GMATPrep Q - eq. triangle inscribed in a circle

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Sat May 10, 2008 1:48 pm
How can I solve the attached problem?
Thanks
M
Attachments
GMATPrepQ4.doc
(62 KiB) Downloaded 254 times

Legendary Member
Posts: 574
Joined: Sun Jun 01, 2008 8:48 am
Location: Bangalore
Thanked: 28 times

by vishubn » Fri Aug 15, 2008 9:15 pm
arc ABC is given to be 24
where arc ABC is formed bt two arc which is again
formed by A and c which is 60 deg

so arc ABC=length arc AB+ length 0f BC
24=12+12
so now its obivous that arc AC is 12 units

so now,

60/360*2*pi*5=12
pi*D=36
r~ 11.4 .. so i choose the answer as C which is close to

hope it helps

Vishu
Last edited by vishubn on Sat Aug 16, 2008 10:06 am, edited 1 time in total.

Legendary Member
Posts: 661
Joined: Tue Jul 08, 2008 12:58 pm
Location: France
Thanked: 48 times

by pepeprepa » Sat Aug 16, 2008 12:46 am
vishubn I am troubled by your answer because you find the right answer and I don't catch your formulas (with the 60/360) and with the fact you use "r" as an answer given they ask us the diameter. Perhaps I just do not manage to follow you and we did the same thing...

Here is how I processed:

Total circumference of the circle is 36.
2*pi*r=36
pi*D=36 given that D=2*r
D=11,4

Thus, diameter is 11,4

Senior | Next Rank: 100 Posts
Posts: 54
Joined: Sat May 10, 2008 1:48 pm

by smehmood » Sat Aug 16, 2008 3:58 am
Why do you say total circumference of the circle is 36?
How can you prove it mathematically?
Thanks
M

Legendary Member
Posts: 661
Joined: Tue Jul 08, 2008 12:58 pm
Location: France
Thanked: 48 times

by pepeprepa » Sat Aug 16, 2008 5:00 am
The thing is clear, the arcs made by the vertex of an equilateral triangle have the same lengths.

Each angle of the equilateral triangle is 60°, so each of the three angles in each angle with origin the center are 120°. So each arc is 120/360=1/3 of the circumference of the triangle

Given we have two arcs which represent 24, each arc is 12.

But it is useless to lose time with that I think.

Legendary Member
Posts: 574
Joined: Sun Jun 01, 2008 8:48 am
Location: Bangalore
Thanked: 28 times

by vishubn » Sat Aug 16, 2008 10:08 am
Thanks and sorry .. that was a typo wat i meant was D which instead i typed r !!

and also 60/360 * 2 * pi *r is the formula to find the length of the arc where s0 is the measurement of one of the angle

Sorry for the confusing post


vishu

Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Sun Oct 12, 2008 9:59 pm

Take a look at this image

by dynamicishwar » Wed Oct 15, 2008 6:14 am
I was wondering if this could be solved faster as it took me atleast 5
5 minutes to solve it. My solution is attached here. If there is a better method. Plz do reply. My geometry skills need tweaking.
Attachments
ProblemwithArcLength.jpg
solution to the problem in my way

Master | Next Rank: 500 Posts
Posts: 446
Joined: Thu Jul 26, 2007 1:07 pm
Thanked: 6 times

by dferm » Thu Oct 16, 2008 10:28 am
I don't understand how you guys are arriving to your answers....
Can someone please offer a clear explanation to this problem...?

Thanks.

Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

by parallel_chase » Thu Oct 16, 2008 1:29 pm
dferm wrote:I don't understand how you guys are arriving to your answers....
Can someone please offer a clear explanation to this problem...?

Thanks.
an equilateral when inscribed in a circle, divides the circle into 3 equal arcs.

the major arc ABC = 24 therefore,

AB + BC = 12+12 [24/2 = 12]

AB + BC + AC or circumference of the circle = 12+12+12 = 36

2*pi*r = 36

r = 18*7/22 = 63/11 = 5.7

diameter or 2r = 2*5.7 = 11.4 ~ 11.

Hence D is the answer.

Hope this helps.
No rest for the Wicked....

Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Mon Oct 13, 2008 7:38 am

by 69aero69 » Fri Oct 17, 2008 3:09 am
parallel_chase,
You're completely right but the answer you wanted to say is (C).
Rgs