RS Aggarwal Solutions Class 9 Ex 13B


RS Aggarwal Class 9 Ex 13B Chapter 13

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

= [13×227×35×35×84]cm3 = 107800 cm3

∴ Curved surface area = [πrh2+r2]=h2+r2

= πr842+352 = πr8281

= 227×35×91 = 10010 cm2

∴ Total surface area = 𝜋r(l + r)

Now,  l = h2+r2 = 842+352

= 7056+1225=8281=91cm

∴ Total surface area = 227×35(91+35)

= (22×5×126)cm2=13860cm2

 

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 = 10262=10036 = 64=8cm

∴ Volume of cone = 13πr2h

= [13×227×8×8×6]cm3 = 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

100𝜋 = 13πr212

r2=100π×3π×12
r2=25
r=25=5cm.

Slant height (l) = h2+r2 = 122+52

l = 144+25=169=13cm.

∴ 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
r=550×722×25cm=7cm

Now, height (h) = l2r2 = (25)2(7)2 = 62549 = 576=24cm

∴ Height of the cone = 24 cm.

Volume of the cone = 13πr2h = [13×227×7×7×24]cm3 = 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 =  372352

=13691225=144=12cm

∴ Height (h) = 12 cm.

Volume of the cone = 13πr2h

= [13×227×35×35×12]cm3 = 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 radius = 35 cm

∴ Curved surface area = 𝜋rl

4070=227×35×l
l=4070110=37cm

∴ 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 = 242+72

= 576+49=625=25m

Now, area of the cloth = 𝜋rl

area=227×7×25=550m2

∴ Length of cloth = Areaofclothwidthofcloth=5502.5=220m

∴ 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

h=13π×1.6×1.6×3.613π×1.2×1.2=6.4cm

∴ 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

And, V2=13π(r)2×3h

V1V2=13π(3r)2×h13π(r)2×3h=31

∴ 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 m and its height, H = 3m

Slant height (l) = 53 m

∴ Area of canvas = (2𝜋RH + 𝜋Rl)

= [(2×227×1052×3)+(227×1052×53)]m2 = (990 + 8745) m2 = 9735 m2

∴ Length canvas = areaofcanvaswidthofcanvasm = 97355=1947m

 

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
r2=14m

Volume of the cone = 13πr2h

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

220=13×227×14×h
h=220×322×2=15m

∴ 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

R2=π×18×18×32×3π×24=1296
R=1296=36cm

∴ Radius of the heap = 36 cm

Slant height (l) = h2+R2 = 242+362 = 576+1296 = 1872=43.27cm

∴ 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 . . . . . . . (ii)

Squaring both sides of equation (i), we have

(2πrh)2=(8x)2

4π2r2h2=64x2 . . . . . . . . . (iii)

From (ii), we have,

πrh2+r2=5x

Squaring both sides,

π2r2(h2+r2)=25x2 . . . . . . . . . .(iv)

4π2r2h2π2r2(h2+r2=6425  [Divide (iii) by (iv)]

h2(h2+r2)=1625
9h2=16r2
r2h2=916
rh=34

∴ 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

radius = 10 cm

Height of the cone (H) = 42 cm

∴ Volume of the pillar = (πr2h+13πr2H)cm3

= πr2(h+13H)cm3

= 227×10×10×(280+13×42)cm3

= 22007×[280+14] = 92400 cm3

∴ Weight of pillar = 92400×7.51000=693kg

 

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 are similar to each other, we have,

rR=h30(since,ΔOCD ΔOAB)

r=Rh30 . . . . . . . . . . . . (1)

Given that the volume of the small cone is 127 of the volume of the given cone,

13πr2h=127×13πR2×30[given]

13π[hR30]2h=181πR2×30[from(1)]
13πh3r3900=181πR2×30
h3=1×30×900×381
h3=1000cm3h=10cm

 

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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)

= (π×6×6×10)cm3(13×π×6×6×10)cm3

= 23π×6×6×10cm3

= (23×3.14×360)cm3=753.6cm3

∴ 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 = 13πtimes400×24π×0.0625×1000 = 51.2 min = 51 min 12 sec.


Practise This Question

In a ABC, AB = 10cm, CAB = 30 and CBA = 60. In DEF, FE = 10cm, DFE = 30, DEF = 60. If the two triangles are congruent, then, which of the following is true?