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ABCD is a quadrilateral with measurements of its sides shown in the above diagram. If AB is parallel to CD What is the area of the quadrilateral?

 

 Jun 21, 2020
 #1
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One way (and I don't claim that this is the easiest way):

 

Draw a line segment from A perpendicular to CD ending at point X on CD.

Draw a line segment from B perpendicular to CD ending at point Y on CD.

 

You now have triange(ADX), rectangle(AXYB), and triangle(BYC).

 

Combine triangle(ADX) and triangle(BYC) so that points A and B coincide and points X and Y coincide, creating a triangle

with sides of lengths 13, 15, and 14.  [CY + DX = CD - 10 = 14]

 

Using Heron's formula to find the area of this new triangle:  s = (13 + 14 + 15)/2 = 21

--->     Area  =  sqrt[ 21(21-13)(21-14)(21-15) ]  =  84.

 

To find the height of this triangle (which is the width of the rectangle):

--->     Area  =  ½·base·height     --->     84  =  ½·(14)·height   --->     height  =  12.

 

Therefore, the area of the rectangle is:  10·12  =  120

and the total area is 84 + 120  =  204

 Jun 21, 2020

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