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.
Let the base and altitude of the right angled triangle be x and y respectively.
Thus the hypotenuse of triangle will be x + 2 cm
(x + 2)2 = y2 + x2 – – – (1)
Also the hypotenuse exceeds twice the length of the altitude by 1 cm
h = 2y + 1
x + 2 = 2y + 1
x = 2y – 1
Putting the value of x in (1) we get
(2y – 1 + 2)2 = y2 + (2y – 1)2
(2y + 1)2 = y2 + 4 y2 – 4y + 1
4y2 + 4y + 1 = 5y2 – 4y + 1 using (a + b)2 = a2 + 2ab + b2
– y2 + 8y = 0
y(y – 8) = 0
y = 8
x = 16 – 1 = 15cm
h = 16 + 1 = 17cm
Thus the base, altitude, hypotenuse of triangle are 15cm, 8cm, 17cm respectively.