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An angle bisector AC divides a trapezoid ABCD into two similar triangles △ABC and △ACD. Find the perimeter of this trapezoid if the leg AB=9 cm and the leg CD=12 cm.

 

The perimeter of △CDE is 55 cm. A rhombus DMFN is inscribed  in this triangle so that vertices M, F, and N lie on the sides 

CD, CE, and DE respectively. Find CD and DE if CF=8 cm and EF=12 cm.

 Jan 19, 2019
 #1
avatar+107405 
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An angle bisector AC divides a trapezoid ABCD into two similar triangles △ABC and △ACD. Find the perimeter of this trapezoid if the leg AB=9 cm and the leg CD=12 cm.

 

Let AC bisect A

 

Then angle CAD = angle CAB

And angle ACD = angle BAC

 

So...triangle ACD is similar to  triangle CAB

 

So

 

CD/AB = AD/BC

 

12/9 = AD/BC

 

But angle ACD = angle CAB = angle CAD = angle ACB

So AD = CD

And BC = AB

So AD = 12

And BC = 9

 

So....the perimeter of the trapezoid is AB + BC + CD + AD = 9 + 9 + 12 + 12 =  42

 

 

cool cool cool

 Jan 19, 2019
 #2
avatar+107405 
+2

Here's a pic :

 

 

Since MCFN is a rhombus....then FN is parallel to CM

 

But FN is also parallel to DC

 

And CE is a transversal cutting parallel segments which makes coresponding angles EFN and ECD equal

 

And angle FEN = angle CED

 

So....by AA congruency, triangle NFE is similar to triangle DCE

 

Then    FN / FE  = CD /CE

 

And FN = CF  = 8 

And FE = 12

And CE = CF + EF =  20

 

Therefore

 

8 /  12 =  CD / 20

2 / 3 = CD /20

CD = 40/3

 

Therefore DE  =  55 - CE - CD =  55 -  20 - 40/3  =  65/3

 

 

cool cool cool

 Jan 19, 2019

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