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.
+336 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
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.
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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
