How to find the area of the black portion?ABCD is a square having each side 14cm.Two quadrants of circle (quarter circle)is drawn inside the square intersecting each other.Now find the area of the...

How to find the area of the black portion?

ABCD is a square having each side 14cm.Two quadrants of circle (quarter circle)is drawn inside the square intersecting each other.Now find the area of the black portion.

The figure is here:-

http://www.flickr.com/photos/78780315@N06/7384954722/

Please explain step by step.

 

Asked on by topperoo

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let the square be as given; let the intersection of the circles inside the square be point E.

The area we need is the area in the sectors DAC and CBD, excluding the intersection of the two sectors. The intersection is the triangle DEC as well as the two segments DEC and CED.

So the area we seek can be found by the areas:

`["sector"DAC-Delta DEC-"seg"CED-"seg"DEC]+["sector"CDB-Delta DEC-"seg"CDE-"seg"DEC]`

Now the area of sector DAC is the same as the area of sector CDB and is `1/4pi(14^2)=49pi` .

The area of `DeltaDEC` is `1/2(14)(7sqrt(3))=49sqrt(3)` .(This is an equilateral triangle since DE=EC=DC=14 (all radii of circles))

The area of segment DEC=area of segment CDE and can be found by:

`1/6pi(14)^2-49sqrt(3)=98/3pi-49sqrt(3)` (The area of a segment is the area of the sector minus the triangle, and we have shown that the triangle is equilateral thus the angle is `60^@` )

Putting this all together, and using symmetry, we get:

`A=2[49pi-49sqrt(3)-2[98/3pi-49sqrt(3)]]`

`=98pi-98sqrt(3)-392/3pi+196sqrt(3)`

`=98sqrt(3)-98/3pi~~67.11cm^2`

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