cm

∴ Volume of the pillar = (πr2h+13πr2H)cm3$(\pi r^{2} h + \frac{1}{3} \pi r^{2} H ) cm^{3}$

= πr2(h+13H)cm3$\pi r^{2} (h + \frac{1}{3} H ) cm^{3}$

= 227×10×10×(280+13×42)cm3$\frac{22}{7} \times 10 \times 10 \times (280 + \frac{1}{3} \times 42 ) cm^{3}$

= 22007×[280+14]$\frac{2200}{7} \times [280 + 14]$ = 92400 cm3

∴ Weight of pillar = 92400×7.51000=693kg$\frac{92400 \times 7.5}{1000} = 693 kg$

Q.15: The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be (1/27) of the volume of the given cone, at what height above the base, the section has been made?

Solution:

Let the smaller cone have radius = ‘r’ cm and height = ‘h’ cm

And, let the radius of the given original cone be R cm

Since, the two triangles, ΔOCDandΔOAB$\Delta OCD and \Delta OAB$ are similar to each other, we have,

rR=h30(since,ΔOCD ΔOAB)$\frac{r}{R} = \frac{h}{30} \: \: \: (since, \Delta OCD ~ \Delta OAB)$

r=Rh30$\Rightarrow r = \frac{Rh}{30}$ . . . . . . . . . . . . (1)

Given that the volume of the small cone is 127$\frac{1}{27}$ of the volume of the given cone,

13πr2h=127×13πR2×30[given]$\frac{1}{3} \pi r^{2} h = \frac{1}{27} \times {1}{3} \pi R^{2} \times 30 [given]$

13π[hR30]2h=181πR2×30[from(1)]$\Rightarrow \frac{1}{3} \pi [ \frac{hR}{30}]^{2} h = \frac{1}{81} \pi R^{2} \times 30 [from (1)]$
13πh3r3900=181πR2×30$\Rightarrow \frac{1}{3} \pi \frac{h^{3} r^{3}}{900} = \frac{1}{81} \pi R^{2} \times 30$
h3=1×30×900×381$\Rightarrow h^{3} = \frac{1 \times 30 \times 900 \times 3}{81}$
h3=1000cm3h=10cm$\Rightarrow h^{3} = 1000 cm^{3} \Rightarrow h = 10 cm$ From the figure,

AC = (OA – OC) = (30 – 10) cm = 20 cm

∴ The required height = 20 cm

Q.16: From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid (take 𝜋 = 3.14).

Solution:

Here, height (h) =10 cm and radius = 6 cm

∴ Volume of the remaining solid = (πr2h)13πr2h)$(\pi r^{2} h) – \frac{1}{3} \pi r^{2} h)$

= (π×6×6×10)cm3(13×π×6×6×10)cm3$(\pi \times 6 \times 6 \times 10) cm^{3} – ( \frac{1}{3} \times \pi \times 6 \times 6 \times 10) cm^{3}$

= 23π×6×6×10cm3$\frac{2}{3} \pi \times 6 \times 6 \times 10 cm^{3}$

= (23×3.14×360)cm3=753.6cm3$(\frac{2}{3} \times 3.14 \times 360) cm^{3} = 753.6 cm^{3}$

∴ Volume of the remaining solid = 753.6 cm3

Q.17: Water flows at a rate of 10 meters per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the surface 40 cm and depth 24 cm?

Solution:

Diameter of the pipe = 5 mm = 0.5 cm

Radius of the pipe = 0.25 cm

Length of the pipe = 10 m = 1000 cm

Volume that flows in 1 minute = [𝜋 × (0.25)2 × 24] cm3

∴ Required time = 13π×(20)2×24π×(0.25)2×1000$\frac{ \frac{1}{3} \pi \times (20)^{2} \times 24}{\pi \times (0.25)^{2} \times 1000}$ = 13πtimes400×24π×0.0625×1000$\frac{ \frac{1}{3} \pi times 400 \times 24}{ \pi \times 0.0625 \times 1000}$ = 51.2 min = 51 min 12 sec.

### Key Features of RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area Ex 13B(13.2)

• These solutions are easy to understand and explained in a simple language.
• Students are advised to solve the RS Aggarwal Maths Solution as it will improve their overall performance level.
• It provides different techniques to solve difficult and tricky questions.
• It provides the solutions to all the questions present in the exercises of RS Aggarwal Maths textbook.

#### Practise This Question

Given a base angle B and the difference of the sides AC and AB (not necessarily in this order), what is the difference in the steps of constructions when

i)                    AC > AB

ii)                    AC < AB