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} = y^{2} + x^{2} – – – (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} = y^{2} + (2y – 1)^{2}

(2y + 1)^{2} = y^{2} + 4 y^{2} – 4y + 1

4y^{2} + 4y + 1 = 5y^{2} – 4y + 1 using (a + b)^{2} = a^{2} + 2ab + b^{2}

– y^{2} + 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.

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