 Try this beautiful Problem on Geometry based on the side of a quadrilateral from AMC 10 A, 2009. You may use sequential hints to solve the problem.

## Side of a Quadrilateral  – AMC-10A, 2009- Problem 12

In quadrilateral $A B C D, A B=5, B C=17, C D=5, D A=9,$ and $B D$ is an integer. What is $B D ?$

,

• $11$
• $12$
• $13$
• $14$
• $15$

### Key Concepts

Geometry

Triangle Inequility

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2009 Problem-12

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

$13$

## Try with Hints

#### First Hint

We have to find out $BD$ . Given that $A B=5, B C=17, C D=5, D A=9,$. Notice that $BD$ divides quadrilateral $A B C D$ into $\triangle ABD$ and $\triangle BCD$. Can you find out using Triangle inequality……?

#### Second Hint

We know that sum of the two sides of a triangle is greater than the Third side and difference between two sides is lesser than the third side

By the triangle inequality we have $BD \text { < }DA+AB=9+5=14$ and $BD \text { > }BC-CD=17-5=12$

#### Third Hint

Therefore We got that $12< B D< 14$, and as we know that $B D$ is an integer, we must have $B D$=$13$