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Since this is an odd-shaped quadrilateral, there's no convenient area formula we can apply here. Instead, let's add an extra line to divide the quadrilateral into two TRIANGULAR regions.
Let's find the area of Region #1 first If we make side AB the base of the triangle, then the height = 10 Area of triangle = (base)(height)/2 = (6)(10)/2 = 30
Now, we'll find the area of Region #2 If we make side AD the base of the triangle, then the height = 4 Area of triangle = (base)(height)/2 = (4)(4)/2 = 8
Another way is to convert it into a rectangle by extending as shown in the sketch. Area of quadrilateral is the area of rectangle-areas of two triangle..
Area of the rectangle = \((11-1)*(7-1)=10*6=60\). Area of the triangle DCF = \(\frac{1}{2}*(11-5)(5-1)=3*4=12\) Area of the triangle BCE = \(\frac{1}{2}*(11-1)(7-5)=10\)
Hence area of quadrilateral ABCD = 60-12-10=38
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Re: What is the area of quadrilateral ABCD?
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