Question

If the perimeter of an isosceles right triangle is $(6+3\sqrt{2})m,$ then the area of the triangle is

• A. $4.5m^2$
• B. $5.4m^2$
• C. $9m^2$
• D. $81m^2$

Answer 1

The correct option is A $4.5m^2$

In an isosceles right angled triangle, the two sides on the right angle are equal.
Let the equal sides be $a$.
Hence, hypotenuse of the isosceles right angled triangle $=\sqrt{a^2+a^2}=\sqrt{2}a$
Given, perimeter of the triangle $=(6+3\sqrt{2})m$
$\Rightarrow a+a+\sqrt{2}a=(6+3\sqrt{2})m$
$\Rightarrow a(2+\sqrt{2})=3(2+\sqrt{2})m$
$\Rightarrow a=3m$
In the isosceles right triangle, the base and height $=a$
Hence, area of the triangle $=\frac{1}{2}\times base\times height$ $=\frac{1}{2}\times 3\times 3=4.5m^2$