Question

The two sides of a right-angled triangle are $x,x+1$ and hypotenuse, the longest side is $\sqrt{1}3$. Find the area of the triangle.

• A. $1$ m${}^2$
• B. $2$ m${}^2$
• C. $3$ m${}^2$
• D. $4$ m${}^2$

Answer 1

Answer: (C)

$3$ m${}^2$

In the right-angled triangle, longest side is the hypotenuse

$\therefore$ Hypotenuse $=\sqrt{13}$ and other two sides are $x$ and $x+1$

Therefore, using Pythagoras Theorem
$x^2+(x+1{)}^2=13$
$⟹x^2+x^2+2x+1-13=0$
$⟹2x^2+2x-12=0$

$⟹x^2+x-6=0$
On factoring, we get
$(x-2)(x+3)=0$
So, $x=2$ or $-3$

Since, the length can't be negative

$\therefore x=2$ and $x+1=3$
Area of the triangle $=$ $\frac{1}{2}\times$ length $\times$ breadth
$=$ $\frac{1}{2}\times 2\times 3$
$=$ $3$ m${}^2$