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.
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 $B D ?$
,
Geometry
quadrilateral
Triangle Inequility
Pre College Mathematics
AMC-10A, 2009 Problem-12
$13$
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......?
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 \(BD \text { > }BC-CD=17-5=12\)
Therefore We got that $12< B D< 14$, and as we know that $B D$ is an integer, we must have $B D$=\(13\)
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.
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 $B D ?$
,
Geometry
quadrilateral
Triangle Inequility
Pre College Mathematics
AMC-10A, 2009 Problem-12
$13$
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......?
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 \(BD \text { > }BC-CD=17-5=12\)
Therefore We got that $12< B D< 14$, and as we know that $B D$ is an integer, we must have $B D$=\(13\)