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AMC 10 Geometry Math Olympiad USA Math Olympiad

Trapezium | AMC 10A ,2009 | Problem No 23

Try this beautiful Problem on Geometry: quadrilateral from AMC 10A, 2009. Problem-12. You may use sequential hints to solve the problem.

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 $A E ?$

Area of trapezium

,

  • $11$
  • $12$
  • $13$
  • $14$
  • $6$

Key Concepts


Geometry

quadrilateral

Similarity

Suggested Book | Source | Answer


Suggested Reading

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

Trapezium

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

Trapezium

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

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