My Math Forum Area problem

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March 19th, 2013, 06:36 AM   #1
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Area problem

I am having problem with this question, I am not 100% that I have worked it out the right way.

Question: An enclosure PQRS is to be made as shown, in the figure ( my diagram ). PQ and RQ are fences of total length of 300m. The other two sides are hedges. The angles at Q and R are right angles and the angle at S is 135 degree. The length of QR is x meters.

a) show that the area, A m^2, of the enclosure is given by: $A=300x-\frac{3x^2}{2}$

I will show my working in steps:
1. $x+y=300\right y=300-x$

So after this I have to find the area so what is did was

2. take my red triangle I have the opposite of length x and a angle of $sin45 \right \frac{\sqrt2}{2}$ I then used this to workout the hypotenuse.

3.hypotenuse $\right x\sqrt x$

4. I then workout the adjacent $(x \sqrt x)^2-x^2=x$

So the area of the red triangle $\frac{x^2}{2}$

5. I then workout the area of the rectangle $(300-x)(x) \right (300x-x^2)$

what did then was take the area of the rectangle and minus the area of the triangle

6. $300x-x^2-\frac{x^2}{2} \right \frac{600x}{2}-\frac{2x^2}{2}-\frac{x^2}{2} \right 300x-\frac{3x^2}{2}$

This gives the right ans, but I don't think what I have done correct, or if I have is there a quicker way, I did spot that I could have used area of trapezium but was not really sure how to apply that.

Could someone please point out anywhere I have gone wrong or even show another way around this question.
Attached Images
 Area problem.png (11.0 KB, 245 views)

 March 19th, 2013, 07:52 AM #2 Senior Member   Joined: Sep 2012 Posts: 200 Thanks: 1 Re: Area problem Right I have over complicated things slight, just realised that because it's a 45 degree triangle adjacent and opposite will be the same length.
 March 19th, 2013, 10:24 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,099 Thanks: 723 Re: Area problem Can't see your diagram (except top part)
 March 19th, 2013, 11:20 AM #4 Senior Member   Joined: Sep 2012 Posts: 200 Thanks: 1 Re: Area problem Strange I will try repost the diagram. I have notice I have smiles in my first post which I did not put in there. Has my post been edit?
 March 19th, 2013, 11:47 AM #5 Senior Member   Joined: Sep 2012 Posts: 200 Thanks: 1 Re: Area problem It seem to be okay to me, I just double clicked on the diagram. Have another go other wise, I will take a photo on the diagram out of my book a post it up. Problem is not to sure how good the quality will be.
March 19th, 2013, 11:54 AM   #6
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Re: Area problem

Here is photo out of book. Not great but best I could do.
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March 19th, 2013, 12:14 PM   #7
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Re: Area problem

[attachment=0cyy7l7y]as.png[/attachmentcyy7l7y]

$\left(PQRS\right)=\left(APS\right)+\left(PBRQ\righ t)=\frac{1}{2}x^2+\left(300-x\right)\cdot x=300x-\frac{1}{2}x^2$.
Attached Images
 as.png (13.2 KB, 205 views)

 March 19th, 2013, 12:59 PM #8 Senior Member   Joined: Sep 2012 Posts: 200 Thanks: 1 Re: Area problem That's what I got the first time I did the question. But it say I should show the area to be $300x-\frac{3x^2}{2}$.
March 19th, 2013, 01:15 PM   #9
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Re: Area problem

[color=#000000]
Quote:
 Originally Posted by taylor_1989_2012 That's what I got the first time I did the question. But it say I should show the area to be $300x-\frac{3x^2}{2}$.
Oh sorry, I gave wrong names to the vertices. My figure is wrong. In that case $(PQRS)=\frac{1}{2}x^2+(300-2x)x=300x-\frac{3}{2}x^2$.[/color]

 March 19th, 2013, 01:54 PM #10 Senior Member   Joined: Sep 2012 Posts: 200 Thanks: 1 Re: Area problem Okay no worries. Could you expand on you method a bit more. I see you have add the areas, where I worked out the whole area of the rectangle and then misused the area for the triangle. Yet will still have the same ans. I think the bit that is confusing is where did you get the (300-2x) from? Thank by the way for the help.

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