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The diagram, K, O, and M are centers of the three semi-circles. Also, OC = 32 and CB = 40. Line l is drawn to touch smaller semi-circles at points S and E so that KS and ME are both perpendicular to l. Determine the area of quadrilateral KSEM.

 

 Jan 1, 2021
 #1
avatar+119 
+1

So guest

A radius of the biggest circle is OB   OB=OC+CB = 32+36 = 68 unitis

AO also equals 68 units.

AC= AO+OC = 68+32=100

So the radius of the middle sized circle is 50 units

KS=50

 -the area of trapezuim KSEM-

= average of the paralell sides * perpendicular height.

= 0.5( ME + KS) * ES

= 0.5( 18+ 50 ) * ES

= 34 * ES

SY=ME=18

KS = 50 = KY+YS = KY+18

so KY = 50-18=32 units.

 

Now KYM is a right angled triangle, t

The hypotenuse is KM = 50+18=68 units

one short side is KY = 32 units

 

KSEM = 34*ES = 34*60 = 2040 units squared. Im pretty sure

But I might have to double check my work but I think I got that right (I asked one of my class mates for help they said I was right)

If you need a more in debt answer just ask

-wolfie

 Jan 1, 2021
 #2
avatar+119 
+2

if I did get it wrong Hopefully someone can help you !

wolfiechan  Jan 1, 2021
 #4
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+1

OC = 32       CB = 40     OC + CB = OB       32 + 40 = 72 smiley

Guest Jan 1, 2021
 #3
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+2

In the diagram, K, O, and M are centers of the three semi-circles. Also, OC = 32 and CB = 40. Line l is drawn to touch smaller semi-circles at points S and E so that KS and ME are both perpendicular to l. Determine the area of quadrilateral KSEM.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

EM = 20       SK = 52         KM = 72          SE = 8√65

 

EM and SK are the bases, KM is a side, and SE is the height of a trapezoid KSEM.

 

[KSEM] = 1/2(EM + SK) * SE = 1/2(20 + 52) * 8√65

 

[KSEM] ≈ 2321.93 square units

 Jan 1, 2021

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