Thread: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?

Hi guys,

Question: The lengths of the sides of a right angled triangle are cm, cm and cm. Calculate the value of x and hence find the area of the triangle

I tried using Pythagoras Theorem but if I were to expand the algebra given, the answers would be completely different? How should I go around to solve it?

Thanks

Originally Posted by FailInMaths

Hi guys,

Question: The lengths of the sides of a right angled triangle are cm, cm and cm. Calculate the value of x and hence find the area of the triangle

I tried using Pythagoras Theorem but if I were to expand the algebra given, the answers would be completely different? How should I go around to solve it?

Thanks

There are two possibilities for the hypotenuse that you need to consider the 6x and the 5x+3 these could both be the hypotenuse, 6x if x>3 and 5x+3 if x<3.

You need to look at both of these cases.

CB

Originally Posted by CaptainBlack

There are two possibilities for the hypotenuse that you need to consider the 6x and the 5x+3 these could both be the hypotenuse, 6x if x>3 and 5x+3 if x<3.

You need to look at both of these cases.

CB

Hmm, the answer stated here is 2. Can I assume that this question is unclear? I mean, like what you said, there are 2 hypotenuse

If the answer is 2 then consider like CB has said as the hypothenuse, that means if we use Phytagoras theorem:

Solve this equation.

Originally Posted by Siron

If the answer is 2 then consider like CB has said as the hypothenuse, that means if we use Phytagoras theorem:

Solve this equation.

Yeah, but I think the question itself is unclear. Without the answer, how would we have know that (5x+3) is the hypotenuse

Originally Posted by FailInMaths

Yeah, but I think the question itself is unclear. Without the answer, how would we have know that (5x+3) is the hypotenuse

By doing the calculations in both cases, you will find only one solution consistent with x>0 and x<3 for the 6x hypotenuse and x>3 for the 5x+3 hypotenuse.

So you compute all four solutions and only one is consistent with the constraints.

CB

Originally Posted by CaptainBlack

By doing the calculations in both cases, you will find only one solution consistent with x>0 and x<3 for the 6x hypotenuse and x>3 for the 5x+3 hypotenuse.

So you compute all four solutions and only one is consistent with the constraints.

CB

Ok, I get it now, thanks