# Question #672c6

Aug 19, 2017

As stated in the problem pyramid V-ABCD is cut from a cube of edge 12 in.,as shown in the figure. The vertex V is the midpoint of an upper edge of the cube. We are to compute the lateral surface of the pyramid.

• The lateral surface of the pyramid composed of 4 triangles
$\Delta V A B , \Delta V D C , \Delta V B C = \Delta V A D$

• Area of $\Delta V A B = \frac{1}{2} A B \times V E = \frac{1}{2} \times 12 \times 12 {\text{in"^2=72"in}}^{2}$

• Length of $V B = \sqrt{E {V}^{2} + E {B}^{2}} = \sqrt{{12}^{2} + {6}^{2}} = 6 \sqrt{5} \text{in}$

• In $\Delta V B C , \angle V B C = {90}^{\circ}$

• Length of $V C = \sqrt{B {V}^{2} + B {C}^{2}} = \sqrt{{\left(6 \sqrt{5}\right)}^{2} + {12}^{2}} = 18 \text{in}$

• Area of $\Delta V B C = \frac{1}{2} B C \times V B = \frac{1}{2} \times 12 \times 6 \sqrt{5} = 36 \sqrt{5} {\text{in}}^{2}$

• So Area of $\Delta V A D = 36 \sqrt{5} {\text{in}}^{2}$

• Now semi perimeter of of $\Delta V D C = \frac{1}{2} \left(18 + 18 + 12\right) = 24 \text{in}$

*So area of $\Delta V D C = \sqrt{24 \left(24 - 18\right) \left(24 - 18\right) \left(24 - 12\right)} {\text{in}}^{2}$

$= \sqrt{24 \times 6 \times 6 \times 12} {\text{in"^2=72sqrt2"in}}^{2}$

• Hence the lateral surface of the pyramid composed of 4 triangles
$= \Delta V A B + \Delta V D C + \Delta V B C + \Delta V A D$
$= 72 + 72 \sqrt{2} + 36 \sqrt{5} + 36 \sqrt{5} = 72 \left(1 + \sqrt{2} + \sqrt{5}\right) {\text{in}}^{2}$