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Re: In the figure above, O and P are the centers of the two circles. If eac [#permalink]
Princ wrote:
OA: B
Area of rhombus =\(\frac{1}{2}*d*\sqrt{4a^2-d^2}\)
where d = length of one of the diagonals of rhombus
a = side of the rhombus
here a=d=r
so Area = \(\frac{1}{2}*r*\sqrt{4r^2-r^2}\)
Area = \(\frac{\sqrt{3}r^2}{2}\)


Derivation of below formula
Area of rhombus =\(\frac{1}{2}*d*\sqrt{4a^2-d^2}\)
Attachment:
rhombus.png
rhombus.png [ 61.27 KiB | Viewed 8183 times ]

Properties of Rhombus used for this derivation
1.Its diagonals divide the figure into 4 congruent triangles.
2.Its diagonals are perpendicular bisectors of eachother

Lets take \(AB=BC=CD=DA=a\)
and one known diagonal \(DB = d\)

In right angle \(\triangle\) CDI,
\(CD=a\) and \(DI=\frac{DB}{2}=\frac{d}{2}\)

Using pythagoras theorem in \(\triangle\)CDI , we get

\(CI =\sqrt{CD^2-DI^2}=\sqrt{a^2-(\frac{d}{2})^2}=\frac{1}{2}\sqrt{4a^2-d^2}\)

Area of \(\triangle\) CDI =\(\frac{1}{2}*DI*CI=\frac{1}{8}*d*\sqrt{4a^2-d^2}\)

As there are 4 such triangle, Area of rhombus = 4 *Area of \(\triangle\) CDI

Area of rhombus =\(4 *\frac{1}{8}*d*\sqrt{4a^2-d^2}=\frac{1}{2}*d*\sqrt{4a^2-d^2}\)

It also imply that other diagonal's length is \(\sqrt{4a^2-d^2}\)
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Re: In the figure above, O and P are the centers of the two circles. If eac [#permalink]
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Bunuel wrote:

In the figure above, O and P are the centers of the two circles. If each circle has radius r, what is the area of the shaded region?


A) \((\frac{\sqrt{2}}{2})(r^2)\)

B) \((\frac{\sqrt{3}}{2})(r^2)\)

C) \((\sqrt{2})(r^2)\)

D) \((\sqrt{3})(r^2)\)

E) \((2\sqrt{3})(r^2)\)


Attachment:
b1pxiJI.jpg


We can see that the shaded region consists of 2 EQUILATERAL TRIANGLES, each with side length = r

It helps (A LOT!) to know the area of an equilateral triangle = (√3)(side²)/4
So, an equilateral triangle with side length r will have area of (√3)(r²)/4

Since the shaded region consists of 2 EQUILATERAL TRIANGLES, the area = (√3)(r²)/4 + (√3)(r²)/4
= (√3)(r²)/2

Answer: B

Cheers,
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In the figure above, O and P are the centers of the two circles. If eac [#permalink]
Approach:

- Notice that every side of shaded region is equal, which is also equal to radius r.

- Now, to find the area of shaded region, divide the region into two halves, we get 2 equilateral triangles of side r each.

- Area of an equilateral triangle with side r: \(\frac{\sqrt{3}}{4} r^2\)

- Multiply the above result by 2 to find the area of shaded region => \(2*\frac{\sqrt{3}}{4} r^2\) = \(\frac{\sqrt{3}}{2} r^2\)

Option B it is!
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Re: In the gure above, O and P are the centers of the two circles. If eac [#permalink]
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Re: In the gure above, O and P are the centers of the two circles. If eac [#permalink]
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