Question

The area of an isosceles triangle is 60 cm^(2) and the length of its equal sides is 13cm. Find its base:


Need answer with proof

Answer 1

So A = 60cm²



AB=BC=13cm

A=1/2 b*h

So currently we only have the area but no height or base, so what you want to do is express height in terms of base.

So using Pythagoras :

a² = b² + c²

13² = h² + (1/2 b)²

h² = 169 - 1/4 b²

h = √(169 - 1/4 b²)

Back to area:

60 = 1/2 b √(169 - 1/4 b²)

We now solve for b:

120 = b√(169 - 1/4 b²)

14,400 = b²(169 - 1/4 b²)

1/4 b⁴ − 169b² + 14,400 = 0

b⁴ - 676b² + 57,600 = 0

(b + 24)(b + 10)(b - 10)(b - 24) = 0

So b = ± 10, 24

However, since this is geometric, we can’t get a negative length so b can either be 10 or 24 cm.

Answer 2

Answer:

10 cm or 24 cm.

Step-by-step explanation:

$A = \dfrac{bh_b}{2}\\\\\implies h_b = \sqrt{a^2 - \dfrac{b^2}{4}}\\\\\\\text{There are 2 solutions for b}\\\\b = 2 \sqrt{2} \dfrac{A}{\sqrt{a^2 \sqrt{a^4 - 4A^2}}}\\\\\\= 2 \sqrt{2} \dfrac{A}{\sqrt{13^2 \sqrt{13^4 - 4\times 60^2}}}\\\\\\= \boxed{10 \: cm \: approx}\\\\\\\textbf{Second Solution:}\\\\b = 2 \sqrt{2} \dfrac{A}{\sqrt{a^2 \sqrt{a^4 - 4A^2}}}\\\\\\= 2 \sqrt{2} \dfrac{A}{\sqrt{13^2 \sqrt{13^4 - 4\times 60^2}}}\\\\\\= \boxed{24 \: cm \: approx}$

Hope it helps.