Thread: Rhombus Problem

Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD

I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.

Originally Posted by Cake

Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD

I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.

$\displaystyle x^2+y^2=10^2=100$

$\displaystyle (10-x)^2+y^2=12^2=144$

$\displaystyle (10+x)^2+y^2=BD^2$

solving the first two equations we get $\displaystyle x=\frac{14}{5}$ and $y=\frac{48}{5}$

plugging into the 3rd equation we get

$\displaystyle \left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$

$\displaystyle \left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$

Originally Posted by romsek

$\displaystyle x^2+y^2=10^2=100$

$\displaystyle (10-x)^2+y^2=12^2=144$

$\displaystyle (10+x)^2+y^2=BD^2$

solving the first two equations we get $\displaystyle x=\frac{14}{5}$ and $y=\frac{48}{5}$

plugging into the 3rd equation we get

$\displaystyle \left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$

$\displaystyle \left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$

Holy cow! That's some math work! Thank you, is there any shorter way than this?

Originally Posted by Cake

Holy cow! That's some math work! Thank you, is there any shorter way than this?

Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle

Originally Posted by bjhopper

Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle

Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?

Originally Posted by Cake

Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?

hrm I should have seen this.

No what you do is notice that

$\displaystyle \left(\frac{BD}{2}\right)^2+\left(\frac{12}{2} \right)^2=10^2$

$\displaystyle \left(\frac{BD}{2}\right)^2=100-36=64 \Rightarrow \frac{BD}{2}=8 \Rightarrow BD=16$

Hello, Cake!

$\displaystyle \text{Given rhombus }ABCD,\;AB = 10\text{ and }AC = 12.\;\text{ Find }AD\text{ and }BD.$

I know that AD = 10 because the sides of a rhombus are all congruent.
I cannot find what BD equals though.
I thought it was 12, but I don't think the diagonals of a rhombus are congruent.
If they were, you'd have a square.

The diagonals of a rhombus are perpendicular and bisect each other.

Hence: .$\displaystyle AO = OC = 6.$

Code:

             A       10        B
              o---------------o
             / *           * /
            /   *6      *   /
           /     *   *     /
       10 /       o       /
         /     *  O*     /
        /   *       *6  /
       / *           * /
      o---------------o
     D                 C

In right triangle $\displaystyle AOB$, we find that $\displaystyle OB = 8.$

Therefore: .$\displaystyle BD = 16.$

Originally Posted by Soroban

Hello, Cake!


The diagonals of a rhombus are perpendicular and bisect each other.

Hence: .$\displaystyle AO = OC = 6.$

Code:

             A       10        B
              o---------------o
             / *           * /
            /   *6      *   /
           /     *   *     /
       10 /       o       /
         /     *  O*     /
        /   *       *6  /
       / *           * /
      o---------------o
     D                 C

In right triangle $\displaystyle AOB$, we find that $\displaystyle OB = 8.$

Therefore: .$\displaystyle BD = 16.$

Makes sense ^_____^
Thank you guys!