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Mama L [17]
4 years ago
11

SURFACE AREAS - PYRAMIDS?

Mathematics
1 answer:
Bond [772]4 years ago
7 0

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.

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see below

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