I'm trying to solve a few "difficult" geometry questions and I was wondering if anyone reading this thread could verify that my solutions are correct.

EAC and DBC both are 30-60-90 triangles.

EC-ED=DC

$\displaystyle 26-13=13$

DC=13

Since that's the hypotenuse of a 30 60 90 triangle...

$\displaystyle 2x=13$

$\displaystyle 13/2=6.5$

$\displaystyle x=6.5$

Now to find BD...

$\displaystyle 1x=6.5$

$\displaystyle x=6.5$

So BD is 6.5 centimetres

Knowing the perimeter is 60 cm I know that each side of the rhombus is 15 cm. Connecting opposite vertices with two diagonals shows that I have 4 right angle triangles.

Since one diagonal is twice as long as the other I can use arctan to find one of the angles in one of the four triangles.

$\displaystyle \arctan\frac{1}{2}=26.57°$

Doubling that gives me 53.14°. So now I have an angle for one vertex.

Since two opposite angles equal to 180°

$\displaystyle 180-53.14=126.86$

The other two angles will be the same as what I got above since a rhombus is an equilateral quadrilateral.

You can find the area of a rhombus using the formula

$\displaystyle area=s^2\sin a$

Where s is the length of a side and a is any interior angle.

$\displaystyle area=15^2\sin53.14$

So the area works out to be 180.02 cm².

To find angle DAE I use trig.

I'll call the center of the rightmost circle T. DTA is a right angle triangle. We know DT is 10 cm and AT is 50 cm.

$\displaystyle \arctan\frac{10}{50}=11.31$

So angle DAE is 11.31°.

I want to find ABO so I'm going to use sine law.

$\displaystyle \frac{\sin11.31}{10}=\frac{\sin ABO}{30}$

ABO works out to be 143.96°. Actually the equation gives you 36.04° because of the "ambiguous case". 36.04° is angle OBC.

If you draw a line from O to B and O to C you get an isosceles triangle. Ergo OBC is equal to OCB.

To find BOC...

$\displaystyle 180-36.04-36.04=107.92$

We're trying to find the length of BC and to do that we can once again use sine law.

We already know BO and CO are both 10 cm and we know all the angles.

$\displaystyle \frac{\sin107.92}{BC}=\frac{sin36.04}{10}$

That works out to be 16.17 centimetres.

The difficult part of this question is finding the length of the metal band that isn't in contact with the two poles.

I just realized I made an error in solving this question so I'm winging it.

ABC and ADO are both 30 60 90 triangles.

AB is $\displaystyle 15\sqrt{3}$

AD is $\displaystyle 5\sqrt{3}$

So to find DB just subract AD from AB.

$\displaystyle 15\sqrt{3}-5\sqrt{3}=10\sqrt{3}$

There are two equal lengths of metal band that are not in contact with the two circles so I double that answer to $\displaystyle 20\sqrt{3}$

Now it is quite easy to find the rest just use $\displaystyle 2\pi r$ for both circles. Since I just need half the circumference for both circles I can drop the 2 from that equation.

$\displaystyle \pi5=15.71$

$\displaystyle \pi15=47.12$

So to find the total length of the metal band just add all those up. $\displaystyle 15.71+47.12+20\sqrt{3}=97.47$

So the length of the metal band is 97.47 centimetres.

The total diameter of the large circle is 12. Let the second largest be $\displaystyle 12-x$ and the smallest $\displaystyle x$.

Bear with me, this is a cumbersome equation.

$\displaystyle \frac{4}{9}\pi(\frac{12}{2})^2=\pi(\frac{12}{2})^2-\pi(\frac{12-x}{2})^2-\pi(\frac{x}{2})^2$

Simplifying that I got $\displaystyle -\frac{1}{2}x^2+6x-16=0$

That's a quadratic equation!

Using the quadratic formula I find that x equals 4 and 8.

so the diameter of the smallest circle is 4 centimetres and the second smallest circles is 8 centimetres.

This is a very poor excuse of a diagram but I think it'll suffice.

OF=H

OE=15-H

CD=X

AB=2x

$\displaystyle OD^2=OF^2+DF^2$

$\displaystyle 100=h^2+\frac{x^2}{4}$

$\displaystyle 400=4h^2+x^2$

$\displaystyle x^2=400-4h^2$

$\displaystyle OB^2=OE^2+EB^2$

$\displaystyle 100=(15-h)^2+x^2$

Substitue $\displaystyle x^2$ with $\displaystyle 400-4h^2$ and expand $\displaystyle 15-h^2$

$\displaystyle 100=225+h^2-30h+400-4h^2$

$\displaystyle 100=625-3h^2-30$

$\displaystyle 3h^2+30h-525=0$

$\displaystyle h^2+10h-175=0$

Woohoo! Anoyther quadratic equation!

Plug that into the quadractic formula leaves me with

$\displaystyle x=8.1$ and a negative value that I don't need.

So CD is 8.1 centimetres.

Since AB is double CD AB is 16.2 centimetres.

Similar question to the one that I just answered.

I know that the diameter of the biggest circle is 10+YZ. Therefore the area is $\displaystyle \pi(\frac{10+YZ}{2})^2$

The area of the unshaded region is $\displaystyle \pi(\frac{YZ}{2})^2$

Since the unshaded and shaded regions have the same area I can set up an equation to solve for YZ.

$\displaystyle \pi(\frac{YZ}{2})^2+\pi(\frac{YZ}{2})^2=\pi(\frac{ 10+YZ}{2})^2$

Simplifying that I got $\displaystyle YZ^2-20x-100=0$

Another quadratic equation.

Using the quadratic formula I get $\displaystyle YZ=24.14, -4.14$

I don't need the negative.

So YZ=24.14 centimetres.

I've reread this post several times and I've found a few errors however it's more than likely some have slipped through. There's also the possibility that I have actually made mistakes when solving this on paper (For the most part I just typed out my work). Once again these are supposed to be "hard" geometry questions and I did have some difficulty solving them so I'd be great if anyone reading the thread could go over any of the questions and see if I've erred.

edit: Imageshack seems to be a bit iffy so I've included a zip file of all the images in case anyone has troubles viewing the images.