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# In the ﬁgure above, O and P are the centers of the two circles. If eac

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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 [ 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]
Top Contributor
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

Cheers,
Brent
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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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