Question

A wooden toy is in the shape of a cylinder with a cone attached to one end
and a hemisphere attached to the other end as shown in the figure. All of
them are of the same radius of 1.5 cm. The total length of the toy is 7 cm
and height of cone is 2 cm. Calculate the volume of the toy.
2 cm
1.5 cm
7 cm
1.5 cm​

Answer 1

Step-by-step explanation:

r = 1.5 m

Height of the cylinder, h = 7 – (1.5 + 2) = 7 – 3.5 = 3.5 m

height of the cone = 2m

Volume of the rocket = Volume of the cone + volume of the cylinder + volume of hemisphere

= 1/3πr2h+ πr2h + 2/3πr3

= 1/3 x 22/7 x (1.5)2 x 2 + 22/7 x (1.5)2 x 3.5 + 2/3 x 22/7 x (1.5)3

= 36.53 m3

Answer 2

The volume of the toy is 36.56 cm³.

Given:

• A wooden toy is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other end as shown in the figure.
• All of them are of the same radius of 1.5 cm. The total length of the toy is 7 cm and the height of the cone is 2 cm.

To find:

• Find the volume of the toy.

Solution:

Formula to be used:

• Volume of hemisphere $\bf V_H = \frac{2}{3} \pi {r}^{3}$
• Volume of Cylinder $\bf V_{Cy} = \pi {r}^{2} \: h$
• Volume of cone: $\bf V_{Co} = \frac{1}{3} \pi {r}^{2} h$

Step 1:

Write the measurement of each shape.

Radius of Hemisphere, Cone and cylinder is (r) = 1.5 cm

Height of cone $\bf h_{co} = 2 \: cm$

Height of cylinder $h_{cy} = 7 - 2 - 1.5$

$\bf h_{cy} = 3.5 \: cm$

Step 2:

Find the volume of toy.

$\bf V_T= V_H+V_{Co}+V_{Cy}$

$V_T= \frac{2}{3} \pi {r}^{3} + \frac{1}{3} \pi {r}^{2} h_{co} + \pi {r}^{2} h_{cy}$

$V_T= \pi {r}^{2} \left( \frac{2}{3} {r} + \frac{1}{3} h_{co} + h_{cy} \right)$

Put the values in the formula.

$V_T= \frac{22}{7} {(1.5)}^{2} \left( \frac{2}{3} (1.5) + \frac{1}{3} \times 2 + 3.5 \right)$

$V_T= \frac{22}{7} \times 2.25 \left( 1+ 0.67 + 3.5 \right)$

$V_T= \frac{22}{7} \times 2.25 \times 5.17$

$V_T= \frac{255.92}{7}$

$\bf V_T= 36.56 \: {cm}^{3}$

Thus,

The volume of toy is 36.56 cm³.

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