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.

Guest Jan 1, 2021

#1**+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

wolfiechan Jan 1, 2021

#3**+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**

Guest Jan 1, 2021