# Triangle and Quadrilateral | AMC-10A, 2005 | Problem 25

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral from AMC 10A, 2005.

## Ratios of the areas of Triangle and Quadrilateral - AMC-10A, 2005- Problem 25

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

- \(\frac{19}{56}\)
- \(\frac{19}{66}\)
- \(\frac{17}{56}\)
- \(\frac{11}{56}\)
- \(\frac{19}{37}\)

**Key Concepts**

Geometry

Triangle

quadrilateral

## Check the Answer

But try the problem first...

Answer: \(\frac{19}{56}\)

AMC-10A (2005) Problem 25

Pre College Mathematics

## Try with Hints

First hint

Given that $AB = 25$, $BC = 39$, and $AC=42$.we have to find out Ratios of the areas of Triangle\(\triangle ADE\) and the quadrilateral \(CBED\).So if we can find out the area the \(\triangle ADE\) and area of the \(\triangle ABC\) ,and subtract \(\triangle ADE\) from \(\triangle ABC\) then we will get area of the region \(CBDE\).Can you find out the area of \(CBDE\)?

Can you find out the required area.....?

Second Hint

Now \(\frac{\triangle ADE}{\triangle ABC}=\frac{AD}{AB}.\frac{AE}{AC}=\frac{19}{25}.\frac{14}{42}=\frac{19}{75}\)

Therefore area of \(BCED\)=area of \(\triangle ABC\)-area of \(\triangle ADE\).Now can you find out Ratios of the areas of Triangle and the quadrilateral?

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

Final Step

\(\frac{[A D E]}{[B C E D]}=\frac{[A D E]}{[A B C]-[A D E]}\)

=\(\frac{1}{[A B C] /[A D E]-1}\)

=\(\frac{1}{75 / 19-1}\)

=\(\frac{19}{56}\)

## Other useful links

- https://www.cheenta.com/circumscribed-circle-amc-10a-2003-problem-17/
- https://www.youtube.com/watch?v=fRj9NuPGrLU&t=282s