Question

of)
A right pyramid on a base 10 m square is
15 m high
a)Find the volume of the pyramid
b) if the top 6m of the pyramid are
removed, what is the volume of the
remaining frustum​

Answer 1

Given :- A right pyramid on a base 10 m square is 15 m high .

To Find :- The volume of the pyramid ?

Answer :-

we know that,

• Volume of pyramid = (1/3) * Base area * height .
• Area of square = (side)² .

since base of the pyramid is a square .

so,

→ Volume of pyramid = (1/3) * (10)² * 15

→ Volume of pyramid = (1/3) * 100 * 15

→ Volume of pyramid = 100 * 5

→ Volume of pyramid = 500 m³ .

now, given that, the top 6m of the pyramid are removed .

so,

→ Ratio of volume of top part and total pyramid = (Ratio of corresponding sides)³

→ Top part removed Volume / Total Volume = (6/15)³

→ V2/500 = (2/5)³

→ V2/500 = 8/125

→ V2 = 32 m³ .

therefore,

→ The volume of the remaining frustum = Total volume - Top volume = 500 - 32 = 468 m³ (Ans.)

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Answer 2

Given :   A right pyramid on a base 10 m side square is 15 m high

To Find :

a)  volume of the pyramid

b) the top 6m of the pyramid are removed then the volume of the

remaining frustum​

Solution:

Volume of pyramid = (1/3) * b  * h

b = base area = 10 * 10 = 100 m²

h = 15 m

=> Volume of pyramid = (1/3) * 100 * 15 = 500 m³

top 6m of the pyramid

top part removed would be similar to original pyramid

Ratio of Volume of  pyramids = ( ratio of corresponding  base sides/height )³

ratio of corresponding height = 6/15  = 2/5

Ratio of Volume of  pyramids = (2/5)³ = 8/125

=> Volume of pyramid cut / 500 = 8/125

=>   Volume of pyramid cut  = 32  m³

volume of the remaining frustum​  = 500 - 32 = 468 m³

Other approach

side of base of upper part cut / 10  = 6/15

=> side of base of upper part cut = 4  m

upper base area = 4 * 4 = 16 m²

Lower base area = 100  m²

Height of  frustum​ = 15 - 6 = 9 m

Volume = (1/3) * ( lower base area + upper base area + √(lower base area * upper base area) ) * height of   frustum

= (1/3) ( 100 + 16 + √100 * 16 ) * 9

= 3 ( 116  + 40)

= 468 m³

volume of the remaining frustum​ = 468 m³  

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