Question

One leg of a right angled triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the triangle.
[3 Marks]

Answer 1

Let the length of the legs of the triangle be $x$ and $x+4$ inches. [0.5 Mark]

Applying Pythagoras' theorem,

$x^2+(x+4{)}^2=(20{)}^2\left[0.5Mark\right]x^2+x^2+8x+16=4002x^2+8x-384=0x^2+4x-192=0\left[1Mark\right]$

Solving the quadratic equation,$x^2+16x-12x-192=0x(x+16)-12(x+16)=0(x+16)(x-12)=0\therefore x=-16\&12$

Length can't be negative, hence
$x=12\text{inches}$

The shortest side(leg) is 12 . [1 Mark]