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

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?

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

Geometry

Area of square

Area of Triangle

But try the problem first...

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

Source

Suggested Reading

AMC-8(2013) Problem 24

Pre College Mathematics

First hint

extend IJ until it hits the extension of AB .

Can you now finish the problem ..........

Second Hint

find the area of the pentagon

can you finish the problem........

Final Step

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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