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
(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,
$x-5=20-5=15 \mathrm{~cm}$
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