Question : In a triangle ${ABC}, {AB}={AC}$ and the perimeter of $\triangle {ABC}$ is $8(2+\sqrt{2}) $ cm. If the length of ${BC}$ is $\sqrt{2}$ times the length of ${AB}$, then find the area of $\triangle {ABC}$.
Option 1: $32 \ cm^2$
Option 2: $28 \ cm^2$
Option 3: $16 \ cm^2$
Option 4: $36 \ cm^2$
Correct Answer: $32 \ cm^2$
Solution :
Given: In a triangle ${ABC}, {AB}={AC}$ and the perimeter of $\triangle {ABC}$ is $8(2+\sqrt{2}) $ cm.
The length of ${BC}$ is $\sqrt{2}$ times the length of ${AB}$.
Let the sides $AB=AC=x$ cm.
⇒ $BC=\sqrt2x$ cm
The perimeter of the triangle,
$x+x+\sqrt2x=8(2+\sqrt2)$
⇒ $x(2+\sqrt2)=8(2+\sqrt2)$
⇒ $x=8$ cm
The sides $AB=AC=8$ cm and $BC=8\sqrt2$ cm.
So, these are the sides of a right-angle triangle.
⇒Area of the triangle = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 8 × 8
= $32 \ cm^2$
Hence, the correct answer is $32 \ cm^2$.
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