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 5x12y=0 BC 5x+12y120=0 CD x3y + 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.

February 22nd, 2016, 12:00 PM  #3 
Member Joined: Sep 2014 From: UK Posts: 82 Thanks: 1  
February 22nd, 2016, 12:02 PM  #4 
Senior Member Joined: Feb 2010 Posts: 701 Thanks: 136  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  
Member Joined: Sep 2014 From: UK Posts: 82 Thanks: 1  Quote:
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  
Member Joined: Sep 2014 From: UK Posts: 82 Thanks: 1  Quote:
 
February 22nd, 2016, 12:46 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,098 Thanks: 1905  As A and C are on the yaxis, I chose E so that its ycoordinate 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  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
$\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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