Question

the ratio of lenght of each equal side and the third side of an isosceles triangle is 3:4 . if the area of triangle is 18√5sq Units. The Third Side Is ?

Answer 1

plz check the answer...

Answer 2

Let the sides of an isosceles triangle be 3x, 3x and 4x
$s = \frac{a + b + c}{2} = \frac{3x + 3x + 4x}{2} = \frac{10x}{2} = 5x \\area \: = 18 \sqrt{5}sq \: units \: \: \: \:...... (given) \\ area = \sqrt{s(s - a)(s - b)(s - c)} \\ 18 \sqrt{5} = \sqrt{5x(5x - 3x)(5x - 3x)(5x - 4x)} \\ 18 \sqrt{5} = \sqrt{5x \times 2x \times 2x \times x} \\ 18 \sqrt{5} = \sqrt{20x ^{4} } \\ 18 \sqrt{5} =2 \sqrt{5} {x}^{2} \\ {x}^{2} = \frac{18 \sqrt{5} }{2 \sqrt{5} } \\ {x}^{2} = 9 \\ x = 3 \\ so \: the \: third \: side \: of \: isosceles \: triangle \: = 4 \times 3 = 12 \: units$