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 = [tex]\sqrt{(\frac{1}{2}b)^{2}+h^{2}}[/tex], where
b is the length of its base and h is the perpendicular height
- Its lateral area = [tex]\frac{1}{2}.p.l[/tex], 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) = [tex]\sqrt{(\frac{1}{2}b)^{2}+h^{2}}[/tex]
∴ l = The slant height of it = [tex]\sqrt{(\frac{1}{2}.24)^{2}+5^{2}}[/tex]
∴ l = [tex]\sqrt{(12)^{2}+25}=\sqrt{144+25}=\sqrt{169}=13[/tex]
∴ l = 13 units
∵ Perimeter of the square = b × 4
∴ The perimeter of the base (p) = 24 × 4 = 96 units
∵ The lateral area = [tex]\frac{1}{2}.p.l[/tex]
∴ The lateral area = [tex]\frac{1}{2}.(96).(13)[/tex]
∴ The lateral area = 624 unit²
* The lateral area is 624 unit²
