Small trickly problem

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matthew042

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Feb 8, 2009
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I have a right triangle and I'm given that the area is 210 and the hypotenuse is 29. I have tried to use the middle height which was 420/29 and have tried ways to attack it like putting the two equations using a(squared) + b(squared) = c(squared) and using the height. Combining that, I couldn't find the perimeter. I have a sense but have no clue how to attack the problem. Please help. Thanks.
 
You have these:

1) \(\displaystyle a^{2} + b^{2} = 29^{2}\)

2) \(\displaystyle \frac{1}{2} \cdot a \cdot b = 210\)

You could use one more piece of information.

Define the semi-perimeter.

\(\displaystyle s = \frac{a+b+29}{2}\)

And you get this for free:

\(\displaystyle 210 = \sqrt{s \cdot (s-a) \cdot (s-b) \cdot (s-29)}\)

Now you can find a solution.

Note: This does NOT lead to a unique answer, as far as a and b are concerned. It does lead to a unique answer as far as the triangle cares.
 
In case that's too fancy, there is an easier way, but it may seem a little unmotivated.

These two again...

1) \(\displaystyle a^{2} + b^{2} = 29^{2}\)

2) \(\displaystyle \frac{1}{2} \cdot a \cdot b = 210\)

Just for fun, let's notice that \(\displaystyle 4 \cdot \frac{1}{2} \cdot a \cdot b = 4 \cdot 210\)

or

\(\displaystyle 2 \cdot a \cdot b = 840\)

Ok, so why did I do that and why should you care?

Contemplate: \(\displaystyle (a+b)^{2} = a^{2} + 2 \cdot a \cdot b + b^2\)
 
Okay, just kidding...

1) \(\displaystyle a^{2} + b^{2} = 29^{2}\)

2) \(\displaystyle \frac{1}{2} \cdot a \cdot b = 210\)

Simply solve directly by substitution.

From 2, we have a = 420/b

Substituting into 1), we get: \(\displaystyle \left[\frac{420}{b}\right]^{2} + b^{2} = 29^{2}\)

And you should be able to solve for BOTH values of b.
 
tkhunny said:
Okay, just kidding...

1) \(\displaystyle a^{2} + b^{2} = 29^{2}\)

2) \(\displaystyle \frac{1}{2} \cdot a \cdot b = 210\)

Simply solve directly by substitution.

From 2, we have a = 420/b

Substituting into 1), we get: \(\displaystyle \left[\frac{420}{b}\right]^{2} + b^{2} = 29^{2}\)

And you should be able to solve for BOTH values of b.
Click to expand...

\(\displaystyle (a + b)^2 = 29^2 + 840\)

\(\displaystyle a + b = 41\)

perimeter = 70[/spoiler:4hmlqlve]
 
I don't really understand how you arrived at that equation. With (a+b)^2, and 840 and am puzzled at how you get that equation. I know how to solve the equation you gave me but i just don't understand how you got that equation. Many thanks to everyone who helped me.
 
Hello, matthew042!

Given: a right triangle with area 210 and hypotenuse 29.
Find the perimeter.
Click to expand...
Code: 
      *
      |  *   29
    b |     *
      |        *
      * - - - - - *
            a

\(\displaystyle \text{We have: }\;\begin{array}{cccc}a^2 + b^2 &=& 29^2 & [1] \\ \frac{1}{2}ab &=& 210 & [2] \end{array}\)

\(\displaystyle \begin{array}{ccccc}\text{From [2]:} & 2ab &=& 840 \\ \text{Add [1]:} & a^2+b^2 &=& 841 \end{array}\)

\(\displaystyle \text{We have: }\:a^2+2ab + b^2 \:=\:1681 \quad\Rightarrow\quad (a+b)^2 \:=\:1681 \quad\Rightarrow\quad a + b \:=\:41\)


\(\displaystyle \text{Therefore: }\:\text{Perimter} \:=\:a+b+c \:=\:41 + 29 \:=\:70\)

 
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