Categories
AMC-8 Geometry Math Olympiad USA Math Olympiad

Area of square and circle | AMC 8, 2011|Problem 25

Try this beautiful problem from AMC 8, 2011 based on the ratio of the area of square and circle.You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Ratio of the area of square and circle.

Area of the star and circle – AMC-8, 2011 – Problem 25


A circle with radius 1 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle’s shaded area to the area between the two squares?

area of square and circle
  • $\frac{3}{2}$
  • $\frac{1}{2}$
  • $1$

Key Concepts


Geometry

Circle

Square

Check the Answer


Answer:$\frac{1}{2}$

AMC-8 (2011) Problem 25

Pre College Mathematics

Try with Hints


Join the diagonals of the smaller square (i.e GEHF)

figure simplified 1

Can you now finish the problem ……….

The circle’s shaded area is the area of the smaller square(i.e. GEHF) subtracted from the area of the circle

and The area between the two squares is Area of the square ABCD – Area of the square EFGH

can you finish the problem……..

figure solution 2

Given that the Radius of the circle with centre O is 1.Therefore The area of the circle is \(\pi (1)^2\)=\(\pi\) sq.unit

The diameter of the circle is 2 i.e \(EF=BC=2\) unit

The area of the big square i.e \(ABCD=2^2=4\) sq.unit

\(OE=OH=1\) i.e \(EH=\sqrt{(1^2+1^2)}=\sqrt 2\)

Therefore the area of the smaller square is \((\sqrt 2)^2=2\)

The circle’s shaded area is the area of the smaller square(i.e. GEHF) subtracted from the area of the circle =\(\pi\) – 2

The area between the two squares is Area of the square ABCD – Area of the square EFGH=4-2=2 sq.unit

The ratio of the circle’s shaded area to the area between the two squares is \( \frac{\pi – 2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \frac{1}{2}\)

Subscribe to Cheenta at Youtube


Leave a Reply

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