 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Shifted regions (Posted on 2019-02-27) A circle is centered at the origin with a radius of 5. Two lines are drawn that divide the circle into four regions, the first is the line y=-1, and the other is x=-3. Find the area of the largest region.

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The largest of the four areas enclosed by the two lines and the circumference is a region larger than, but containing, the circle's first quadrant. The coordinate axes divide this largest region into four parts, which we label, going counterclockwise: I, II, III, IV.

We will add their areas. But first we need to understand segments.

Below the y=-1 line is cap "segment1" with central angle (the opening angle from the origin to the two cap corners) theta1:

theta1 = 2 x arccos(1/5)

To the left of the x=-3 line is another, smaller cap segment, segment3, with central angle:

theta3 = 2 x arccos(3/5)

The area of a segment is (1/2) R^2 [theta - sin(theta)], with theta in radians and here, R=5.

The way to get this formula is to get the area of the sector (A=theta R^2) and subtract two triangles (of total area (R^2/2) sin(theta)) to find the area of the cap segments. Our areas I and III are 1/2 the difference between the area of a half circle and the "cap" segments computed above.

Area I = (1/2) [ (1/2) A_circle - A_segment1]

= (R^2/4) [pi-theta1+sin(theta1)]

Area II = (1/4) A_circle  [as it's the first quadrant]

Area III = (1/2) [ (1/2) A_circle - A_segment3]

= (R^2/4) [pi-theta3+sin(theta3)]

Area IV = rectangle 3 x 1 = 3

theta 1,3 = 156.93, 106.25 in degrees (approx.)

Area  = Area I + II + III + IV

Area = 41.645 (approx.)

Edited on February 28, 2019, 11:06 am
 Posted by Steven Lord on 2019-02-27 14:12:36 Please log in:

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