The length of the hypotenuse of a right angle triangle exceed the length of its base by 1 cm and exceed twice the length of altitude by 3 cm. Find the length of a each side of triangle.
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let length of base =x.
thus length of hypotenuse=x+1...............eq.-1
length of altitude=√[(x+1)²-x²]
=√x²+1+2x-x²
=√2x+1
hypotenuse exceeds twice the altitude by 3cm
thus hypotenuse =2√(2x+1)+3.....................eq.-2
equating eq.1 and eq.2
x+1=2√(2x+1)+3
x-2=2√(2x+1)
square both sides
x²+4-4x=4(2x+1)
x²-4x+4=8x+4
x²-12x=0
x=12,0
side cannot be 0
so sides are base=x=12cm
hypotenuse=x+1=13cm
altitude=√[2x+1]
=√24+1=√25=5cm
thus length of hypotenuse=x+1...............eq.-1
length of altitude=√[(x+1)²-x²]
=√x²+1+2x-x²
=√2x+1
hypotenuse exceeds twice the altitude by 3cm
thus hypotenuse =2√(2x+1)+3.....................eq.-2
equating eq.1 and eq.2
x+1=2√(2x+1)+3
x-2=2√(2x+1)
square both sides
x²+4-4x=4(2x+1)
x²-4x+4=8x+4
x²-12x=0
x=12,0
side cannot be 0
so sides are base=x=12cm
hypotenuse=x+1=13cm
altitude=√[2x+1]
=√24+1=√25=5cm
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