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

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

Answer 2

Le the two equal sides of the triangle be 3x each and the unequal be 4x.
----{Because ratio given is 3:4}

Area of isosceles Δ = 18√5 unit²

But, area of anyΔ by herons formula = √s(s-a)(s-b)(s-c),
where, a, b, and c are the three sides of the Δ and s = (a+b+c)÷2.

In this case, a = b = 3x and c = 4x and we have to find 4x.
And also, s = (3x + 3x + 4x) ÷ 2 = 10x ÷ 2 = 5x.

Thus, area(Δ) = √5x(5x-3x)(5x-3x)(5x-4x)

= √5x(2x)(2x)(x) 

= √(2x²)² × 5

= 2x²√5

According To The Question.....

2x²√5 = 18√5

2x² = 18 × √5 ÷ √5 = 18 × 1 =18

2x² = 18

x² = 18 ÷ 2 = 9

x² = 9

x = √9 = √(3)² = 3

x = 3......

Now, the two equal sides = 3x = 3(3) = 9 units
and the unequal side = 4x = 4(3) = 12 units

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