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

• Sep 19th 2011, 06:32 AM
FailInMaths
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 \$\displaystyle (2x+1)\$ cm, \$\displaystyle 6x\$cm and \$\displaystyle (5x+3)\$ 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
• Sep 19th 2011, 06:44 AM
CaptainBlack
Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Quote:

Originally Posted by FailInMaths
Hi guys,

Question: The lengths of the sides of a right angled triangle are \$\displaystyle (2x+1)\$ cm, \$\displaystyle 6x\$cm and \$\displaystyle (5x+3)\$ 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
• Sep 19th 2011, 06:50 AM
FailInMaths
Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Quote:

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
• Sep 19th 2011, 07:41 AM
Siron
Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
If the answer is 2 then consider like CB has said \$\displaystyle 5x+3\$ as the hypothenuse, that means if we use Phytagoras theorem:
\$\displaystyle (5x+3)^2=(6x)^2+(2x+1)^2\$
Solve this equation.
• Sep 19th 2011, 08:25 AM
FailInMaths
Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Quote:

Originally Posted by Siron
If the answer is 2 then consider like CB has said \$\displaystyle 5x+3\$ as the hypothenuse, that means if we use Phytagoras theorem:
\$\displaystyle (5x+3)^2=(6x)^2+(2x+1)^2\$
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
• Sep 19th 2011, 01:34 PM
CaptainBlack
Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Quote:

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
• Sep 20th 2011, 03:33 AM
FailInMaths
Re: The lengths of the sides of a right angled triangle are.... (Pythagoras Theorem)?
Quote:

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