Question

How do you find the area of an isosceles right angle triangle with a perimeter of 40 units?

Answer 1

Hint: An isosceles right-angle triangle means that two of the sides of the right-angled triangle are equal, we are given that the perimeter of this triangle is equal 40 units, that is, the sum of the three sides of the triangle is equal to 40 units and we can find the relation between the three sides of the triangle using Pythagoras theorem. Using the above-mentioned information, we can formulate two equations and find the required quantity.

Complete step-by-step answer:
Let the base and the height of the isosceles right-angle triangle be of the length x units and the length of the hypotenuse be y units.
By Pythagoras theorem, we get –
 $
  {x^2} + {x^2} = {y^2} \\
   \Rightarrow 2{x^2} = {y^2} \\
   \Rightarrow y = \sqrt {2{x^2}} \\
   \Rightarrow y = \pm x\sqrt 2 \;
  $
The negative value is rejected as length cannot be negative, so $ y = x\sqrt 2 $
Now, the perimeter is equal to 40 units, so –
 $
  x + x + y = 40 \\
  2x + x\sqrt 2 = 40 \\
   \Rightarrow x = \dfrac{{40}}{{2 + \sqrt 2 }}units \;
  $
Thus, the base and the height of the triangle is equal to $ \dfrac{{40}}{{2 + \sqrt 2 }} $ .
The area of a right-angled triangle is given as –
 $
  Area = \dfrac{1}{2} \times base \times height \\
   \Rightarrow Area = \dfrac{1}{2} \times {(\dfrac{{40}}{{2 + \sqrt 2 }})^2} \\
   \Rightarrow Area = \dfrac{1}{2} \times (\dfrac{{1600}}{{4 + 2 + 4\sqrt 2 }}) \\
   \Rightarrow Area = \dfrac{{800}}{{6 + 4\sqrt 2 }} \\
   \Rightarrow Area = \dfrac{{400}}{{3 + 2\sqrt 2 }}sq.units \;
  $
Hence the area of the given isosceles right-angle triangle is $ \dfrac{{400}}{{3 + 2\sqrt 2 }}unit{s^2} $ .
So, the correct answer is “ $ \dfrac{{400}}{{3 + 2\sqrt 2 }}unit{s^2} $ .”.

Note: The three sides of a right-angled triangle are related by the Pythagoras theorem, according to which, the square of the hypotenuse is equal to the sum of the square of the base and the perpendicular. From this theorem, we clearly see that the hypotenuse is greater than both the base and the perpendicular, so it can’t be equal to any of them that is why we take the base and height equal.