Question

A jewel box is in the shape of a cuboid of dimensions 30cmx15 cmx10cm surmounted by a half part of a cylinder as shown in the figure.Find the volume and TSA of the box.the

Answer 1

7151.79$cm^{3}$ is the volume of the box and 1607.14 $cm^{2}$ is the TSA of the box.

Step-by-step explanation:

Let the length, breadth, and height of the box be l, b, and $h_{1}$ respectively

while r and $h_{2}$ as the radius and height of the cylinder:

As we know,

Volume of the box = volume of the cuboid + 1/2 (volume of the cylinder)

= (l * b * $h_{1}$) + 1/2 (π$r^{2} h_{1}$) cu. units

= (30 * 15 * 10) + 1/2(22/7 * 15/2 * 15/2 * 30)

= 7151.79cu. units.

Therefore, the volume of the box is equal to 7151.79$cm^{3}$

Now,

T.S.A. of the box = C.S.A. of the cuboid+ 1/2 (C.S.A. of the cylinder)

= 2(l + b)$h_{1}$ + 1/2(2πr$h_{2}$)

= 2( 45 * 10) + 1/2(2 * 22/7 * 15/2 * 30)

Therefore, TSA of the box = 1607.14$cm^{2}$

Learn more: TSA of the box

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Answer 2

Answer:

7151.79cm^{3}cm3 is the volume of the box and 1607.14 cm^{2}cm2 is the TSA of the box.

Step-by-step explanation:

Let the length, breadth, and height of the box be l, b, and h_{1}h1 respectively

while r and h_{2}h2 as the radius and height of the cylinder:

As we know,

Volume of the box = volume of the cuboid + 1/2 (volume of the cylinder)

= (l * b * h_{1}h1 ) + 1/2 (πr^{2} h_{1}r2h1 ) cu. units

= (30 * 15 * 10) + 1/2(22/7 * 15/2 * 15/2 * 30)

= 7151.79cu. units.

Therefore, the volume of the box is equal to 7151.79cm^{3}cm3

Now,

T.S.A. of the box = C.S.A. of the cuboid+ 1/2 (C.S.A. of the cylinder)

= 2(l + b)h_{1}h1 + 1/2(2πrh_{2}h2 )

= 2( 45 * 10) + 1/2(2 * 22/7 * 15/2 * 30)

Therefore, TSA of the box = 1607.14cm^{2}cm2