Question

A frustum of regular square pyramid has height 2cm. the lateral faces of the pyramid are equilateral triangles of side 3√2 cm, then what is the volume of the frustum? A. 52 3 cm3 B. 44cm3 C. 52cm3 D. 104/3cm3

Answer 1

ANSWER

(a) Since all the edges of a square pyramid are of length 12 cm, lateral faces of the pyramid are equilateral triangles having each edge of 12 cm.

Thus, area of one lateral face =

4

3

×12×12=36

3

sq.cm

(b) An edge =12 cm=a

Slant height =l=

2

3

×1=6

3

cm

Surface area of pyramid

=a

2

+4×

4

3

a

2

=12

2

+

4

3

×12

2

=144×144

3

=144(1+

3

)sq.cm

(c) Length of edge of pyramid, a=12 cm

New length of edge, a

′

=24 cm

New surface area of the pyramid

=(a

′

)

2

+4×

4

3

(a

′

)

2

=24

2

+4×

4

3

×24

2

=576+576

3

=576(1+

3

)sq.cm

Previous surface are, S=144(1+

3

)sq.cm

New surface area, S

′

=576(1+

3

)sq.cm

⇒S

′

=4S

Thus, if the length of the sides of the pyramid is doubled, the surface area will be 4 times the previous one