Question

The length of a hypotenuse of a right angle triangle exceeds the length of its base by $2$ cm and exceeds twice the length of the altitude by $1$ cm. Find the length of each side of the triangle (in cm).

• A. $6,8,10$
• B. $7,24,25$
• C. $8,15,17$
• D. $7,40,41$

Answer 1

The correct option is C $8,15,17$

Let the lenght of hypotenuse of the triangle be $x$cm
Lenght of base of the triangle be $y$cm
Lenght of altitude of the triangle be $z$cm
According to given equation,
$x=y+2$ and $x=2z+1$
$=>y=x-2$ and
$=>z=\frac{x-1}{2}$
By Pythagoras Theorem
$(AC{)}^2=(AB{)}^2+(AC{)}^2$
$x^2=y^2+z^2$ (i)
putting values of $y$ and $z$ in equation (i) we get,
$x^2=(x-2{)}^2+(\frac{x-1}{2}{)}^2$
$=>x^2=\frac{4{(x-2)}^2+{(x-1)}^2}{4}$
$=>4x^2=4(x^2+4-4x)+x^2+1-2x$
$=>4x^2=4x^2+16-16x+x^2+1-2x$
$=>x^2-18x+17=0$
$=>x^2-17x-x+17=0$
$=>x(x-17)-1(x-17)=0$
$=>(x-17)(x-1)=0$
Therefore,
$x=17$ or $x=1$
$x=1$ is not possible as base =$x-2=1-2=-1cm$ which cannot be in negative.
Therefore,$x=17$
Putting the value in $y$
$y=x-2$
$y=17-2=15cm$ and
$z=\frac{x-1}{2}$
$z=\frac{17-1}{2}$
$z=8cm$
Therefore,
lenght of the sides of the triangle is $17cm,15cm,8cm$.