In the diagram to the right, equilateral triangle ADE is

Question

Say the length of a side of the square, which also equal to a side of the equilateral triangle is 1.

The are of the square = 1^2 = 1.

The radius of the circle inscribed in an equilateral triangle is  r=a*\frac{\sqrt{3}}{6} , where a is the length of the side of the triangle. So, in our case  r=\frac{\sqrt{3}}{6} .

The area of the circle =  \pi{r^2}=\pi{(\frac{\sqrt{3}}{6})^2}=\frac{\pi}{12} .

The ratio of the areas of square and circle =  \frac{1}{(\frac{\pi}{12})}=\frac{12}{\pi} .

Answer: A.

Alternatively, one can estimate that the ratio is approximately 1/4. Only A fits, other options are not even close.

Bunuel · Answer

It's a well known property. Check Triangles chapter of Math Book: math-triangles-87197.html

pipe19 · Answer

I came across the same problem. I have a question regarding the explanation that is given in the book for how to solve it without estimating.

They say that if we draw a line from the center of the circle O to D it bisects the angle of the triangle and creates another 30:60:90 triangle.

Why does the line OD bisect the angle? Is there some kind of rule when a line bisects an angle?

Thankful for any suggestions.

PareshGmat · Answer

Refer diagram below

If I get you correct, you are talking about the red lines drawn bisecting the 60 deg angle to 30 deg

Yes ; that's correct because the CG of inscribed circle = CG of equilateral triangle

The red line drawn is joining the midpoint of the opposite side; so it will bisect; That's the rule

pipe19 · Answer

Yes, exactly! So because the center of the circle = center of triangle the line bisects the angle?

PareshGmat · Answer

For this problem, we can summarize as follows

1. The inscribe circle touches the midpoints of all sides of the equilateral triangle
2. The lines in red are the medians of the equilateral triangle
3. Medians also bisect the angle 60 of equilateral triangle to angle 30
4. In an equilateral triangle, the median, angle bisector, and altitude are equal. They intersect at center of the inscribed and circumscribed circle.

PareshGmat · Answer

Agree.. This is correct

syahasa2 · Answer

Say r=1 and O the center of the circle. Then because of ADE=60, OD biscepts ADE to 30+30.
Therefore we have a triangle ODF with OF=1 and FD= √3 . (OF is the line that biscepts ED to its half).
We know that 30-60-90 triangle have sides 1-√3 - 2.

Ac=πr^2=π*1=π
Asq=ΕD^2= (2 √3 ) ^2 = 12

Ac/Asq=π/12

satya2029 · Answer

Let the side of the square is X, and AP is the perpendicular distance from vertex A to side ED.
Side of the equilateral triangle is also X according to the figure
Half of an equilateral triangle(AEP) is 30-60-90 triangle, and sides are in the ratio of a:a√3:2a
a=X/2(half of the side of the equilateral triangle)
AP=√3X/2
Since it is inscribed circle, centriod and the center of the circle will be same. AP will be median, and centroid will divide it in 2:1(2Y:1Y)
3Y=AP=√3X/2
Y=X/2√3(Radius of the circle)
area of the circle/area of the square=pi*Y^2/X^2
=(pi*X^2/12)/X^2
=pi/12
A:)

Fdambro294 · Answer

Rule 1: the In-radius of an Inscribed Circle inside an Equilateral Triangle = (1/3) * (Height of Equilateral Triangle)

Since the question asks for a Ratio answer, we can choose Numbers to fit the Diagram.

Let Side of Equilateral Triangle = 1

The Side of the Equilateral Triangle on the Bottom shares the Side of the Square ------> thus Side of Square = 1

Area of Square = (1)^2 = 1

In-radius of Inscribed Circle:

Height of the Equilateral Triangle is given by = (side) * sqrt(3) * (1/2)

(1/3) * (Height of Equilateral Triangle) = Radius of Inscribed Circle

given side =1:

(1/3) * (1) * sqrt(3) * (1/2) = sqrt(3)/6 = Radius of Inscribed Circle

Area of Inscribed Circle = (pi) * (r)^2 = (sqrt(3) / 6)^2 * (pi) = 3(pi) / 36 = (pi)/12

Ratio of the: (Area of Circle) / (Area of Square) =   /   = (pi) /12

-A-

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

In the diagram to the right, equilateral triangle ADE is

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In the diagram to the right, equilateral triangle ADE is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards