Question

Find the perimeter of an isosceles right-angled $\Delta $ having an area of $\Delta $ is 500 square centimetres.

Answer 1

Hint: In the Isosceles triangle, two sides are equal, but given the isosceles triangle is right-angled triangle so its base and height will be equal. Here, in this question, we need to determine the perimeter of the triangle whose area is 500 square centimetres. For this, we will use the formula to find the area of isosceles as $\Delta \,i.e. = \dfrac{1}{2} \times base \times height$ and then use perimeter formula for $\Delta $, i.e. sum of all its sides.


Complete step by step solution:
Let $\Delta ABC$ be an Isosceles triangle. In an Isosceles triangle, two sides are equal, and it is given that it is a right-angled isosceles triangle right angled at B.
Let the length of the side $AB = a\,cm$
Length of side $BC = a\,cm$
Length of side $CA = b\,cm$
Here we have taken the base and height of isosceles as $'a'$ and length of hypotenuse as $'b'$.
From Right$\Delta ABC$
Area of Isosceles right-angled triangle $ = 500\,c{m^2}$(Given)
Since the area of a right-angled triangle$ = \dfrac{1}{2} \times base \times height$
$ \Rightarrow 500 = \dfrac{1}{2} \times a \times a$
\[
   \Rightarrow \dfrac{{{a^2}}}{2} = 500 \\
   \Rightarrow {a^2} = 500 \times 2 \\
   \Rightarrow {a^2} = 1000 \\
 \]
On taking square root both sides, we get
\[\sqrt {{a^2}} = \sqrt {1000} \]
\[
   \Rightarrow a = \sqrt {10 \times 10 \times 10} \\
  a = 10\sqrt {10} \,cm \\
 \]
Now in $\Delta ABC$
Use Pythagoras theorem
\[
  {(AC)^2} = {(BC)^2} + {(AB)^2} \\
  {b^2} = {(10\sqrt {10} )^2} + {(10\sqrt {10} )^2} \\
  {b^2} = \left( {100 \times 10} \right) + \left( {100 \times 10} \right) \\
  {b^2} = 1000 + 1000 \\
  {b^2} = 2000 - - - - (i) \\
 \]
Taking square root both sides of the equation (i),
\[
  \sqrt {{b^2}} = \sqrt {2000} \\
  b = \sqrt {20 \times 10 \times 10} \\
   = 10\sqrt {20} \\
   = 10\sqrt {2 \times 2 \times 5} \\
   = 20\sqrt 5 \\
 \]

As we know that perimeter of triangle$ = $sum of all three sides of the triangle
$
  P = a + a + b \\
   = 10\sqrt {10} + 10\sqrt {10} + 20\sqrt 5 \\
   = 10 \times 3.16 + 10 \times 3.16 + 20 \times 2.2360 \\
   = 31.6 + 31.6 + 44.72 \\
   = 107.92{\text{ cm}} \\
 $

Hence, the perimeter of the isosceles triangle whose area is 500 square centi-meters is 107.72 centi-meters.


Note: Students must keep in mind that in the question when the triangle is isosceles right-angled triangle then, they should use Pythagoras theorem in that case as they would not be able to find the area of the triangle directly by substituting the values in the formula of area of a triangle.