How to solve this Mensuration sum of Circle? Please explain step by step. And remember I am in class 10.In the figure given in the link ,ABC is an equilateral triangle of side 8 cm.A,B and C are...

How to solve this Mensuration sum of Circle?

Please explain step by step.

And remember I am in class 10.

In the figure given in the link ,ABC is an equilateral triangle of side 8 cm.A,B and C are the centres of the circular arcs of equal radius .Find the area of the shaded region(region DEF) correct upto 2 decimal places.

Don't use the sectorial area formula.I am only in class 10.

The link:

http://www.flickr.com/photos/78780315@N06/7616085296/in/photostream/

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The area of the shaded section is going to be the area of the triangle minus the areas of the three sectors.  Since it is an equilateral triangle, each sector is one sixth of the area of the circle with radius 4, which means the area of all three sectors is one half the area of the circle.  

The area of a triangle is `1/2 bh`.  The base is 8 cm and the height can be found from the Pythagorean theorem, since we know the hypotenuse.  

`h^2+4^2=8^2`

`h=\sqrt{64-16}`

`h=sqrt{48}`

This means that the area of the shaded area is

`A=1/2 8sqrt{48} - {pi 4^2}/2`

`=2.58 cm^2` 

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