Question

A pyramid has a rectangular base with edges of length 10 and 24. The vertex of the pyramid is directly 13 units above the center of the base. What is the surface area?

Answer 1

Answer:

Step-by-step explanation:

It is given that  A pyramid has a rectangular base with edges of length 10 and 24. The vertex of the pyramid is directly 13 units above the center of the base.. then The total surface area = area of rectangular base + area of 2 isosceles triangles with a base of 24 units + area of 2 isosceles triangles with a base of 10 units.

Area of rectangular base = $24{\times}10=240 sq units$

The slant height of isosceles triangles with a base of 24 units =$(\frac{10}{2})^{2}+(13)^{2})^\frac{1}{2}=(25+169)^{\frac{1}{2}}=13.928units$.

The area of 2 isosceles triangles with a base of 24 units=$\frac{2{\times}24{\times}13.928}{2}=334.281 sq units$

The slant height of isosceles triangles with a base of 10 units =$((12)^{2}+(13)^{2})^\frac{1}{2}=(144+169)^\frac{1}{2}=17.691 units$

The area of 2 isosceles triangles with a base of 10 units=$\frac{2{\times}10{\times}17.691}{2}=176.918 sq units$

The total surface area of the pyramid = 240 + 334.281 + 176.918 = 591.97 sq units.