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

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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

quadrilateral

Triangle Inequility

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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 $B D$=$13$

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