# Need Help finding perimeter of square and rhombus

• Jul 26th 2009, 04:54 PM
sunshii17
Need Help finding perimeter of square and rhombus
1. In square SQUA, the lengths of diagonal QA is 24 cm. In rhombus RHOM the length of diagonal MH is 24 cm and the measure of angle RHO = 60. Which quadrilateral has the greater perimeter? Show all the work to support your conclusion.

I'm taking an online geometry class right now and I having a really hard time understanding it. If someone could please explain to me how to solve this problem, that would be great!

Thanks
• Jul 26th 2009, 05:39 PM
sa-ri-ga-ma
Quote:

Originally Posted by sunshii17
1. In square SQUA, the lengths of diagonal QA is 24 cm. In rhombus RHOM the length of diagonal MH is 24 cm and the measure of angle RHO = 60. Which quadrilateral has the greater perimeter? Show all the work to support your conclusion.

If x is the side pf the square and d is the diagonal, using Pythagoras theorem, you can write 2*x^2 = d^2. D is given, Find x and perimeter which is 4x.

In the rhombus let o be the point of intersection of diagonals. In triangle RHO angle RHO = 30 degrees. If x is the side of the rhombus, OH = x*cos30. OH is given. Find x and perimeter which is again 4x.
• Jul 27th 2009, 02:52 AM
Perimeter of square and rhombus
Hello sunshii17

Welcome to Math Help Forum!
Quote:

Originally Posted by sunshii17
1. In square SQUA, the lengths of diagonal QA is 24 cm. In rhombus RHOM the length of diagonal MH is 24 cm and the measure of angle RHO = 60. Which quadrilateral has the greater perimeter? Show all the work to support your conclusion.

I'm taking an online geometry class right now and I having a really hard time understanding it. If someone could please explain to me how to solve this problem, that would be great!

Thanks

You don't need to do detailed calculations; all you need to know is a little about cosines.

Draw the two diagrams carefully, joining the diagonals in each case. Suppose the diagonals in each diagram meet at the point $\displaystyle X$.

Then (since the diagonals of a rhombus meet at right angles), all the angles in each diagram at $\displaystyle X$ are right angles, and so each figure - the square and the rhombus - consists of four identical right-angled triangles. In the case of the square, the other angles in one of the triangles (for instance, $\displaystyle \triangle SQX$) are each $\displaystyle 45^o$. In the case of the rhombus the other angles in one of the triangles (for instance, $\displaystyle \triangle RHX$) are $\displaystyle \angle H=30^o$ and $\displaystyle \angle R = 60^o$.

So for the rhombus, we have:

$\displaystyle \cos 30^o = \frac{XH}{RH} \Rightarrow RH = \frac{12}{\cos 30^o}$

And for the square:

$\displaystyle \cos 45^o = \frac{XQ}{SQ}\Rightarrow SQ =\frac{12}{\cos 45^o}$

Now $\displaystyle \cos 30^o > \cos 45^o$ and so $\displaystyle RH < SQ$. This is because, when you calculate $\displaystyle RH$, you divide the $\displaystyle 12$ by a larger number than when you calculate $\displaystyle SQ$.

So the perimeter of the rhombus ($\displaystyle = 4\,RH$) is less than the perimeter of the square ($\displaystyle = 4\,SQ$).