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The diagram, K, O, and M are centers of the three semi-circles. Also, OC = 32 and CB = 36. 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.

 Oct 3, 2018
 #1
avatar+102441 
+2

 

You will need to follow all this with the diagram  wink

 

Since KS abd ME are both perpendicular to l,  it follows that KS and ME are parallel to each other. 

 

So

KSEM is a trapezium where the parallel sides are KM and ME and the prependicular height is ES

 

Now it is given that the diameter of the little circle is 36 so its radius is 18 so ME=18

 

A radius of the biggest circle is OB   Now 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

 

So now we have the area of trapezuim KSEM

= average of the paralell sides * perpendicular height.

= 0.5( ME + KS) * ES

= 0.5( 18+ 50 ) * ES

= 34 * ES

 

So how do I find the length of ES ?

 

I am going to introduce a new point Y. It will be on the line KS such that SYME is a rectangle.

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

So the other short side, KY= \(\sqrt{68^2-32^3}=60\;units\)

 

KY=SE which is the height of the trapezium KSEM

so area of KSEM = 34*ES = 34*60 = 2040 units squared.

 Oct 4, 2018
 #2
avatar+464 
+2

thank you so much

 Oct 4, 2018

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