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Trapezoid ABCD has area of 45 . Its diagonals intersect at M. Find the area of AMB if AB is parallel to CD and AB=2CD.

 Aug 2, 2022
 #1
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45 = (1/2) (height) (2CD + CD)

 

90  =  (height) ( 3CD)

 

30  = (height) ( CD)

 

30  / CD  = height  of trapezoid

 

AMB  and BMC  are similar triangles

Since  AB = 2CD

Then height of  AMB  = 2 (height of  BMC)

 

So  there are three equal parts to the height of the trapezoid and  the height of AMB is two of them

 

So  height AMB    =  (2/3) (30) /( CD)  =  20 / CD

 

Area of AMB  = (1/2) (AB) [height (AMB) ]    =   (1/2) (2CD) [ 20/CD]  =  20

 

 

cool cool cool

 Aug 2, 2022
 #2
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Thanks!

 Aug 2, 2022

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