co-ordinate gprep

[Suyash]

6)Equilateral triangle ABC is inscribed in a circle.If length of the arc ABC is 24,what is approximate diameter of the circle?

a 5 b 8 c 11 d 15 e19.
Oa 11

How to do this Ron?

[shrenik]

here we have to consider two arcs,one is arc BAO and other is BOC (assuming centre of circle O)

then equation is 2*(120/360)*2*(22/7)*r=24

where r=radius of circle and 120 is the angle subtended by the arc at centre.

after calculation 2r=11.45which is approx 11

6)Equilateral triangle ABC is inscribed in a circle.If length of the arc ABC is 24,what is approximate diameter of the circle?

a 5 b 8 c 11 d 15 e19.
Oa 11

How to do this Ron?

[RonPurewal]

6)Equilateral triangle ABC is inscribed in a circle.If length of the arc ABC is 24,what is approximate diameter of the circle?

a 5 b 8 c 11 d 15 e19.
Oa 11

How to do this Ron?

ok.

first comment: this problem is absolutely perfect for estimation. the answer choices are extremely spread out, so, if you don't clearly know how to solve the problem, just draw a decent picture and take your best guess.

in your diagram, the arc ABC should be two-thirds of the circle's circumference. therefore, one-third of the circumference (arc AB, or arc BC) is 12. if you've drawn a good picture, then, if you imagine straightening out that 12, it's awfully close to the diameter of the circle. therefore, you'd go with choice c.

--

the real way to do the problem:
two-thirds of the circle's circumference is 24.
therefore, the circle's circumference is 36.
36 = pi * diameter
diameter = 36 / pi
= 36 / (a lil more than 3)
= a lil less than 12

[rajibgmat]

[RonPurewal]

that works too.

in your diagram, you should probably trace out the arcs in color, rather than pointing to them with arrows; especially for larger arcs, it's unclear exactly where the arrow is pointing. (you can't really point at two-thirds of a whole circle - i.e., a 240° arc - with a single arrow, cool colors notwithstanding.)[/list]

[zarak_khan]

Hi Ron,

This is how I solved this problem:

< ABC = 60 since this is an equilateral triangle
Inscribed angle at center of circle must be twice < ABC = 120

Now, 120 deg corresponds to 24 and 360 deg corresponds to 2*pi*r where r = radius

--> 120 / 360 = 24 / (2*Pi*r)
--> 1/3 = 12/(pi*r)
--> pi*r = 36
--> r = 36/pi
--> 2r = diameter = 72/pi = approx. 22

Can you please see where my mistake is?

Thanks!!

[RonPurewal]

zarak

your mistake is here:

120 deg corresponds to 24

nope.

120 degrees would be any of these arcs:
AB
AC
BC

... but, in this problem, we are given arc ABC. that is two-thirds of the way around the circle (draw a picture if you don't see why).
so 240 degrees, not 120 degrees, corresponds to an arc length of 24.

[stucked]

Ron you said, the arc ABC should be two-thirds of the circle's circumference. Are we assuming this, as not stated in the question but diagrams are not drawn to scale in the exam?
or it is a principle that for any such inscribed triangles, they make 2 3rd of circumference?

[RonPurewal]

The triangle is equilateral (all 3 sides the same, all 3 angles the same).
If you inscribe an equilateral triangle in a circle, it will divide the circle into three equal arcs. (In general, symmetry creates further symmetry.)

[RonPurewal]

The same is certainly not true for random triangles, which could split a circle into absolutely any combination of 3 arcs.

For instance, if I want a triangle that divides a circle into arcs of 14º, 87º, and 259º, then I can just (a) place points on the circle that divide it into those arcs, and then (b) connect those points to make a triangle.
Same thing for any collection of arcs at all.

[GMATstudent2015]

I don't understand how 120 is obtained for the central angle. I know that central angles are twice inscribed angles. However in Rajib's diagram, the 60 degree angles are divided into 2, so 30 degrees . Or do we sum the inscribed 30 degrees at each end and then multiply by 2?

[RonPurewal]

I know that central angles are twice inscribed angles

yes, this is one possible way to find the 120º angles.

e.g., (let's say that the center of the circle is called 'O', per the usual convention)
we have angle BAC = 60º.
thus angle BOC, which cuts off the same arc, must be 120º.

[RonPurewal]

on the other hand, there's also a much less 'fancy' solution: just use the fact that there are 180º in the three angles of a triangle.

• by symmetry, each of the 60º angles is cut into two 30º angles.

• thus, each of triangles AOB, BOC, COA has two 30º angles and a central angle. thus each central angle is 180 – 2(30) = 120º.

it's important to realize that solutions like this one exist. GMAT geometry problems are engineered so that they can be solved with BASIC knowledge (like 'there are 180º in a triangle').
sure, it could help to know 'fancy' things (such as the inscribed-angle rule). but just make sure you have an absolutely flawless grip on the basics.

[dmendelzon]

I just stumbled across this question and my approach was slightly different. I actually tried to use the Pythagorean to approximate the value of the middle of the circle.

[*]ABC = 24
[*]Calculate each side = 24/3
[*]Each side = 8

[*]Find the right triangle --> divide the base by 2 = 8/2 = 4
[*]Find the height = 8^2 - 4^2 = height^2
[*]Calculate height = 64 - 16 = 48
[*]Square root of 48 = approx 7 (7^2 is 49)

The above would mean that a little more than 7 would be the total length. In that case would be 11.

I only recommend this approach if you are under time pressure and need a quick and dirty way to resolve.

[Sage Pearce-Higgins]

In this approach, what do you mean by 'each side'? If you're talking about the triangle, then it's not correct. After all 24 is the length of the arc ABC (that's the edge of the circle).

Actually, this problem is easier than it looks - you don't need to worry about the triangle, just to notice that 24 represents 2/3 of the circumference of the circle, and then it's just a question of working out the approximate diameter from the circumference.