Question

find the base of isosceles triangle whose area is 12 CM square and one of equal side is 5cm

Answer 1

Answer:

6 or 8 cm

Step-by-step explanation:

Let say base of triangle = B cm

Perpendicular to base = H

5^2 = H^2 + (B/2)^2

Area of triangle = (1/2)×B×H

(1/2)×B×H = 12

H = 24/B

5^2 = (24/B)^2 + (B/2)^2

25 = 576/B^2 + B^2/4

Multiplying by 4B^2both sides

100B^2 = 576×4 + B^4

B^4 - 100B^2 + 2304 = 0

B^4 - 64B^2 - 36B^2 + 2304 = 0

B^2(B^2 - 64) -36(B^2-64) = 0

(B^2 -36) (B^2-64) = 0

B^2 = 36

B = 6

B^2 = 64

B = 8

So base of triangle can be 6 or 8 cm