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# What percent of the area of the rhombus ABCD is the area of the circle

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What percent of the area of the rhombus ABCD is the area of the circle  [#permalink]

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07 Mar 2019, 15:49
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75% (hard)

Question Stats:

18% (01:41) correct 82% (01:53) wrong based on 17 sessions

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GMATH practice exercise (Quant Class 19)

What percent of the area of the rhombus ABCD is the area of the circle that is inscribed in it?

(1) Angle BCD is equal to 60 degrees
(2) AB = 4

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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Re: What percent of the area of the rhombus ABCD is the area of the circle  [#permalink]

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07 Mar 2019, 20:01
Can someone plz explain how A is the answer?

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Re: What percent of the area of the rhombus ABCD is the area of the circle  [#permalink]

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07 Mar 2019, 20:19
2
What percent of the area of the rhombus ABCD is the area of the circle that is inscribed in it?

(1) Angle BCD is equal to 60 degrees
Join the diagonals as shown.. Triangle COB is 30-60-90 and sides are in ratio 1:$$\sqrt{3}$$:2...
so CO = $$a\sqrt{3}$$, and BO = a, when side BC = 2a.... so AREA of rhombus = 2a*2$$a\sqrt{3}$$
NEXT take triangle COX, by taking x on tangent .. so again Triangle COX is 30-60-90 and sides are in ratio 1:$$\sqrt{3}$$:2...
Now CO =2*common ratio =$$a\sqrt{3}$$, so radius of circle = $$\frac{CO}{2}=a\sqrt{3}$$/2
Thus area of circle = $$\pi*(a\sqrt{3}/2)^2=\pi*3a^2/4$$..
Fraction = $$\frac{\pi*3a^2}{4*2a*2a*sqrt(3)}$$... a will get cancelled out and we can find the fraction
Sufficient

(2) AB = 4
We do not the angles, we cannot say anything..
If all angles are 90, then the fraction will be different and if 60 and 120 then different
Insufficient

A
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What percent of the area of the rhombus ABCD is the area of the circle  [#permalink]

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08 Mar 2019, 06:39
fskilnik wrote:
GMATH practice exercise (Quant Class 19)

What percent of the area of the rhombus ABCD is the area of the circle that is inscribed in it?

(1) Angle BCD is equal to 60 degrees
(2) AB = 4

$$? = {{{S_{{\rm{circle}}}}} \over {{S_{{\rm{rhombus}}}}}}$$

$$\left( 1 \right)\,\,\,\left\{ \matrix{ BCD = {60^ \circ } \hfill \cr ABCD\,\,{\rm{rhombus}} \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\Delta \,CBD\,\,{\rm{and}}\,\,\Delta ABD\,\,{\rm{equilaterals}}\,\,\,\left( * \right)\,\,\,\,\,$$

$$?\,\,\mathop = \limits^{\left( * \right)} \,\,\frac{{{S_{{\text{circle}}}}}}{{2 \cdot {S_{\Delta {\text{ABD}}}}}}\,\,\mathop = \limits^{\left( * \right)} \,\,\frac{{\pi \cdot O{E^2}}}{{2 \cdot \left( {\frac{1}{2} \cdot BD \cdot \frac{{BD \cdot \sqrt 3 }}{2}} \right)}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,? = {\left( {\frac{{OE}}{{BD}}} \right)^2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\boxed{\,? = \frac{{OE}}{{BD}}\,}$$

$$\left\{ \matrix{ \,\Delta OEB\,\,\,\,\left[ {{{30}^ \circ },{{60}^ \circ },{{90}^ \circ }} \right] \hfill \cr \,{1 \over 2}BD = OB\,\,\,{\rm{hyp}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left[ {{{30}^ \circ },{{60}^ \circ },{{90}^ \circ }} \right]} \,\,\,\,OE = {{OB \cdot \sqrt 3 } \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {{OE} \over {BD}} = {{OE} \over {2 \cdot OB}} = {1 \over 2}\left( {{{\sqrt 3 } \over 2}} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,{\rm{Insuff}}.\,\,\,\,\left( {{\rm{geometric}}\,\,{\rm{bifurcation}}\,{\rm{,}}\,\,{\rm{see}}\,\,{\rm{images}}} \right)$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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What percent of the area of the rhombus ABCD is the area of the circle   [#permalink] 08 Mar 2019, 06:39
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