The perimeter of an isosceles triangle is 14 | vedclass

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(B) Let the lateral side be $5x$ and the base be $4x$.Since it is an isosceles triangle,the two lateral sides are equal.Perimeter $= 5x + 5x + 4x = 14

Vedclass · Accepted Answer

$2 \sqrt{21} \ cm^2$

Vedclass · Answer

(B) Let the lateral side be $5x$ and the base be $4x$.Since it is an isosceles triangle,the two lateral sides are equal.Perimeter $= 5x + 5x + 4x = 14x$.Given perimeter $= 14 \ cm$,so $14x = 14 \Rightarrow x = 1$.Thus,the sides are $5 \ cm, 5 \ cm$,and $4 \ cm$.Using the formula for the area of an isosceles triangle: $	ext{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$,where $a = 5$ and $b = 4$.$	ext{Area} = \frac{4}{4} \sqrt{4(5)^2 - (4)^2} = \sqrt{100 - 16} = \sqrt{84} = 2\sqrt{21} \ cm^2$.

The perimeter of an isosceles triangle is $14 \ cm$. The ratio of the lateral side to the base is $5:4$. The area of the triangle is:

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