﻿ Volume of Pyramids and Cones Worksheet | Problems & Solutions

# Volume of Pyramids and Cones Worksheet

Volume of Pyramids and Cones Worksheet
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1.
3 iron bars of dimensions 1.25 m × 1.25 m × 5 m are melted to mould conical pieces of base radius 10 cm and height 20 cm. Find out the number of pieces that can be made taking $\pi$ = 3.
 a. 3906 b. 85 c. 117 d. 11719

#### Solution:

Number of conical pieces = Volume of the iron bars Volume of one cone
[Formula.

Volume of the iron bars = 3(1.25 × 1.25 × 5) = 23.438 m³
[Simplify.]

Volume of one cone = 1 / 3π (102) × 20 = 2000 cm3
[Volume of a cone = 1 / 3π r2 h.]

Number of cones that can be moulded = 23.438 × 100 × 100 × 100 2000 = 11719
[1 m3 = 1000 litres.]
[Substitute and simplify.]

2.
A pile of sand is in the form of cone. The circumference of its base is measured as 540 cm. The slant height is measured as 150 cm. This has to be transported in a lorry. The sand is loaded on to the lorry at the rate of 4500 c.c/min. How long will it take to load the entire sand on the lorry? [Take $\pi$ = 3.]
 a. 224 min b. 221 min c. 222 min d. 216 min

#### Solution:

Circumference of the base of the cone = 540 cm
[Given.]

Radius of the cone (r) = 5402 × 3 = 90 cm
[Radius of the cone (r) = c / , where c is the circumference. ]

Slant height = 150 cm
[Given.]

Height of the cone(h) = 1502 -902 = 120 cm
[Height of the cone = l2 -r2.]

Volume of the sand = 1 / 3× 3 × (90)2 × 120 = 972000 cm3
[Volume of the cone = 1 / 3πr2 h.]

[Given.]

[Formula.]

Time taken to load = 972000 / 4500= 216 min.
[Substitute in step 7 and simplify.]

3.
121 litres of grains is stored on a paddy field in the form of a cone. The height of the vertex of the cone from the ground is 250 cm. Find the circumference of the base of the cone. [Take $\pi$ = 3. ]
 a. 137 cm b. 132 cm c. 157 cm d. 144 cm

#### Solution:

Volume of the grains = 121 litres = 121 ×1000 c.c = 121000 c.c

[Given.]
[1 litre = 1000 c.c .]

Height of the cone = 250 cm
[Given.]

1 / 3× 3 × r2 × 250 = 121000
[Volume of a cone = 1 / 3π r2h.]

r2 = 121000 / 250
[Divide each side by 250.]

r = 22 cm
[Simplify.]

Circumference of the base of the cone = 2 × 3 × 22 = 132 cm
[Circumference = 2 π r.]

4.
A metallic solid in the form of a square pyramid of base 16 cm and height 2 cm is melted to make conic pieces of base radius 3 cm and height 3 cm. How many pieces can be made approximately? [Take $\pi$ = 3.]
 a. 35 b. 25 c. 28 d. 6

#### Solution:

Number of conic pieces = Volume of the pyramid Volume of one cone
[Formula.]

Volume of the pyramid = 1 / 3× 162 × 2 = 170.67 c.c
[Volume of pyramid = 1 / 3b2h.]

Volume of one cone = 1 / 3× 3 × 32 × 3 = 27 c.c
[Volume of one cone = 1 / 3π r2h.]

Number of cones that can be made = 170.67 / 27= 6 (approximately)
[Substitute in step 1.]

5.
A tent is in the form of a square pyramid with a base length of 21 m. The slant height of the pyramid is 16 m. If the rate of making the tent is $60 per cubicmeter, then what is the cost of making the tent?  a.$106496.40 b. $106507.40 c.$106457.40 d. $106485.40 #### Solution: Base length = 21 m [Given.] Slant height = 16 m. Height of the tent = 162-(212)2 = 12.07 m [Height of tent = l2-a24.] Volume of the tent = 1 / 3× 212 × 12.07 = 1774.29 m3 [Volume of the pyramid = 1 / 3a2h.] Rate of making =$60 per m3
[Given.]

Cost of making the tent = Volume of the tent × Rate of making
[Formula.]

Cost of making the tent = 1774.29 × 60 = \$106457.40
[Substitute in step 6 and simplify.]

6.
A cone has a base radius of $r$ and height $h$. If $x$ is the measure of the height of a similar cone with half the volume of the original cone, then what is the relation between $x$ and $h$?
 a. $h$3 = 2 $x$3 b. $x$ = $\sqrt{\frac{2h}{3}}$ c. $x$ = $\frac{h}{2}$ d. $x$ = $\sqrt{h}$

#### Solution:

AB = r, OB = h, OD = x, CD = rxh.
[Δ ABO ~ Δ CDO.]

Volume of the original cone = 1 / 3 π r2 h
[Formula.]

Volume of the second cone = 1 / 3 π CD2 x
[From the figure, height = x.]

Volume of the second cone =13h² π r2 x3
[From step 1, CD = r × xh.]

Volume of first cone = 2 (Volume of the second cone).
[According to the data.]

1 / 3π r2 h = 2 [ 13h² π r2 x3]
[From steps 2 and 4.]

h = 2 x³h²
[Simplify.]

The relation between h and x is h 3 = 2 x3.
[Multiply each side by h2.]

7.
What is the ratio of volumes of a cone of radius 28 cm and height 42 cm to a cylinder of radius 7 cm and height 14 cm? [Take $\pi$ = 3.]
 a. 16 : 1 b. 22 : 1 c. 21 : 1 d. 24 : 1

#### Solution:

Volume of the cone = 1 / 3× 3 × (28)2 × 42 = 32928
[Volume of the cone = 1 / 3π r2h.]

Volume of the cylinder = 3 × (7)2 × 14 = 2058
[Volume of the cylinder = π r2h.]

Ratio of the volumes = Volume of the cone Volume of the cylinder
[Formula.]

= 329282058 = 161
[From steps 1 and 2.]

Ratio of the volumes = 16 : 1

8.
A tank is in the shape of a cone with a depth of 20 cm and maximum inner radius of 30 cm. Water is pumped into it at a rate of 900 c.c per minute. How long it takes for the vessel to get filled up? [ Take $\pi$ = 3. ]
 a. 20 minutes b. 74 minutes c. 52 minutes d. 36 minutes

#### Solution:

Time taken to fill up the tank = Volume of the vessel Rate of pumping
[Formula.]

Volume of the vessel = 1 / 3 × 3 × 302 × 20 = 18000 cm3
[Volume of the cone = 1 / 3π r2 h.]

Rate of pumping = 900 c.c per minute
[Given.]

Time taken to fill up the tank = 18000 / 900 = 20 minutes
[From steps 2 and 3.]

9.
The height of 3 ounces of liquid in a conical cup is half of the height of the cup. How many ounces of the liquid will be required to fill the entire conical cup?

 a. 24 ounces b. 6 ounces c. 29 ounces d. 12 ounces

#### Solution:

Volume of the cup, V = 1 / 3π R2 H
[Volume of the cone = 1 / 3π r2 h.]

Volume of the part of the conical cup containing liquid in it, v = 1 / 3 πr2h

Consider ΔABC and ΔDEC, Rr=Hh
[As ΔABC and ΔDEC are similar triangles.]

Rr=HH2
[Given, h = H2.]

Rr=2
[Simplify.]

Vv = volume of the total cupvolume of the part of the cup containing liquid

Vv=13πR2H13πr2h
[Formula.]

V3=(Rr)2×HH2
[Volume of the liquid = 3 ounces.]

V3=4×2
[From steps 4 and 5.]

V = 3 × 4 × 2
[Multiply with 3 on both sides.]

V = 24 ounces
[Simplify.]

So, the volume of the liquid required to fill the entire conical cup = 24 ounces

10.
A metallic cone of radius 8 cm and height 18 cm is melted and made into identical spheres each of radius 2 cm. How many spheres can be made?
 a. 36 b. 24 c. 48 d. 72

#### Solution:

Number of spheres = Volume of cone Volume of each sphere
[Formula.]

Volume of a cone = 1 / 3× 22 / 7× 8 × 8 × 18 = 8448 / 7
[Volume of a cone = 1 / 3π r2h.]

Volume of each sphere = 4 / 3× 22 / 7× (2)3 = 704 / 21
[Volume of sphere = 4 / 3π r3.]

Number of spheres = 8448 / 7× 21 / 704= 36
[Substitute in step 1 and simplify.]

Therefore, 36 spheres can be made.