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 February 22nd, 2016, 11:06 AM #1 Member   Joined: Sep 2014 From: UK Posts: 82 Thanks: 1 Area of a quadrilateral? I got a quadrilateral ABCD, (0,0), (12,5), (0,10) and (-6,8 ). The previous part of the questions asked me to sketch it, find the gradients of each side, lengths and the equation of each side. i) Gradients m AB = 5/12 m BC = -5/12 m CD = 1/3 m DA = -4/3 ii) Lengths AB = 13 BC = 13 CD = Sqrt of 40 DA = 10 iii) Equations AB 5x-12y=0 BC 5x+12y-120=0 CD x-3y + 30 DA = 4x +3y = 0 iv asks me to find the area. My question is, can I find the area of the whole shape at once using a formula? Because it doesn't seem like I can use area of a rhombus or parallelogram. Yet I have noticed that ABC is an isosceles and ACD is a scalene, and by working out the area of them separately and then combining together I could find the total area. I just don't know if they want me to find it this way and if this is the quickest/best approach to this question. How do you think I go about finding the area?
 February 22nd, 2016, 11:21 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 If E is (-6, 5), ABCE is a kite with area AC*BE/2, which equals the area of ABCD. Thanks from Tangeton
February 22nd, 2016, 12:00 PM   #3
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Quote:
 Originally Posted by skipjack If E is (-6, 5), ABCE is a kite with area AC*BE/2, which equals the area of ABCD.
Do you just guess some random space where this equilateral is going to become a kite? Could you explain what process you do to get E?

February 22nd, 2016, 12:02 PM   #4
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Quote:
 Originally Posted by Tangeton I got a quadrilateral ABCD, (0,0), (12,5), (0,10) and (-6,8 ).
Sketch a picture. Cut the diagram into two triangles with a common side AC = 10.

Draw in the heights for each of these triangles. One will go from B perpendicular to AC and the other from D perpendicular to AC. Now calculate the areas of the two triangles. The total area is 90.

February 22nd, 2016, 12:11 PM   #5
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Quote:
 Originally Posted by skipjack If E is (-6, 5), ABCE is a kite with area AC*BE/2, which equals the area of ABCD.
Do you just guess some random space where this equilateral is going to become a kite? Could you explain what process you do to get E?

Edit: Nvm. I guess I do have to just look how all triangles have same area so doesn't matter where D is as long as its on x = -6. (Explaining to myself...)

February 22nd, 2016, 12:11 PM   #6
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Quote:
 Originally Posted by mrtwhs Sketch a picture. Cut the diagram into two triangles with a common side AC = 10. Draw in the heights for each of these triangles. One will go from B perpendicular to AC and the other from D perpendicular to AC. Now calculate the areas of the two triangles. The total area is 90.
Thanks, that's what I done

February 22nd, 2016, 12:46 PM   #7
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 Originally Posted by Tangeton Do you just guess some random space where this equilateral is going to become a kite? Could you explain what process you do to get E?
As A and C are on the y-axis, I chose E so that its y-coordinate equals that of B. The formula I used works when the diagonals are perpendicular (so ABCE needn't be a kite).

February 22nd, 2016, 10:30 PM   #8
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Quote:
 Originally Posted by Tangeton I got a quadrilateral ABCD, (0,0), (12,5), (0,10) and (-6,8 ). iv asks me to find the area. My question is, can I find the area of the whole shape at once using a formula?
Yes.

$\displaystyle Area \ \ of \ \ Polygon \ \ P_1P_2P_3P_4 \ = \tfrac{1}{2}(x_1y_2 \ + \ x_2y_3 \ + \ x_3y_4 \ + \ x_4y_1 \ - y_1x_2 \ - \ y_2x_3 \ - y_3x_4 \ - \ y_4x_1)$

The sign of the area is positive or negative according as to whether
$\displaystyle P_1P_2P_3P_4 \$ is a counterclockwise or clockwise polygon.

Source: CRC Standard Mathematical Tables 24th Edition

For instance, let $\displaystyle ( x_1, y_1) = (0,0).$

Let $\displaystyle \ ( x_2, y_2) = (-6,8).$

Let $\displaystyle \ ( x_3, y_3) = (0,0).$

Let $\displaystyle \ ( x_4, y_4) = (12,5).$

 February 24th, 2016, 11:50 AM #9 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85 I graphed the points and drew a vertical line segment from (0,0) to (0,10) to create two triangles. I used the semiperimeter formula for the area of each triangle. One triangle had an irrational perimeter, but with a calculator that wasn't a problem. Alternatively, you could do what I learned in school, which is to draw the smallest possible rectangle that includes the whole polygon. This rectangle has an area of 180. Then use area = base*height/2 to calculate the areas of the four triangles that are part of the rectangle but not the polygon. The triangles have areas of 24, 6, 30, and 30. 180 - 24 -6 - 30 - 30 = 90. Last edited by EvanJ; February 24th, 2016 at 11:54 AM.

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