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.
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=x−12
Applying Pythagoras theorem on ∆ ABC,
(AC)2=(AB)2+(BC)2
x2=y2+z2 — (1)
Putting corresponding values of y and z in equation (1), we get
x2=(x−2)2+(x−12)2
x2=4(x−2)2+(x−1)24
⇒ 4x2=4(x2+4–4x)+x2+1–2x
⇒ 4x2=4x2+16–16x+x2+1–2x
⇒ x2–18x+17=0
⇒ x2–17x–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.
∴x=17 cm
y=x–2=17–2=15 cm
and z=x−12=17−12=8 cm
∴ Length of sides of triangle are 17 cm, 15 cm and 8 cm.