Question 161674: 1) there is a circle with a 24" diameter.
2) There is a line drawn across the circle six inches from the bottom.
3) What is the area under the line?
4) figure this problem out without using the formula to find the area under a curve.
5 Good luck!
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! area under the line should be the area of the circle minus the area of the triangle minus the area of the circle above the triangle.
that's what i think anyway.
working on that assumption i get the following:
radius of the circle is 12.
area of the circle is pr^2, where p stands for the greek letter pie which equals 3.14159.......
-----
so area of the circle = p*(12)^2 = 452.389...
-----
since the line is drawn 6 inches from the bottom of the circle, i assume it's perpendicular to the radius of the circle at that point and that means that 6 inches from that point on the radius gets you to the center of the circle. in other words, that line is perpendicular to and cuts the radius in half at that point.
-----
if that assumption holds, then we can create a triangle from the center of the circle to the intersection points of that line with the circle on both ends.
this creates an isosceles triangle that has the radius of the circle as the 2 equal legs and the line 6 inches from the bottom of the circle as its base.
the altitude of that triangle is the radius that cuts perpendicular to the line going across (base of the triangle) and is bisected by that line (6 inches above and 6 inches below). this makes the height of the triangle 6 inches.
-----
the angle at each end of this triangle will be 30 degrees because the sine of those angles will be the altitude of the triangle divided by the radius of the triangle (opposite / hypotenuse).
altitude is 6 inches, hypotenuse is 12 inches, sine is 6/12 = .5 so angle is 30 degrees.
-----
this makes the sum of the angles both right triangles make with the center of the circle equal to 120 degrees (60 degrees for each right triangle).
-----
if we create the diameter of the circle parallel to the line we cut across the circle, then the angle between the diameter and the triangles will be 180 - 120 = 60 degrees in total, 30 degrees on each side.
this has to do with supplementary angles and the fact that opposite internal angles of parallel lines are equal.
-----
what we have then is an area of the circle above the triangle that is equal to 240 degrees / 360 degrees times the area of the circle.
-----
we can now solve for the area above the triangle and the area of the triangle and what is left is the area under the base of the triangles which is what we are looking for.
-----
area of the circle above the triangle is 240/360 * p * r^2 which equals
2/3 * p * 144 = 301.5928947
-----
area of the triangle is 1/2 base * height.
we need the base.
tangent of 30 degrees is opposite over adjacent.
opposite is 6
adjacent is unknown.
if tan 30 = o/a, then a = o/tan30 so we need to find 6/tan30 to get the base of one of the right triangles which will be 1/2 the base of the isosceles triangle we are looking for (the line across the circle 6 inches from the bottom).
-----
(6/tan30) = 10.39230485.
-----
area of the isosceles triangle is 1/2 base * height = 10.39230485 * 6 = 62.35382907
-----
area of the triangle plus portion of circle above the triangle equals 363.9467238.
-----
area at the bottom of the circle up to the bottom of the triangle is area of the circle minus this which equals 88.4426183.
----
area we are looking for at the bottom of the circle is 88.4426183.
-----
that's what i think anyway.
logically it seems correct.
you may wish to confirm with others before accepting as gospel.
tough to do without pictures.
don't know how to add pictures here.
if the above confused you, try this.
-----
draw a circle.
let A be the center of the circle.
draw a diameter across the circle.
call the ends of the diameter B and C respectively.
you have a diameter BC that goes from B to A to C across the circle left to right horizontally.
draw a radius perpendicular to the diameter at A and drop it vertically down to touch the circle at D.
the radius is AD.
draw another line midpoint between A and D and perpendicular to AD. this line will be parallel to BC (the diameter) and will be 6 inches from the center of the circle and 6 inches from the bottom of the circle.
make the endpoints of this line E and F so the line is EF. this is the line drawn across the circle six inches from the bottom.
call the intersection of AD and EF point G.
draw a line from A to E, and draw a line from A to F.
you now have the triangles mentioned above.
the isosoceles triangle is triangle EAF with EA and AF being the equal sides because they are also radii of the circle.
the 2 right triangles formed are AEG and AFG.
angle AEG and angle AFG are 30 degrees.
angle EAG and angle FAG are 60 degrees.
angle EAF within the isosceles triangle is 120 degrees (EAG plus FAG).
angle FAC and angle EAB are each 30 degrees.
area of the circle formed above the triangle is enclosed by the circumference of the circle from E to B to C to F.
angle EAF external to the isosceles triangle is 240 degrees. this is 180 plus angle BAE plus angle FAC.
if you can draw that then you should be able to see what i'm talking about.
if not, send me an email and i'll forward you a picture.
-----
|
|
|
