Question

One leg of a right angled triangle exceeds the other leg by 2 inches. The hypotenuse is 10 inches. The length of the shorter leg of the triangle is .

• A. 4 inches
• B. 6 inches
• C. 8 inches

Answer 1

The correct option is B 6 inches

Let the length of the legs of the triangle be $xand(x+2)inches.$

Applying Pythagoras theorem,

$x^2+(x+2{)}^2=(10{)}^2$

$x^2+x^2+4x+4=100$

$2x^2+4x-96=0$

$x^2+2x-48=0$

$x^2+8x-6x-48=0$

$x(x+8)-6(x+8)=0$

$(x+8)(x-6)=0$

$\therefore x=-8,6$

Since length can't be negative, hence
$x=6inches$