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| Question 965826:  Find the area of quadrilateral ABCD whose sides are 9m 40m 28m 15m and angle ABC=90°
 Answer by Alan3354(69443)
      (Show Source): 
You can put this solution on YOUR website! Find the area of quadrilateral ABCD whose sides are 9m 40m 28m 15m and angle ABC=90° -------------
 Assuming AB = 9 and BC = 40:
 Put A at the Origin, B at (0,9), C at (40,9)
 Point D is the intersection of 2 circles, one centered at A with r = 15 and one centered at C with r = 28.
 --> x^2 + y^2 = 225 and
 (x-40)^2 + (y-9)^2 = 784
 Find the intersection of the 2 circles:
 x^2 + y^2 = 225
 x^2 + y^2 - 80x - 18y = -897
 ------------------------------------- subtract
 80x + 18y = 1122 is the equation of the line passing thru the 2 intersections of the circles.
 --> y = (561 - 40x)/9
 x^2 + y^2 = 225
 Sub for y
 x^2 + ((561 - 40x)/9)^2 = 225
 81x^2 + (561 - 40x)^2 = 225*81 = 18225
 81x^2 + 1600x^2 - 44880x + 314721 = 18225
 1681x^2 - 44880x + 296496 = 0
 
 
 | Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |  | Quadratic equation  (in our case  ) has the following solutons: 
 
  
 For these solutions to exist, the discriminant
  should not be a negative number. 
 First, we need to compute the discriminant
  :  . 
 Discriminant d=20575296 is greater than zero. That means that there are two solutions:
  . 
 
  
  
 Quadratic expression
  can be factored: 
  Again, the answer is: 14.6983938132064, 12.
Here's your graph:
 
  |  ==============
 x = 14.6984
 --> y = -2.9929
 D is (14.6984,-2.9929)
 ======================
 Write the points in order.  I'll use (14.7,-3) for D
 A...B...C...D....A
 0...0..40..14.7..0
 0...9...9..-3....0
 Add the diagonal products starting at the upper left.
 0+0-120+0 = 120
 Add the diagonal products starting at the lower left.
 0+360+132.3+0 = 492.3
 ------------
 Find the difference = 492.3 - 120 = 372.3
 The area is 1/2 that
 Area = 186.15 sq units
 
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