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

10 square root of 313+24 square root of 194+240

Explanation:

its right

i submitted it to rsm

Answer 2

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{*}10=240 sq units`

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

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

The slant height of isosceles triangles with a base of 10 units =
`((12)^(2)+(13)^(2))^(1)/(2)=(144+169)^(1)/(2)=17.691 units`

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

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