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Side of a Quadrilateral | AMC 10A, 2009 | Problem No 12

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



Triangle Inequility

Suggested Book | Source | Answer

Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2009 Problem-12

Check the answer here, but try the problem first


Try with Hints

First Hint

finding the side of a quadrilateral

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

side of a quadrilateral

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

finding the length of BD

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

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