Question

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.

Answer 1

Consider ∆ ABC,

Let, the length of hypotenuse of the triangle be x cm

Length of base of triangle = y cm

Length of altitude of triangle = z cm

According to given equation,

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

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

Applying Pythagoras theorem on ∆ ABC,

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

$x^2=y^2+z^2$ — (1)

Putting corresponding values of y and z in equation (1), we get

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

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

⇒ $4x^2=4(x^2+4–4x)+x^2+1–2x$​​​​​​​

⇒ $4x^2=4x^2+16–16x+x^2+1–2x$​​​​​​​

⇒ $x^2–18x+17=0$

⇒ $x^2–17x–x+17=0$

⇒ $x(x–17)–1(x–17)=0$

⇒ $(x–1)(x–17)=0$

⇒ $x=1,17$

$x=1$, cannot be possible, because base = x – 2 = 1 – 2 = –1 ~cm\), cannot be negative.

$\therefore x=17cm$

$y=x–2=17–2=15cm$

and $z=\frac{x-1}{2}=\frac{17-1}{2}=8cm$ ​​​​​​​

∴ Length of sides of triangle are 17 cm, 15 cm and 8 cm.