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?
Geometry
Area of square
Area of Triangle
Answer:$\frac{1}{3}$
AMC-8(2013) Problem 24
Pre College Mathematics
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........
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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