Question

A rectangular pyramid is shown below.

7 in.

7 in. -

_L-

10 in.

10 in

Which diagram is a net for the pyramid?

---

7 in.

7 in.

--.

17 in.

10 in.

17 in.

10 in.

10 in.

10 in.

7 in.

10 in.

10 in.

77 in.

10 in.

17 in.

What is the lateral area of the rectangular pyramid?

square inches

Answer 1

I can't read the net diagram from the text you pasted. Could you either upload a clear picture of the pyramid/net or else tell me exactly which three numbers are which (the base side lengths, the pyramid height, and/or the lateral edge or slant heights)?

While you do that, here is the formula and how to compute the lateral area so you can check quickly once you confirm the measures.

Notation:
- Let the base be a rectangle with sides a and b.
- Let h be the vertical height of the pyramid (apex above the base center).
- Let l be the lateral edge length (distance from apex to any base vertex) if that is given.
- Let s_a be the slant height of a triangular face whose base is side a (the distance from the midpoint of that base side to the apex).
- Let s_b be the slant height for base side b.

Formulas:
- s_a = sqrt(h^2 + (b/2)^2)
- s_b = sqrt(h^2 + (a/2)^2)
(If you are given l instead of h, then s_a = sqrt(l^2 - (a/2)^2) and s_b = sqrt(l^2 - (b/2)^2), because l^2 = h^2 + (a/2)^2 + (b/2)^2.)
- Lateral area LA = 2 * (1/2 a s_a) + 2 * (1/2 b s_b) = a*s_a + b*s_b.

Example (if the base is 7 in by 10 in and the lateral edge is 10 in):
- a = 7, b = 10, l = 10
- s_a = sqrt(10^2 - (7/2)^2) = sqrt(100 - 12.25) = sqrt(87.75) ≈ 9.370
- s_b = sqrt(100 - (10/2)^2) = sqrt(100 - 25) = sqrt(75) ≈ 8.660
- LA = 7·9.370 + 10·8.660 ≈ 65.59 + 86.60 ≈ 152.19 in^2.

Send the diagram or confirm which of a, b, h, l are 7, 10, 17, etc., and I'll compute the exact lateral area.