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

Try this beautiful problem from Geometry: Ratios of the areas of Triangle and Quadrilateral from AMC-10A. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral

## 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 $BCED$?

• $\frac{19}{56}$
• $\frac{19}{66}$
• $\frac{17}{56}$
• $\frac{11}{56}$
• $\frac{19}{37}$

Geometry

Triangle

quadrilateral

## Check the Answer

Answer: $\frac{19}{56}$

AMC-10A (2005) Problem 25

Pre College Mathematics

## Try with Hints

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…..?

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……..

Now $\frac{\triangle ADE}{quad.BCED}$=$\frac{\triangle ADE}{{\triangle ABC}-{\triangle ADE}}$=$\frac{1}{\frac{\triangle ABC}{\triangle ADE}-1}=\frac{1}{\frac{75}{19}-1}=\frac{19}{56}$