Question

Question 5
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer 1

Let the base of the right triangle be $xcm$.
Its altitude = $(x-7)cm$
From Pythagoras theorem, we have
$Bas{e}^2+Altitud{e}^2=Hypotenus{e}^2$
$\therefore x^2+(x-7{)}^2={13}^2$
$\Rightarrow x^2+x^2+49-14x=169$
$\Rightarrow 2x^2-14x-120=0$
$\Rightarrow x^2-7x-60=0$
$\Rightarrow x^2-12x+5x-60=0$
(Splitting the middle term -7x as -12x+5x because $-12x\times 5x=-60x^2$)
$\Rightarrow x(x-12)+5(x-12)=0$
$\Rightarrow (x-12)(x+5)=0$
Either $x-12=0orx+5=0$,
$\Rightarrow x=12orx=-5$
Since sides are positive, $x$ can only be $12$.
Therefore, the base of the given triangle is $12cm$ and the altitude of this triangle will be $(12-7)cm=5cm$.