# Math Help - Rhombus Problem

1. ## 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.

2. ## Re: Rhombus Problem

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.

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

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

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

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

plugging into the 3rd equation we get

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

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

3. ## Re: Rhombus Problem

Originally Posted by romsek

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

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

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

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

plugging into the 3rd equation we get

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

$\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$
Holy cow! That's some math work! Thank you, is there any shorter way than this?

4. ## Re: Rhombus Problem

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

5. ## Re: Rhombus Problem

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?

6. ## Re: Rhombus Problem

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

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

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

7. ## Re: Rhombus Problem

Hello, Cake!

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

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: . $AO = OC = 6.$

Code:
             A       10        B
o---------------o
/ *           * /
/   *6      *   /
/     *   *     /
10 /       o       /
/     *  O*     /
/   *       *6  /
/ *           * /
o---------------o
D                 C
In right triangle $AOB$, we find that $OB = 8.$

Therefore: . $BD = 16.$

8. ## Re: Rhombus Problem

Originally Posted by Soroban
Hello, Cake!

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

Hence: . $AO = OC = 6.$

Code:
             A       10        B
o---------------o
/ *           * /
/   *6      *   /
/     *   *     /
10 /       o       /
/     *  O*     /
/   *       *6  /
/ *           * /
o---------------o
D                 C
In right triangle $AOB$, we find that $OB = 8.$

Therefore: . $BD = 16.$

Makes sense ^_____^
Thank you guys!