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Trapezoid $ABCD$ is inscribed in the semicircle with diameter $\overline{AB}$, as shown below.  Find the radius of the semicircle.  Find the area of ABCD.

 

PQDC is a square.

 

 Dec 22, 2023
 #1
avatar+128732 
+1

radius =  (9 + 16 + 9) /  2  =   17

 

Let O  be the center of the circle

 

QO =  8  OD  = 17

 

Triangle QOD is right  with OD=  the hypotenuse

 

So

 

DQ = sqrt ( 17^2 - 8^2) =  sqrt ( 225) =  15

 

So ABCD is a trapezoid  with height  =  15    and bases of 34  and 16

 

[ ABCD ] =  (1/2) ( 15) ( 34 + 16)  =   375

 

cool cool cool

 Dec 22, 2023

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