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AMC-8 Geometry Math Olympiad USA Math Olympiad

Ratio of the area of Square and Pentagon | AMC 8, 2013

Try this beautiful problem from Geometry: Ratio of the area between Square and Pentagon. . You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: Ratio of the area between Square and Pentagon.

Ratio of the area between Square and Pentagon – AMC-8, 2013 – Problem 24


Squares ABCD ,EFGH and GHIJ are equal in area .Points C and D are the mid points of the side IH and HE ,respectively.what is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?

ratio of area of square and pentagon

  • $\frac{1}{4}$
  • $\frac{1}{3}$
  • $\frac{3}{8}$

Key Concepts


Geometry

Area of square

Area of Triangle

Check the Answer


Answer:$\frac{1}{3}$

AMC-8(2013) Problem 24

Pre College Mathematics

Try with Hints


extend  IJ until it hits the extension of  AB .

Can you now finish the problem ……….

find the area of the pentagon

can you finish the problem……..

solution figure

First let L=2 (where L is the side length of the squares) for simplicity. We can extend  IJ until it hits the extension of  AB . Call this point  X.

Then clearly length of AX=3 unit & length of XJ = 4 unit .

Therefore area of \(\triangle AXJ= (\frac{1}{2} \times AX \times XJ)=(\frac{1}{2} \times 4 \times 3)=6\) sq.unit

And area of Rectangle BXIC= \(( 1 \times 2)\)=2 sq.unit

Therefore the of the pentagon ABCIJ=6-2=4 sq.unit

The combined area of three given squares be \( (3 \times 2^2)\)=12 sq.unit

Now, the ratio of the shaded area(pentagon) to the combined area of the three squares is \(\frac{4}{12}=\frac{1}{3}\)

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