# area of triangles 2

• May 26th 2010, 03:47 AM
Tessarina
area of triangles 2
a rhombus has an area of 50 cm^2 and an internal angle of size 63 degrees. Find the lenth of its sides.

• May 26th 2010, 05:32 AM
skeeter
Quote:

Originally Posted by Tessarina
a rhombus has an area of 50 cm^2 and an internal angle of size 63 degrees. Find the lenth of its sides.

... use the parallelogram area formula provided in your previous post.
• May 26th 2010, 02:12 PM
Tessarina
i worked out the other one, but with this one i just can't seem to get the right answer.

i think iam doing something wrong,
• May 26th 2010, 02:50 PM
skeeter
Quote:

Originally Posted by Tessarina
i worked out the other one, but with this one i just can't seem to get the right answer.

i think iam doing something wrong,

can't fix it if we don't know what you tried ...
• May 27th 2010, 12:30 AM
Tessarina
could you just show me what working out you would do?
• May 27th 2010, 02:30 AM
Prove It
Try using the formula

$A = ab\sin{\theta}$.

Like Skeeter said, show us what working you have done. We are not here to do your work for you.
• May 27th 2010, 02:57 AM
Tessarina
this is the working out i did:

A= 50cm^2

1/2 x a x c x sin63= 50

acsin63= 100

-and thats all i got to.
• May 27th 2010, 03:01 AM
Prove It
Quote:

Originally Posted by Tessarina
this is the working out i did:

A= 50cm^2

1/2 x a x c x sin63= 50

acsin63= 100

-and thats all i got to.

You're almost there. Just remember that in a rhombus, all sides are equal.
• May 27th 2010, 03:09 AM
Tessarina
then do i work it out by going--
2asin63=100
2a=100/sin63
2a=112.23
a=56.11

but then if i double that to get the area of the whole rhombus i get
112.23

what am i doing wrong?
• May 27th 2010, 03:13 AM
Prove It
Quote:

Originally Posted by Tessarina
then do i work it out by going--
2asin63=100
2a=100/sin63
2a=112.23
a=56.11

but then if i double that to get the area of the whole rhombus i get
112.23

what am i doing wrong?

It should actually be

$a^2\sin{\theta} = 100$.

BTW you only halve when you are finding the area of a triangle. Since it's half of a parallelogram.
• May 27th 2010, 03:18 AM
Tessarina
then it would be-

a^2= 100/sin63
a= 10.59

so if that is the length of one side, how do i then work out he are of the entire rhombus?
• May 27th 2010, 03:20 AM
Prove It
Quote:

Originally Posted by Tessarina
then it would be-

a^2= 100/sin63
a= 10.59

so if that is the length of one side, how do i then work out he are of the entire rhombus?

Oops, it should have actually been

$a^2\sin{63^{\circ}} = 50$.

You already know the area of the rhombus, it's $50\,\textrm{cm}^2$. You're trying to find the side length, i.e. $a$.
• May 27th 2010, 03:25 AM
Tessarina
I got it!
Thank you so much.

i realised then that i was looking at the wrong answer in the back of the book, thats why it wasnt making sense.

Thank you :)