Question

The length of the hypotenuse of a right angle triangle exceeds the length of the base by $2$ cm and exceeds twice the length of the attitude by $1$ cm. The perimeter of the triangle is:

• A. $18$ cm
• B. $17$ cm
• C. $25$ cm
• D. $40$ cm

Answer 1

The correct option is D $40$ cm

Let length of the hypotenuse be $x$.
length $=\frac{x-1}{2}$
and breadth $=x-2$
Applying Pythagoras theorem
$(x-2{)}^2+\frac{(x-1{)}^2}{4}=x^2$
$\frac{(x-1{)}^2}{4}=x^2-(x-2{)}^2$
On solving,
$16(x-1)=(x-1{)}^2$
$x^2-2x+1=16x-16$
$x^2-18x+17=0$
$(x-1)(x-17)=0$
Taking $x=17$, we get the rest as
altitude $=8cm$ Base $=15cm$

$\therefore$ perimeter of triangle $=17+8+15=40cm$