Question

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

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

Answer 1

Answer: (A)

$4.5m^2$

In an isosceles right angled triangle assume the side opposite to right angle is b and the other two sides are a
$Perimeterofatriangle=Sumofthreesides$
$P=a+a+b$
$P=6+3\sqrt{2}$
Equation no 1
$6+3\sqrt{2}=2a+b$
Using Pythagoras Theorem to determine side b
$a^2+a^2=b^2$
$2a^2=b^2$
Equation no 2
$b=\sqrt{2}\times a$
Substituting the value of equation 2 in equation 1
$2a+\sqrt{2}\times a=6+3\sqrt{2}$
$a(2+\sqrt{2})=3+(2+\sqrt{2})$
Therefore,
$a=3$
$Areaoftriangle=A$
$A=\frac{1}{2}\times base\times height$
$A=\frac{1}{2}\times a\times a$
Substituting the value of $a=3$
$A=\frac{9}{2}$
$A=4.5$
The answer to the question is Option A.