Question

a right rectangular pyramid with a height of 12 centimeters is shown. what is the surface area of the pyramid to the nearest square centimeter? type the correct answer in the box. use numerals instead of words. the surface area of the rectangular pyramid is about square centimeters.

Explanation:

Step1: Calculate base area

The base is a rectangle with length $l = 10$ cm and width $w=8$ cm. The area of the base $A_{base}=l\times w = 10\times8=80$ $cm^{2}$.

Step2: Calculate area of two pairs of triangular faces

For the pair of triangles with base 10 cm:
First, find the slant - height $s_1$. Using the Pythagorean theorem, $s_1=\sqrt{(\frac{8}{2})^{2}+12^{2}}=\sqrt{16 + 144}=\sqrt{160}=4\sqrt{10}$ cm.
The area of one such triangle is $A_{1}=\frac{1}{2}\times10\times4\sqrt{10}=20\sqrt{10}$ $cm^{2}$, and the area of the pair is $2A_{1}=40\sqrt{10}$ $cm^{2}$.
For the pair of triangles with base 8 cm:
Find the slant - height $s_2$. Using the Pythagorean theorem, $s_2=\sqrt{(\frac{10}{2})^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$ cm.
The area of one such triangle is $A_{2}=\frac{1}{2}\times8\times13 = 52$ $cm^{2}$, and the area of the pair is $2A_{2}=104$ $cm^{2}$.

Step3: Calculate total surface area

$A = A_{base}+2A_{1}+2A_{2}=80 + 40\sqrt{10}+104$.
$40\sqrt{10}\approx40\times3.162 = 126.48$.
$A=80 + 126.48+104=310.48\approx310$ $cm^{2}$.

Answer:

310