Question

If the perimeter of an isosceles right triangle is (6 + $3\sqrt{2}$) m, then what is the area of the triangle?
(a) 4.5 ${{m}^{2}}$
(b) 5.4 ${{m}^{2}}$
(c) 9 ${{m}^{2}}$
(d) 81 ${{m}^{2}}$

Answer 1

Hint: To solve this problem, one must be aware about the basic properties of triangles (isosceles right triangle in this case). We will require the formulas for the perimeter of an isosceles right triangle and its area. The perimeter is given by 2x + y (where x is the length of sides which have equal length and y is the length of the other side). The area of the isosceles right triangle is given by $\dfrac{1}{2}{{x}^{2}}$. We will use this to solve this problem in hand.

Complete step-by-step answer:
Before beginning the problem, we will address a few important things to refer to. We will refer to the right triangle ABC below with AB = BC = x, AC = y and angle ABC = 90 degrees.

Also, we can use the Pythagoras theorem on this triangle, thus, we get,
$A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
${{x}^{2}}+{{x}^{2}}={{y}^{2}}$
${{y}^{2}}=2{{x}^{2}}$
y = $\sqrt{2}x$ -- (1)
Now, the perimeter of this triangle is (6 + $3\sqrt{2}$), thus, we have,
Perimeter of ABC = x + x + y
Perimeter of ABC = 2x + $\sqrt{2}x$
(6 + $3\sqrt{2}$) = 2x + $\sqrt{2}x$
x = $\dfrac{6+3\sqrt{2}}{2+\sqrt{2}}$
Dividing the numerator and denominator by $\sqrt{2}$, we get,
x = $\dfrac{3\sqrt{2}+3}{\sqrt{2}+1}$ -- (A)
Now, the area is given by $\dfrac{1}{2}{{x}^{2}}$. Thus, we have,
= $\dfrac{1}{2}{{x}^{2}}$
= $\dfrac{1}{2}{{\left( \dfrac{3\sqrt{2}+3}{\sqrt{2}+1} \right)}^{2}}$
By rationalization, we get,
= $\dfrac{1}{2}{{\left( \left( \dfrac{3\sqrt{2}+3}{\sqrt{2}+1} \right)\left( \dfrac{\sqrt{2}-1}{\sqrt{2}-1} \right) \right)}^{2}}$
= $\dfrac{1}{2}{{\left( 6-3\sqrt{2}+3\sqrt{2}-3 \right)}^{2}}$
= $\dfrac{1}{2}{{\left( 6-3 \right)}^{2}}$
= 4.5
Hence, the correct answer is (a) 4.5 ${{m}^{2}}$.

Note: In the problem, we were given that the triangle is the right isosceles triangle. However, if we were given to find the area of any triangle with no special properties, we use the heron’s formula given by $\sqrt{s(s-a)(s-b)(s-c)}$. Here, s is half the length of the perimeter and a, b and c are the lengths of the respective sides of the triangle.