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Try this beautiful problem from AMC-8, 2008 based on the Ratio of the area of the triangle and square.

In square ABCE, AF=2FE and CD=2DE .what is the ratio of the area of \(\triangle BFD\) to the area of square ABCE?

- $\frac{7}{20}$
- $\frac{5}{18}$
- $\frac{11}{20}$

Geometry

Triangle

Square

But try the problem first...

Answer:$\frac{5}{18}$

Source

Suggested Reading

AMC-8 (2008) Problem 23

Pre College Mathematics

First hint

Area of the square =\((side)^2\)

Area of triangle =\(\frac{1}{2} \times base \times height\)

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

Second Hint

The area of the \(\triangle BFD\)=(The area of the square ABCE- The area of \(\triangle ABF\) -The area of\( \triangle BCD\) -Area of the\( \triangle EFD) \)

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

Final Step

Let us assume that the side length of the given square is 6 unit

Then clearly AB=BC=CE=EA=6 unit & AF=4 unit,EF=2 unit, CD=4 unit

Total area of the square is \(6^2\)=36 sq.unit

Area of the \(\triangle ABF=\frac{1}{2}\times AB \times AF= \frac{1}{2}\times 6 \times 4= 12\) sq.unit

Area of the \(\triangle BCD=\frac{1}{2}\times BC \times CD= \frac{1}{2}\times 6 \times 4= 12\) sq.unit

Area of the \(\triangle EFD=\frac{1}{2}\times EF \times ED= \frac{1}{2}\times 2 \times 2= 2\) sq.unit

The area of the \(\triangle BFD\)=(The area of the square ABCE- The area of \(\triangle ABF\) -The area of\( \triangle BCD\) -Area of the\( \triangle EFD)=(36-12-12-2)=10 \)sq.unit

the ratio of the area of \(\triangle BFD\) to the area of square ABCE=\(\frac{10}{36}=\frac{5}{18}\)

- https://www.cheenta.com/radius-of-a-semi-circle-amc-82017-problem-22/
- https://www.youtube.com/watch?v=e2oyUpYCeBI

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