2 perpendicular chords of a circle ab and cd meet at m. if am=6m, bm = 4m and cm= 12m, then find the area of circle

Question

Plz give sol with figure

Answer 1

draw a circle with a vertical chord near the left side and a horizontal chord near the bottom of the circle

ab is the vertical chord, from top to bottom, and cd is the horizontal chord from right to left

they intersect in point m

am=6, bm=4, cm=12 and therefore dm=2 because the product of the segments of the chords are equal

6*4=12*?

24=12*?

?=2, which is dm

find the center of the circle and call it point o

from point o, draw a perpendicular to cd and call the point of intersection point x, and draw a perpendicular to ab and call the point of intersection point y

these perpendicular lines bisect each chord

the vertical chord is 10 and the horizontal chord is 14

so half the vertical chord is 5 and half the horizontal chord is 7

draw these lines

the vertical line from point o to chord cd at x is 1 because ym is 1 because 6-5=1

the horizontal line from point o to chord ab at y is 5 because 12-7=5

draw a radius from the center, point o, to point c

ox is one leg, xc is another leg, and oc is the hypotenuse of a right triangle

the hypotenuse is a radius don't forget

use the Pythagorean Theorem to find oc

1^2+7^2=oc^2

1+49=oc^2

50=oc^2

oc=√50 m, the radius of the circle

now find the area of the circle

A=∏r^2

A=(3.14159)(√50)^2

A=(3.14159)(50)

A=157.0795 square meters

 

 

Answer 2

You draw a circle and those two chords as it is said, then connect point a  and c, and c and b. You have will have a triangle abc. By using properties of inscribed triangle you can find the area of the circle.