Question

8. Find the surface area of a triangular pyramid if the area of each phase is 16 cm² and each of its lateral faces has an area of 9 cm².

Answer 1

The base of a right pyramid is an equilateral triangle with area 16√3 cm². area of one of its lateral faces is 30 cm²

To find : height (in cm) i

Solution:

The base of a right pyramid is an equilateral triangle with area 16√3 cm²

Area of an equilateral triangle = (√3 / 4) Side ²

(√3 / 4) Side ² = 16√3

=> Side ² = 64

=> Side = 8 cm

Distance of center of triangle to mid point of Side for Equilateral triangle = side/ 2√3 = 8/ 2√3 = 4/√3

Let say Height = H

Slant Height = √((4/√3)² + H²)

Area of lateral surface = (1/2) * base * Slat Height

= (1/2) * 8 * Slant height = 30

=> Slant height = 15/2

15/2 = √((4/√3)² + H²)

=> 225/4 = 16/3 + H²

=> 611/12 = H²

=> H = 7.135 cm

Explanation: