Area of a rectangle using right triangles

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luissi

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Oct 25, 2007
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XYZW is a rectangle with:

YZ = 10cm
YP = 8cm
ZP = 6cm

Determine the area of XYZW.


rectangle.jpg


So, what I figured is that because it is a rectangle, I could use the pythagorean theorem, but the problem is, I don't have the necessary side lenghts and I can't think of any other way to do it...
 
Let y=XY=WZ

Let x=XP, then PW = 10-x

So, you have:

\(\displaystyle \L\\x^{2}+y^{2}=64\)

\(\displaystyle \L\\(10-x)^{2}+y^{2}=36\)

Solve for x and y.


EDIT: Thanks, SK. Typo/brain fart
 
galactus said:
Let y=XY=WZ

Let x=XP, then PW = 10-x

So, you have:

\(\displaystyle \L\\x^{2}+y^{2}=100\)..............shouldn't this be = 64

\(\displaystyle \L\\(10-x)^{2}+y^{2}=36\)

Solve for x and y.
Click to expand...
 
okay, I got the answer!
its area is 20 cm squared
thanks so much for helping me :D
 
yeah, I just figured that out after trying to plug the numbers into the Pythagorean Theorem

... I don't know what I'm doing wrong, but I'll just keep trying with the equalitites I was given and see
 
Hello, luissi!

XYZW is a rectangle with: YZ = 10cm, YP = 8cm, ZP = 6cm

Determine the area of XYZW.
Click to expand...
Code: 
                      P
    X *---------------*-------* W
      |             * :*      |
      |           * θ : *     |
      |      8  *     :  * 6  |
      |       *      h:   *   |
      |     *         :    *  |
      |   *           :     * |
      | *             :    θ *|
    Y *---------------*-------* X
      : - - - - - 10  - - - - :

Note that \(\displaystyle \Delta YPZ\) is a right triangle . . . and we want \(\displaystyle h.\)

Note that \(\displaystyle h\) divides \(\displaystyle \Delta YPZ\) into two similar right triangles.

In the triangle at the right: \(\displaystyle \:\frac{h}{6} \:=\:\sin\theta\)
In the triangle at the left: \(\displaystyle \:\frac{h}{8} \:=\:\cos\theta\)

Square the equations: \(\displaystyle \;\L\begin{array}{ccc}\frac{h^2}{36} &=&\sin^2\!\theta \\ \frac{h^2}{64} & = & \cos^2\!\theta\end{array}\)

. . and add: .\(\displaystyle \L\frac{h^2}{36}\,+\,\frac{h^2}{64}\)\(\displaystyle \;=\;\sin^2\!\theta\,+\,\cos^2\!\theta \;=\;1\)

Multiply by 576: \(\displaystyle \:16h^2\,+\,9h^2\:=\:576\;\;\Rightarrow\;\;25h^2\:=\:576\;\;\Rightarrow\;\;h^2\:=\:\frac{576}{25}\)

Therefore: \(\displaystyle \:h \:=\:\frac{24}{5}\)

 
Why all that work? Or am I missing something?

Since the 2 legs of the right triangle are 6 and 8,
then the rectangle's area is simply 6 * 8 = 48 (twice the triangle's).
And the rectangle's height is simply 48/10 = 4.8 or 24/5.
 
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