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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.

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Solution

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=x2 and z=x12​​​​​​​

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=(x2)2+(x12)2

x2=4(x2)2+(x1)24

4x2=4(x2+44x)+x2+12x​​​​​​​

4x2=4x2+1616x+x2+12x​​​​​​​

x218x+17=0

x217xx+17=0

x(x17)1(x17)=0

(x1)(x17)=0

x=1,17

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

x=17 cm

y=x2=172=15 cm

and z=x12=1712=8 cm ​​​​​​​

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


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