Question

72. Find the altitude AD of an isosceles triangle angle ABC with sides
2a, 2a and a, if AB = AC -​

Answer 1

let ABC be a Δ with AB=AC

AB=2a AC=2a BC=a

using herons formula

perimeter of Δ = sum of all its sides

                         = 2a + 2a+a

                          = 5a

semi perimeter(s) =5a/2

herons formula= $\sqrt{(s-a)(s-b)(s-c)$

                         =$\sqrt5a/2(5a/2-a/2)(5a/2-a/2)(5a/2-a)$

                           = $\sqrt5a/2(a/2)(a/2)(3a/2)$

                            = a²/4√15

now we know that area of triangle = $\frac12$×b×h

⇒ a²/4√15=$\frac12$×a×h

⇒h=(2a²×√15)÷4a

⇒h=√15×a²/2

∴altitude ofΔ is (√15 × a²)/2