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
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(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
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