What is the NO-SHORTCUT approach for learning great Mathematics?
Learn More

August 9, 2020

How to Pursue Mathematics after High School?

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

Try this beautiful Problem on Geometry from Circular arc from (AMC 10 A, 2012).

Circular arc - AMC-10A, 2012- Problem 18


The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2 \pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2 .$ What is the area enclosed by the curve?

Circular Arc

,

  • $2 \pi+6$
  • $2 \pi+4 \sqrt{3}$
  • $3 \pi+4$
  • $2 \pi+3 \sqrt{3}+2$
  • $\pi+6 \sqrt{3}$

Key Concepts


Geometry

Circle

Hexagon

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2012

Check the answer here, but try the problem first

$\pi+6 \sqrt{3}$

Try with Hints


First Hint

Circular arc

We have to find out the area enclosed by the curve. but we can not find out eassily. Now we join the centre of 9 congruent circles. it will form a hexagon.

Circular arc with hexagon

Can you find out the required area of the closed curve? Can you finish the problem?

Second Hint

Circular arc with hexagon

The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

Third Hint

 hexagon

The areas of the three sectors outside the hexaqon=$2 \pi$ .

the areas of the three sectors inside the hexagon but outside the figure $(\pi)$

Therefore the area enclosed by the curve is $\pi+6 \sqrt{3}$

Subscribe to Cheenta at Youtube


What to do to shape your Career in Mathematics after 12th? 

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

  • What are some of the best colleges for Mathematics that you can aim to apply for after high school?
  • How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
  • What are the best universities for MS, MMath, and Ph.D. Programs in India?
  • What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
  • How can you pursue a Ph.D. in Mathematics outside India?
  • What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Want to Explore Advanced Mathematics at Cheenta?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

To Explore and Experience Advanced Mathematics at Cheenta
Register here

5 comments on “Circular arc | AMC 10A ,2012 | Problem No 18”

  1. I have A Big DOUBT:
    In Problem:It should be 6 congruent circular arcs, instead of 9 congruent circular arcs,
    The closed curve in the figure is made up of 6 congruent circular arcs each of length 2π/3, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
    Pl Check If Problem needs Correction OR AM I Wrong?

    1. I realised the problem is All-Right.
      It is made of circular sectors of circle of radius of 1;
      Three Outside sectors Of Hexagon: each of reflex angles (4/3)*π;
      Three Inside Sectors Of Hexagon each of obtuse angle (2/3)* π;
      Area of HeHexagon =6X (1/2)*2*√3=6*√3;
      Three ouside sectors are equivalent to two circle of radiuus 1; To be added in area of hexagon
      AND Three inside sectors are equivalent to one circle of radius 1; to be substracted in the area of hexagon;
      Area of Figure provided= 6*√3+π* (1)^2=6*√3+1

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com