Question

A regular square pyramid is placed in a cube so that the base of the pyramid and that of the cube coincide. The vertex of the pyramid lies on the face of the cube opposite to the base, as shown in the below. An edge of the cube is $7$ inches. How many square inches (approximately) are in the positive difference between the surface area of the cube and the surface area of the pyramid?

• A. $452.6$
• B. $158.6$
• C. $294$
• D. $135.4$

Answer 1

The correct option is D $135.4$

Lets assume the length of the sides of cube be $2a$.

Surface area of Cube $=6\times (2a{)}^2=24a^2$

Area of base of Cube $=$ Area of base of Pyramid $=4a^2$

Height of pyramid $=$ Side of Cube, i.e., $2a$

The perpendicular distance between tip of pyramid and the side of base side of pyramid (slant height) is

$l^2=a^2+(2a{)}^2=5a^2$

$l=\sqrt{5}{a}^2$

Surface Area of Pyramid $=4a^2+4\times \frac{1}{2}\times \left(2a\right)\times \sqrt{5}{a}^2$

$=4(1+\sqrt{5})a$

Difference between Area of Cube and pyramid

$=24a^2-4(1+\sqrt{5})a^2=4(5-\sqrt{5})a$

Now, it is given that side of cube is $7$ inches

Therefore $2a=7$

$\Rightarrow a=\frac{7}{2}$

Difference in area $=4(5-\sqrt{5})a^2=4(5-\sqrt{5}){\left(\frac{7}{2}\right)}^2$ inches${}^2$

$=49(5-\sqrt{5})=135.43$ inches${}^2$