Question

The length of a hypotenuse of a right triangle exceeds the length of its base by $2cm.$ and exceeds twice the length of the altitude by $1cm$. 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

Answer: (C)

$8,15,17$

Let the hypotenuse of the triangle be $xcm$

Let the base of the triangle be $ycm$

Let the altitude of the triangle be $zcm$

According to the given equation

$x=y+2$ and $x=2z+1$

$y=x-2$ and $z=\frac{x-1}{2}$

Applying Pythagoras theorem

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

$x^2=y^2+z^2$

$\Rightarrow x^2=(x-2{)}^2+(\frac{x-1}{2}{)}^2$

$\Rightarrow x^2=\frac{4(x-2{)}^2+(x-1{)}^2}{4}$

$\Rightarrow 4x^2=4(x^2+4-4x)+x^2+1-2x$

$\Rightarrow 4x^2=4x^2+16-16x+x^2+1-2x$

$\Rightarrow x^2-18x+17=0$

$\Rightarrow x^2-17x-x+17=0$

$\Rightarrow x(x-17)-1(x-17)=0$

$\Rightarrow (x-1)(x-17)=0$

$x=1$ or $x=17$

If $x=1$ then $y=1-2=-1$ that is not possible

So $x=17$

$y=17-2=15$

$z=\frac{17-1}{2}=8$

$\therefore$ Length of sides of triangle are $17cm,15cm,8cm$