+0

# i need help!

0
54
1

AF = 21, AE = 27, BC = 30, CD = 10.  Find the area of quadrilateral ABCD. Jan 12, 2020

#1
+1

First:  notice that triangle(DCE) is similar to triangle(BCF) by AA.

Since BC = 3·DC:  if DE = x, then BF = 3x

and if CE = y, then CF = 3y.

Since triangle(ABE) is a right triangle:  (21 + 3x)2  =  (27)2 + (30 + y)2

--->            441 + 126x + 9x2  =  729 + 900 + 60y + y2

--->        9x2 - y2 + 126x - 60y  =  1188                                          (Equation #1)

Since triangle(ADF) is a right triangle:  (27 + x)2  =  (21)2 + (3y + 10)2

--->              729 + 54x + x2  =  441 + 9y2 + 60y + 100

--->      x2 - 9y2 + 54x - 60y  =  -188                                               (Equation #2)

Subtracting Equation #2 from Equation #1:     8x2 + 8y2 + 72x  =  1376                         (Equation #3)

We need to get rid of the y2-term; let's look at triangle(CDE):  x2 + y2  =  100     --->     y2  =  100 - x2

Substituting this value into Equation #3:  8x2 + 8y2 + 72x  =  1376     --->     8x2 + 8(100 - x2) + 72x  =  1376

--->     8x2 + 800 - 8x2 + 72x  =  1376     --->     800 + 72x  =  1376     --->     72x  =  576     --->     x  =  8

Since  x2 + y2  =  100  and  x  =  8     --->     y  =  6.

You can now determine all the length of all the sides; divide the region into non-overlapping triangles and find the appropriate areas.

Jan 13, 2020