1.

The length of the hypotenuse of a right 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»

Let the Hypotenuse be a cm

Base = (a - 2) cm

ALTITUDE = (a - 1) / 2

By pythagores theorem,

a^2 = (a-2)^2 + [(a-1)/2]^2

=> a^2 = a^2 + 4 - 4a + ( a^2 - 2a + 1) / 4

On multiplying the whole EQUATION by 4, we get

=> 4a^2 = 4a^2 + 16 - 16a + a^2 - 2a + 1

=> 0 = 16 - 16a + a^2 - 2a + 1

=> a^2 - 18a + 17 = 0

=> a^2 - a - 17A + 17 = 0

=> a(a-1) - 17(a - 1) = 0

=> (a-17)(a-1) = 0

a = 17 and 1

But if Hypotenuse is 1, altitude will be 0, that is not possible.

So, a = 17

Hypotenuse = 17 cm

Base = 17 - 2 = 15 cm

Altitude = ( 17 - 1) /2 = 8 cm


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