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

## Area of the star and circle – AMC- 8, 2008 – Problem 23

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

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

### Key Concepts

Geometry

Triangle

Square

But try the problem first…

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

Source

AMC-8 (2008) Problem 23

Pre College Mathematics

## Try with Hints

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=$\frac{10}{36}=\frac{5}{18}$