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

Try this problem from Geometry: Ratios of the areas of Triangle and Quadrilateral from AMC-10A, 2005 You may use sequential hints to solve the problem.

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

But try the problem first…

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

Source

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

$$\frac{[A D E]}{[B C E D]}=\frac{[A D E]}{[A B C]-[A D E]}$$
=$$\frac{1}{[A B C] /[A D E]-1}$$
=$$\frac{1}{75 / 19-1}$$

=$$\frac{19}{56}$$