# Rhombus Properties - Problem 2

###### Transcript

The special properties of diagonals of a rhombus are going to help us find angles 1 and 2 and the value of x. notice in this diagram that 1 and 2 don’t have the little degree sign next to it. So that tells you that it’s not the measure of an angle, that’s just a name. So let’s start off by finding angles 1 and 2.

Well, how can I figure out what angle 1 is? If I look at my diagram here, I see that in a rhombus, the diagonals are perpendicular bisectors of each other, which means angle 1 has to be a right angle. So I’m going to erase the 1, I’m going to label it a right angle and I’m going to say angle 1 is 90 degrees, since I know these diagonals form a right angle.

Next to find 2 I know that 30 degrees and 2 must be complementary. Which means 2 plus 30 is 90 so 2 must be 60 degrees. So we’re almost done with the problem, now we need to find x.

Well, I see that x is in the labels of those two segments. Since we have a bisected diagonal, we can say that 2x minus 3 must equal 4x minus 11. So let’s solve that equation. 2x minus 3 equals 4x minus 11. So this is just an algebraic problem. We’re going to subtract 2x on both sides, keeping our negative, we’re going to get 2x minus 11 and we’re going to add 11 to both sides here. Excuse me, 11 minus 3 is 8 and if we divide both sides by 2, we see that x is 4.

Now you might want to be careful because sometimes your teacher might ask you what is the length of the entire diagonal, or what is the length of one of those bisected segments? In which case you’d have to substitute back in and add them up but in this problem all you’re asked to find is the value of x.

And the keys were, diagonals of a rhombus are perpendicular and they bisect each other into two congruent segments.