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# Trapezium | AMC 10A ,2009 | Problem No 23

Try this beautiful Problem on Geometry from Trapezium  from (AMC 10 A, 2009).

## Trapezium - AMC-10A, 2009- Problem 23

Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. What is $A E ?$

,

• $11$
• $12$
• $13$
• $14$
• $6$

Geometry

Similarity

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2009 Problem-23

#### Check the answer here, but try the problem first

$6$

## Try with Hints

#### First Hint

Given that Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. we have to find out the length of $AE$.

Now if we can show that $\triangle AEB$ and $\triangle DEC$ are similar then we can find out $AE$?

Can you find out?

#### Second Hint

Given that area of $\triangle AED$ and area of $\triangle BEC$ are equal. Now area of $\triangle ABD$ = area of $\triangle AED$ + $\triangle ABE$

Area of $\triangle ABC$ = area of $\triangle AEB$ + $\triangle BEC$

Therefore area of $\triangle ABD$= area of $\triangle ABC$ [as area of $\triangle AED$ and area of $\triangle BEC$ are equal]

Since triangles $A B D$ and $A B C$ share a base, they also have the same height and thus $\overline{A B} || \overline{C D}$ and $\triangle A E B \sim \triangle C E D$ with a ratio of 3: 4

Can you finish the problem?

#### Third Hint

Therefore $A E=\frac{3}{7} \times A C,$ so $A E=\frac{3}{7} \times 14=6$

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