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?
Geometry
Triangle
Square
Answer:$\frac{5}{18}$
AMC-8 (2008) Problem 23
Pre College Mathematics
Area of the square =\((side)^2\)
Area of triangle =\(\frac{1}{2} \times base \times height\)
Can you now finish the problem ..........
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........
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=\(\frac{10}{36}=\frac{5}{18}\)
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