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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 has and . Diagonals and intersect at , and and have equal areas. What is

,

Geometry

Similarity

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2009 Problem-23

## Try with Hints

#### First Hint

Given that Convex quadrilateral has and . Diagonals and intersect at , and and have equal areas. we have to find out the length of .

Now if we can show that and are similar then we can find out ?

Can you find out?

#### Second Hint

Given that area of and area of are equal. Now area of = area of +

Area of = area of +

Therefore area of = area of [as area of and area of are equal]

Since triangles and share a base, they also have the same height and thus and with a ratio of 3: 4

Can you finish the problem?

Therefore so

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

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

Convex quadrilateral has and . Diagonals and intersect at , and and have equal areas. What is

,

Geometry

Similarity

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2009 Problem-23

## Try with Hints

#### First Hint

Given that Convex quadrilateral has and . Diagonals and intersect at , and and have equal areas. we have to find out the length of .

Now if we can show that and are similar then we can find out ?

Can you find out?

#### Second Hint

Given that area of and area of are equal. Now area of = area of +

Area of = area of +

Therefore area of = area of [as area of and area of are equal]

Since triangles and share a base, they also have the same height and thus and with a ratio of 3: 4

Can you finish the problem?

Therefore so

## Subscribe to Cheenta at Youtube

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