Question

What is the lateral area of a regular square pyramid if the base edges are of length 24 and perpendicular height is 5?

Answer 1

Answer:

The lateral area is 624 unit²

Step-by-step explanation:

* Lets explain how to solve the problem

- The regular square pyramid has a square base and four congruent

 triangles

- The slant height of it = $\sqrt{(\frac{1}{2}b)^{2}+h^{2}}$, where

  b is the length of its base and h is the perpendicular height

- Its lateral area = $\frac{1}{2}.p.l$, p is the perimeter of the base

 and l is the slant height

* Lets solve the problem

∵ The base of the pyramid is a square with side length 24 units

∵ Its perpendicular height is 5 units

∵ The slant height (l) = $\sqrt{(\frac{1}{2}b)^{2}+h^{2}}$

∴ l = The slant height of it = $\sqrt{(\frac{1}{2}.24)^{2}+5^{2}}$

∴ l = $\sqrt{(12)^{2}+25}=\sqrt{144+25}=\sqrt{169}=13$

∴ l = 13 units

∵ Perimeter of the square = b × 4

∴ The perimeter of the base (p) = 24 × 4 = 96 units

∵ The lateral area = $\frac{1}{2}.p.l$

∴ The lateral area = $\frac{1}{2}.(96).(13)$

∴ The lateral area = 624 unit²

* The lateral area is 624 unit²