Posted by **Cathy** on Saturday, December 1, 2012 at 7:03am.

Two perpendicular chords divide a circle with a radius of 13 cm into 4 parts. If the perpendicular distances of both chords are 5 cm each from the center of the circle, find the area of the smallest part.

Please include solution. Thanks.

- solid mensuration -
**Aliza**, Saturday, December 1, 2012 at 7:08am
4 and 1/2

- solid mensuration -
**Aliza**, Saturday, December 1, 2012 at 7:09am
Oh wait. . im sorry but i think its wrong . sorry :(

- solid mensuration -
**Reiny**, Saturday, December 1, 2012 at 9:51am
Start with one chord which is 5 units from the centre

Notice the perpendicular height of 5 cm hits the midpoint of the chord.

By Pythagoras, we can see that the chord must be 24 cm long (5^2 + 12^2 = 13^2)

Now pretend you are rotating the chord around the centre until you have reached the perpendicualr postition. Clearly the new chord must also be 24

Notice the square 5 by 5 created by the two chords and the perpendiculars from the centre ?

Clearly that leaves our small region to be a 1/4 sector of a circle with radius 7

area of smaller sector = (1/4)π(7^2) = 49π/4

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