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 =
, where
b is the length of its base and h is the perpendicular height
- Its lateral area =
, 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}}](https://tex.z-dn.net/?f=%5Csqrt%7B%28%5Cfrac%7B1%7D%7B2%7Db%29%5E%7B2%7D%2Bh%5E%7B2%7D%7D)
∴ l = The slant height of it = ![\sqrt{(\frac{1}{2}.24)^{2}+5^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%28%5Cfrac%7B1%7D%7B2%7D.24%29%5E%7B2%7D%2B5%5E%7B2%7D%7D)
∴ l = ![\sqrt{(12)^{2}+25}=\sqrt{144+25}=\sqrt{169}=13](https://tex.z-dn.net/?f=%5Csqrt%7B%2812%29%5E%7B2%7D%2B25%7D%3D%5Csqrt%7B144%2B25%7D%3D%5Csqrt%7B169%7D%3D13)
∴ 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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D.p.l)
∴ The lateral area = ![\frac{1}{2}.(96).(13)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D.%2896%29.%2813%29)
∴ The lateral area = 624 unit²
* The lateral area is 624 unit²