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Difficulty:
95%
(hard)
Question Stats:
35% (02:15) correct
65%
(02:44)
wrong
based on 84
sessions
History
Date
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Not Attempted Yet
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 !!
Untitled.png [ 17.23 KiB | Viewed 4204 times ]
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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File comment: Solution
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Circles2.PNG [ 80.71 KiB | Viewed 3894 times ]
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Untitled.png [ 17.23 KiB | Viewed 4204 times ]
02 Sep 2018, 06:50
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Mani2879
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.
Attachments
02Set18_Mani.gif [ 3.1 KiB | Viewed 3991 times ]
\(? = {S_{^ \circ }} = \,\pi {r^2}\,\,\,\,\)
\(\left( 1 \right)\,\,\,\,\frac{{{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\,\,\, = \,\,\,\frac{{\frac{1}{2}\pi {r^2}}}{{{S_{\Delta \,ABD}}}}\,\,\,\mathop = \limits^{{\text{post}}\,\,{\text{above!}}} \,\,\,\frac{{\pi {r^2}}}{{4\,\,\,{S_{\Delta \,ABC}}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{S_{^ \circ }} = \pi {r^2}\, = 4\,\,\,\left( {{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}} \right) = 4\,\,\,\left( {{S_{\,{\text{sector}}\,\,ACB}}} \right)\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\theta \,\,{\text{ = }}\,\,{\text{9}}{{\text{0}}^ \circ }\)
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.
\(\left( 1 \right)\,\,\,\,\frac{{{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}}}{{{S_{\Delta \,ABC}}}}\,\,\, = \,\,\,\frac{{\frac{1}{2}\pi {r^2}}}{{{S_{\Delta \,ABD}}}}\,\,\,\mathop = \limits^{{\text{post}}\,\,{\text{above!}}} \,\,\,\frac{{\pi {r^2}}}{{4\,\,\,{S_{\Delta \,ABC}}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{S_{^ \circ }} = \pi {r^2}\, = 4\,\,\,\left( {{S_{\,{\text{blue}}}} + {S_{\Delta \,ABC}}} \right) = 4\,\,\,\left( {{S_{\,{\text{sector}}\,\,ACB}}} \right)\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\theta \,\,{\text{ = }}\,\,{\text{9}}{{\text{0}}^ \circ }\)
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.
Attachments
02Set18_3p.gif [ 19.87 KiB | Viewed 3821 times ]
