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What is the area of the circle with center C, shown above?

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What is the area of the circle with center C, shown above?  [#permalink]

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Updated on: 02 Sep 2018, 01:42
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What is the area of the circle with center C, shown above?

(1) The ratio of the area of the region enclosed by segment AC, segment CB, and minor arc AB to the area of triangle ABC is equal to the ratio of the ratio of half the area of circle C to the area of triangle ABD.

(2) The ratio of the perimeter of triangle ABC to the area of triangle ACD is $$(2+\sqrt{2}):6$$

Hi,
I didnt understand how could they conclude that the areas of triangle ABC And ACD are equal ? I do understand that both triangles are congruent (By AAS ),but in that case Wouldn't Statement 2 be sufficient to answer the question ?

What am I missing here ?

Any inputs will be appreciated !!

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Originally posted by Mani2879 on 01 Sep 2018, 11:33.
Last edited by Bunuel on 02 Sep 2018, 01:42, edited 1 time in total.
Renamed the topic and edited the question.
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Re: What is the area of the circle with center C, shown above?  [#permalink]

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02 Sep 2018, 05:50
1
Mani2879 wrote:

I didn´t understand how could they conclude that the areas of triangle ABC And ACD are equal.

Hi, Mani!

Please follow the figure attached. Through it, we have:

$${S_{\Delta \,ABC\,}} = \,\,\,\frac{1}{2}\left( {BC \cdot h} \right) = \frac{1}{2}rh = \frac{1}{2}\left( {CD \cdot h} \right) = {S_{\Delta \,ACD\,}}$$

As far as your other question is concerned (why statement (2) is insufficient), I will post (below) my full solution, I hope the (subtle) bifurcation clears your doubt!

If you don´t understand my rationale, feel free to ask details!

Regards,
fskilnik.
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02Set18_Mani.gif [ 3.1 KiB | Viewed 494 times ]

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Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount!

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What is the area of the circle with center C, shown above?  [#permalink]

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Updated on: 02 Sep 2018, 12:48
1
1
$$? = {S_{^ \circ }} = \,\pi {r^2}\,\,\,\,$$

$$\left( 1 \right)\,\,\,\,\frac{{{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\,\,\, = \,\,\,\frac{{\frac{1}{2}\pi {r^2}}}{{{S_{\Delta \,ABD}}}}\,\,\,\mathop = \limits^{{\text{post}}\,\,{\text{above!}}} \,\,\,\frac{{\pi {r^2}}}{{4\,\,\,{S_{\Delta \,ABC}}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{S_{^ \circ }} = \pi {r^2}\, = 4\,\,\,\left( {{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}} \right) = 4\,\,\,\left( {{S_{\,{\text{sector}}\,\,ACB}}} \right)\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\theta \,\,{\text{ = }}\,\,{\text{9}}{{\text{0}}^ \circ }$$

Note that statement (1) is equivalent to the geometric configuration in which triangle ABD is right and isosceles, but the radius is "free" , that´s why the bifurcation is trivial:

> Take r = 1, so that ? = pi
> Take r = 2, so that ? = 4*pi

$$\left( 2 \right)\,\,\,\,\frac{{2 + \sqrt 2 }}{6} = \frac{{{\text{peri}}{{\text{m}}_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\,$$

In the figure shown, we present two possible geometric configurations, in which we have:

(2a) Triangle ABD right and Isosceles (case explored in (1) above) :

$$\frac{{2 + \sqrt 2 }}{6} = \frac{{{\text{peri}}{{\text{m}}_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\, = \frac{{2r + r\sqrt 2 }}{{\frac{1}{2}\left( {r \cdot r} \right)}}\,\,\,\, \Rightarrow \,\,\, \ldots \,\,\, \Rightarrow \,\,\,r = 12\,\,\,\left( {viable!} \right)\,\,\,\, \Rightarrow \,\,\,\,?\,\,\, = \,\,144\pi$$

(2b) Triangle ADC is equilateral and, consequently, triangle ABD is a 30-60-90 triangle:

$$\frac{{2 + \sqrt 2 }}{6} = \frac{{{\text{peri}}{{\text{m}}_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\, = \frac{{2r + r\sqrt 3 }}{{\frac{1}{2}\left[ {r \cdot \frac{{r\sqrt 3 }}{2}} \right]}}\,\,\,\,\, \Rightarrow \,\,\, \ldots \,\,\, \Rightarrow \,\,\,r > 0,\,\,\,r \ne 12\,\,\,\left( {viable!} \right)\,\,\,\, \Rightarrow \,\,\,\,?\,\,\, \ne \,\,\,144\pi$$

(1+2) Now we know we necessarily have the first case in the bifurcation shown in statement (2), hence (1+2) is SUFFICIENT.

This solution follows the notations and rationale taught in the GMATH method.
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_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)
Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount!

Originally posted by fskilnik on 02 Sep 2018, 06:23.
Last edited by fskilnik on 02 Sep 2018, 12:48, edited 5 times in total.
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What is the area of the circle with center C, shown above?  [#permalink]

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02 Sep 2018, 06:38
1
What is the area of the circle with center C, shown above?

(1) The ratio of the area of the region enclosed by segment AC, segment CB, and minor arc AB to the area of triangle ABC is equal to the ratio of the ratio of half the area of circle C to the area of triangle ABD.

See the attached figure..
a, b, c and d are the areas of the region as shown in the figure..
So $$\frac{(a+b)}{b}=\frac{(a+b+c+d)}{(b+c)}.........ab+ac+b^2+bc=ab+b^2+bc+bd...... ac=bd.....$$

Now areas b and area c have same base r and height is same so b=c
So $$ac=cd$$ or a$$=d$$.....
Since a=d and b=c we can say $$a+b=c+d.$$.
This is possible only if AC divides the semi circle in equal parts and thus ABD is isosceles right angled triangle
$$AB=AD=\frac{BD}{√2}=\frac{2r}{√2}=r√2$$
But we do not know anything about r
Insufficient

(2) The ratio of the perimeter of triangle ABC to the area of triangle ACD is $$(2+\sqrt{2}):6$$
$$\frac{(AB+BC+AC)}{c}=\frac{(2+√2)}{6}.......\frac{(AB+r+r)}{c}=\frac{(2+√2)}{6}$$
Insuff

Combined
$$\frac{(AB+2r)}{c}=\frac{(2+√2)}{6}...................................\frac{(r√2+2r)}{(1/2*AD*BD)}=\frac{(2+√2)}{6}.$$....

$$\frac{r(2+√2)}{1/2*r*r}=\frac{(2+√2)}{6}............\frac{1}{2}*r=6$$.......r=12
thus the area = $$\pi*12^2=144\pi$$
Sufficient

C
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What is the area of the circle with center C, shown above? &nbs [#permalink] 02 Sep 2018, 06:38
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