| Last visit was: 20 Sep 2026, 03:20 |
It is currently 20 Sep 2026, 03:20 |
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:15) correct
65%
(02:44)
wrong
based on 84
sessions
History
Date
Time
Result
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 4357 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 !!
Attachment:
File comment: Solution
Circles2.PNG [ 80.71 KiB | Viewed 4032 times ]
Circles2.PNG [ 80.71 KiB | Viewed 4032 times ]
Attachment:
File comment: Question
Circles1.PNG [ 32.78 KiB | Viewed 4033 times ]
Circles1.PNG [ 32.78 KiB | Viewed 4033 times ]
Attachment:
Untitled.png [ 17.23 KiB | Viewed 4357 times ]
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
02 Sep 2018, 06:50
Kudos
Bookmarks
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 4136 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 3970 times ]
02 Sep 2018, 07:38
Kudos
Bookmarks
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
(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
Attachments
PicsArt_09-02-07.45.52.jpg [ 9.67 KiB | Viewed 3981 times ]
