Question:
The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer. The lengths of these sides are
(a) 10 cm, 15 cm
(b) 15 cm, 20 cm
(c) 12 cm, 17 cm
(d)13 cm, 18 cm
Solution:
(b) 15 cm, 20 cm
It is given that the length of hypotenuse is 25 cm.
Let the other two sides be $x \mathrm{~cm}$ and $(x-5) \mathrm{cm}$.
Applying Pythagoras theorem, we get:
$25^{2}=x^{2}+(x-5)^{2}$
$\Rightarrow 625=x^{2}+x^{2}+25-10 x$
$\Rightarrow 2 x^{2}-10 x-600=0$
$\Rightarrow x^{2}-5 x-300=0$
$\Rightarrow x^{2}-20 x+15 x-300=0$
$\Rightarrow x(x-20)+15(x-20)=0$
$\Rightarrow(x-20)(x+15)=0$
$\Rightarrow x-20=0$ or $x+15=0$
$\Rightarrow x=20$ or $x=-15$
Side of a triangle cannot be negative.
Therefore, $x=20 \mathrm{~cm}$
Now,
$x-5=20-5=15 \mathrm{~cm}$