SOLUTION: TWO PARALLEL CHORDS PQ AND MN ARE 3 cm APART ON THE SAME SIDE OF THE CIRCLE WHERE PQ=7 cm and MN = 14cm. CALCULATE THE RADIUS OF THE CIRCLE

Algebra ->  Circles -> SOLUTION: TWO PARALLEL CHORDS PQ AND MN ARE 3 cm APART ON THE SAME SIDE OF THE CIRCLE WHERE PQ=7 cm and MN = 14cm. CALCULATE THE RADIUS OF THE CIRCLE       Log On


   

Question 407869: TWO PARALLEL CHORDS PQ AND MN ARE 3 cm APART ON THE SAME SIDE OF THE CIRCLE WHERE PQ=7 cm and MN = 14cm. CALCULATE THE RADIUS OF THE CIRCLE
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Let MN and PQ be the parallel chords with O as center of the circle
given:
+MN=14cm and PQ=7cm with distance 3cm between the parallel chords MN and PQ.
Let X and Y be the mid+points of MN and PQ.
Then OXY is a unique line which is the perpendicular+bisector to both MN and PQ
This is the theorem of the line joining the center of the circle and the mid point of the chord is perpendicular to the chord .
Or else, a line joining the center of the circle cuts chord perpendicularly, then it bisects the chord.

So, XN+=7cm and YQ+=+3.5cm
XY+=+3cm ..........given
XN+=7cm and YQ+=+3.5
Let OX+=+x,
now from the right angled triangle, OXN , by Pythagoras theorem,
OX%5E2%2BXN%5E2+=+ON%5E2+=+r%5E2, where r is the radius of the circle,
or
x%5E2+%2B+7%5E2+=+r%5E2.........................(1)

Similarly from right triangle OYQ,

OY%5E2%2BYQ%5E2=r%5E2 or

%283%2Bx%29%5E2%2B3.5%5E2+=+r%5E2...................(2).
From (1) and (2), left sides must be equal because right sides are equal:

x%5E2%2B7%5E2=%28x%2B3%29%5E2%2B%283.5%29%5E2.solve for x
x%5E2%2B49=x%5E2%2B6x%2B9%2B12.25
49-21.22+=+6x
6x+=+27.75
x=27.5%2F6+=+4.625+cm.
Therefore, from(1).
%284.625%29%5E2%2B7%5E2+=+r%5E2
r+=+sqrt%28%284.625%29%5E2%2B7%5E2%29+=+8.380012097cm
r+=+8.38cm is the radius of the circle.