Question

A solid toy is in the form of a hemisphere surmounted by a a right circular cone . The height of the cone is 4cm and the diameter of the base is 8cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volume of cylinder and toy.

Answer 1

Hi friend,

DIAMETER = 8CM

RADIUS = D/2 = 8/2 = 4CM

HEIGHT OF CONICAL PART OF THE TOY = 4CM.

VOLUME OF TOY = VOLUME OF HEMISPHERICAL PART OF THE TOY + VOLUME OF CYLINDRICAL PART OF THE TOY.

= 2/3πR³ + 1/3πR²H

PUT VALUES AND SOLVE U WILL GET THE TOTAL VOLUME OF THE TOY..

Answer 2

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Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere as shown in the above figure.

The radius BO of the hemisphere (as well as of the cone) =( ½) × 4 cm = 2 cm.

So, volume of the toy = (⅔) πr3 + (⅓) πr2h

$\tt = (⅔) × 3.14 × 23 + (⅓)× 3.14 × 22 × 2$

$\tt = 25.12 cm^3$

Now, let the right circular cylinder EFGH circumscribe the given solid.

The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is

$\tt EH = AO + OP = (2 + 2) cm = 4 cm$

So, the volume required = volume of the right circular cylinder – the volume of the toy

$\tt = (3.14 × 22 × 4 – 25.12) cm^3$

$\tt = 25.12 cm^3$

Hence, the required difference between the two volumes = 25.12 cm^3

Hope it's Helpful.....:)