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.
3 answers
4 and 1/2
Oh wait. . im sorry but i think its wrong . sorry :(
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
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
1 answer