Question

ABCD is a quadrilateral where AB is parallel to DC. If x = (4y)/3 and y = (3z)/8 find x, y, and z.

Answer 1

\[\angle A+\angle D=180^\circ\\\angle B + \angle C=180^\circ\]

Try to solve it now :)

Answer 2

I tried but I'm getting z as more than 90 degrees.

Answer 3

z is an acute angle isn't it?

Answer 4

z + y = 180
and
\[y = \frac{ 3z }{ 8 }\]
\[z + \frac{ 3z }{ 8 } = 180\]
\[\frac{ 8z + 3z }{ 8 } = 180\]
\[\frac{ 11z }{ 8 } = 180\]
\[z = \frac{ 180\times8 }{ 11 }\]

Answer 5

I don't think that's right.

Answer 6

Go ahead, you are right. Try to put z instead of y in your graph ;)

Answer 7

What do you mean 'in your graph'?

Answer 8

if y = 3/8 of z, it will be smaller than z. Are you sure you have y and z labelled correctly?

Answer 9

Yes. You think it could be a printing mistake?

Answer 10

No clue, but if z is acute, and y is 3/8 of z, then there's no way y can be obtuse.

Answer 11

Yeah, I never thought of that.

Answer 12

Sorry, graphic*. I mean that you have the wrong graphic or wrong data.

Answer 13

The answers given are 96, 96 and 84.

Answer 14

Try it the other way around. z = 3y/8

Answer 15

Then you get 11y/8 = 180

Answer 16

\(y = \frac{180 \times 8}{11}\)

Answer 17

That gives an angle of 131, meaning z would be 49. Doesn't work.

Answer 18

Well, it works, but not to the answers you're given. :)

Answer 19

yes.

Answer 20

I already tried all that.

Answer 21

Well, thanks for your help anyway. I'll ask my maths teacher.

Answer 22

No problem. :)