1. ## Parallelograms & Quadrilaterals

Hey Guys!

For NUMBER 9)

I'm trying to find the answer but I don't know how to set up this equation.

Since Angle A = Angle C

I am setting up the equation as 2x/3 = x/2 + 20

I take out the fractions with 6 and I get 4x = 3x + 20
then I get x = 20

Doesn't seem to work though. The answers are m∠A = m∠C = 80°;m∠B = m∠D = 100°

2. ## Re: Parallelograms & Quadrilaterals

Hey Cake.

Are you assuming your shape is a parallelogram (or a rhombus)? If so then you can use the fact that angle A = angle C and angle B = angle D where total sum of angles is 360.

So basically you have 2*angle(A) + 2*angle(B) = 360 where

2x/3 = x/2 + 20
4x = 3x + 120 (multiply everything by 6)
x = 120

This implies angle(A) = angle(C) = 2*120/3 = 80.

Since 2*angle(A) + 2*angle(B) = 360 you have angle(B) = (360 - 2*80)/2 = angle(D) = 100.

3. ## Re: Parallelograms & Quadrilaterals

I see now.

so angle B is basically (360 - 2(angle A or C, which is 80)) then you divide by 2?

Sorry, why do you divide by two? Is it because there are two angles (angle B and D)?

4. ## Re: Parallelograms & Quadrilaterals

Basically because the angle is equal to its opposite. In other words B = D and A = C (in terms of angles) of the interior of the rhombus.

I divide by two because you have a factor of 2 in front of the variables. I have 2x = blah so I need to use 2x/2 = blah/2 = x to solve for x.

5. ## Re: Parallelograms & Quadrilaterals

Originally Posted by Cake
I am setting up the equation as 2x/3 = x/2 + 20

I take out the fractions with 6 and I get 4x = 3x + 20
then I get x = 20
Should be: 4x = 3x + 120 ; ok?

6. ## Re: Parallelograms & Quadrilaterals

Originally Posted by chiro
Basically because the angle is equal to its opposite. In other words B = D and A = C (in terms of angles) of the interior of the rhombus.

I divide by two because you have a factor of 2 in front of the variables. I have 2x = blah so I need to use 2x/2 = blah/2 = x to solve for x.
Perfect, thanks.