Thread: Geometry (Pythagoras)

If anyone can help me with this questions that would be great!!

The hypotenuse of a right triangle is 3.00ft longer than one of the sides, which in turn is 5.00ft longer than the other side. How long are the sides and the hypotenuse?

Thanks a lot!

Originally Posted by stephie

If anyone can help me with this questions that would be great!!

The hypotenuse of a right triangle is 3.00ft longer than one of the sides, which in turn is 5.00ft longer than the other side. How long are the sides and the hypotenuse?

Thanks a lot!

Let's call the smallest side "x"

Therefore we have: x^2 + (x + 5)^2 = ( (x + 5) + 3)^2

Or in other words: x^2 + (x + 5)^2 = (x + 8)^2

Therefore: x^2 + x^2 + 10x + 25 = x^2 + 16x + 64

Subtract both sides by x^2 to get: x^2 + 10x + 25 = 16x + 64

Take 25 from both sides: x^2 + 10x = 16x + 39

Put it all on one side: x^2 - 6x - 39 = 0

Quadratic equation: x = [ -b +/- SQRT( b^2 - 4ac ) ] / ( 2a )

Substitute: x = [ -(-6) +/- SQRT( [-6]^2 - 4[1][-39] ) ] / ( 2[1] )

Which eventually becomes x ~ 9.92820323 and x ~ -3.92820323

x can't be negative, so: x = 9.92820323 = 3 + 4[ SQRT(3) ]

So one side is 3 + 4[ SQRT(3) ]

The other is 8 + 4[ SQRT(3) ]

And the hypotenuse is: 11 + 4[ SQRT(3) ]

I came up with a different answer for this:

Let x = the side that the hyp is 3 more than
Let x-5 = the other side

x^2 + (x - 5)^2 = (x + 3)^2

resolves to:

x^2 - 16x + 16

= (16 +/- sqrt(16^2 - 4(16))) / 2

= 8 +/- 4sqrt(3)

x = approx. 14.93 or x = approx 1.07

if x = 14.93 then the sides are 14.93 and 9.93 and the hyp is 17.93

14.93^2 + 9.93^2 = 17.93^2 = approx 321.5 checks

if x = 1.07 then the sides are 1.07 and -3.93 ....impossible

Originally Posted by Quick

The other is 8 + 4[ SQRT(3) ]

Originally Posted by spiritualfields

= 8 +/- 4sqrt(3)

We came up with the same answer

thanks a lot everyone. You have been a great help!

We came up with the same answer

Sure did. Not sure how I missed that. It was late and I was tired, I guess.