Question

SURFACE AREAS - PYRAMIDS?

Answer 1

The total surface area consists of the base area and the surface area (the area of all lateral faces). The base is a hexagon with side a=3 cm.

$A_{hexagon}=6A_{triangle}=6\cdot \dfrac{1}{2} \cdot \dfrac{a^2\sqrt{3}}{4} =\dfrac{3}{4} \cdot 3^2\cdot \sqrt{3}=\dfrac{27\sqrt{3}}{4}$ sq. cm.

Each lateral face is an isosceles triangle with height h=12 cm and base a=3 cm. The area of one such lateral triangle is

$A_{lateral}=\dfrac{1}{2}\cdot h\cdot a= \dfrac{1}{2}\cdot 12\cdot 3=18$ sq. cm.

Then the total surface area is:

$A_{total}= \dfrac{27\sqrt{3}}{4} +6\cdot 18=\dfrac{27\sqrt{3}}{4} +108=11.7+108=119.7$ sq. cm.

Answer: 119.7 sq. cm.