SOLUTION: How many square inches are there in the lateral surface of a regular octagonal pyramid with 12 and 18 inches of basal and lateral edges? Round off the answer to the nearest integer

Algebra ->  Polygons -> SOLUTION: How many square inches are there in the lateral surface of a regular octagonal pyramid with 12 and 18 inches of basal and lateral edges? Round off the answer to the nearest integer      Log On


   

Question 1198305: How many square inches are there in the lateral surface of a regular octagonal pyramid with 12 and 18 inches of basal and lateral edges? Round off the answer to the nearest integer.
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Diagram

AB = 12 is the side length along the base.

AE = EB = lateral edge length = 18 inches

Focus on right triangle ADE
AD = (AB)/2 = 12/2 = 6 inches
This works because I've defined D as the midpoint of AB.

We'll use the pythagorean theorem to find the length of segment DE (aka slant height).
a%5E2+%2B+b%5E2+=+c%5E2

b+=+sqrt%28c%5E2-a%5E2%29

DE+=+sqrt%28%28AE%29%5E2-%28AD%29%5E2%29

DE+=+sqrt%28%2818%29%5E2-%286%29%5E2%29

DE+=+sqrt%28288%29

Let's now find the area of isosceles triangle ABE.
base = AB = 12
height = DE = sqrt%28288%29


area+=+%281%2F2%29%2Abase%2Aheight

area+=+%281%2F2%29%2AAB%2ADE

area+=+%281%2F2%29%2A12%2Asqrt%28288%29

area+=+6%2Asqrt%28288%29
Multiply this area by 8 since there are 8 congruent triangular panels along the lateral surface area (LSA).



LSA+=+8%2A%286%2Asqrt%28288%29%29

LSA+=+48%2Asqrt%28288%29

LSA+=+814.587011926902
This value is approximate.

When rounding to the nearest whole number, we get the final answer of 815 square inches