Question

Find the perimeter of an isosceles right-angled triangle having area of $5000{{m}^{2}}$ $\left( \text{ use }\sqrt{2}=1.41 \right)$

Answer 1

Hint: Now we know that area of any triangle is given by $\dfrac{1}{2}\times base\times height$ . Now we have an isosceles right-angle triangle. Hence its base and height would be the same. Now we are given that the area of the triangle is $5000{{m}^{2}}$. Hence we will get the length of two equal sides of the isosceles triangle. Now we can find the third side using the Pythagoras theorem. Now once we have all three sides we will add them to find the perimeter of the triangle.

Complete step-by-step solution:
Now let us consider a right-angled isosceles triangle ABC such that $\angle B={{90}^{\circ }}$.

Now we know that in an isosceles triangle two angles are equal.
Now we have $\angle B={{90}^{\circ }}$
Hence $\angle A=\angle C$.
This means that $AB = BC. $
Now we know that area of the triangle is given by formula $\dfrac{1}{2}\times base\times height$.
Now in triangle ABC, we have base = BC and height = AC
Hence area of triangle ABC = $\dfrac{1}{2}\times BC\times AC$
But we have $BC = AC. $
Let us take $BC = AC = x.$
Hence we get Area of triangle ABC is $\dfrac{1}{2}\times {{x}^{2}}$
Now we are given that area of triangle is $5000{{m}^{2}}$
Hence we have
$\begin{align}
  & \dfrac{1}{2}\times {{x}^{2}}=5000 \\
 & \Rightarrow {{x}^{2}}=10000 \\
 & \therefore x=100 \\
\end{align}$
Now we have $AC = BC = 100 $………………. (1)
Now let us use Pythagoras theorem in the right angle triangle ABC.
Hence we get $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
From equation (1) we get.
$\begin{align}
  & {{100}^{2}}+{{100}^{2}}=A{{C}^{2}} \\
 & \Rightarrow A{{C}^{2}}=10000+10000 \\
 & \Rightarrow A{{C}^{2}}=20000 \\
 & \Rightarrow AC=\sqrt{2\times 10000} \\
 & \therefore AC=100\sqrt{2} \\
\end{align}$
Now we are given that $\sqrt{2}=1.41$
Hence we have AC = $100\times 1.41=141$.
Now $AC = 141$………………………. (2)
Now we know that perimeter of a triangle nothing but the sum of the length of sides
Hence we get perimeter of triangle = $AB + BC + AC.$
Hence we have perimeter = $100 + 100 + 141 = 341$
Hence perimeter of triangle ABC is 341m.

Note: Now note that in right isosceles angle triangle ABC if $\angle B={{90}^{\circ }}$ then always we will get $\angle A=\angle C$. Now, this is because in an isosceles triangle we know that two angles are equal. Now let us say we have \[\angle A=\angle B\] or \[\angle B=\angle C\]. This means \[\angle A={{90}^{\circ }}\] or $\angle B={{90}^{\circ }}$. But if this is true we won’t get a triangle as the sum of angles of triangles is ${{180}^{\circ }}$ as we have $\angle C+\angle B={{180}^{\circ }}$ or $\angle A+\angle B={{180}^{\circ }}$. Which means one angle is always 0. Hence this will never form a triangle. This means the only possible case is $\angle A=\angle C$