# Difficult geometry questions

• Apr 28th 2009, 09:46 PM
iamanoobatmath
Difficult geometry questions
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
$26-13=13$
DC=13
Since that's the hypotenuse of a 30 60 90 triangle...
$2x=13$
$13/2=6.5$
$x=6.5$
Now to find BD...
$1x=6.5$
$x=6.5$
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.

$\arctan\frac{1}{2}=26.57°$
Doubling that gives me 53.14°. So now I have an angle for one vertex.
Since two opposite angles equal to 180°
$180-53.14=126.86$
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
$area=s^2\sin a$
Where s is the length of a side and a is any interior angle.
$area=15^2\sin53.14$
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.
$\arctan\frac{10}{50}=11.31$
So angle DAE is 11.31°.
I want to find ABO so I'm going to use sine law.
$\frac{\sin11.31}{10}=\frac{\sin ABO}{30}$
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...
$180-36.04-36.04=107.92$
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.
$\frac{\sin107.92}{BC}=\frac{sin36.04}{10}$
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.
http://img514.imageshack.us/img514/6156/asdasdasdp.jpg
ABC and ADO are both 30 60 90 triangles.
AB is $15\sqrt{3}$
AD is $5\sqrt{3}$
So to find DB just subract AD from AB.
$15\sqrt{3}-5\sqrt{3}=10\sqrt{3}$
There are two equal lengths of metal band that are not in contact with the two circles so I double that answer to $20\sqrt{3}$
Now it is quite easy to find the rest just use $2\pi r$ for both circles. Since I just need half the circumference for both circles I can drop the 2 from that equation.
$\pi5=15.71$
$\pi15=47.12$
So to find the total length of the metal band just add all those up. $15.71+47.12+20\sqrt{3}=97.47$
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 $12-x$ and the smallest $x$.
Bear with me, this is a cumbersome equation.
$\frac{4}{9}\pi(\frac{12}{2})^2=\pi(\frac{12}{2})^2-\pi(\frac{12-x}{2})^2-\pi(\frac{x}{2})^2$
Simplifying that I got $-\frac{1}{2}x^2+6x-16=0$
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.
http://bayimg.com/image/bapjfaabf.jpg
OF=H
OE=15-H
CD=X
AB=2x
$OD^2=OF^2+DF^2$
$100=h^2+\frac{x^2}{4}$
$400=4h^2+x^2$
$x^2=400-4h^2$

$OB^2=OE^2+EB^2$
$100=(15-h)^2+x^2$
Substitue $x^2$ with $400-4h^2$ and expand $15-h^2$
$100=225+h^2-30h+400-4h^2$
$100=625-3h^2-30$
$3h^2+30h-525=0$
$h^2+10h-175=0$

Plug that into the quadractic formula leaves me with
$x=8.1$ 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 $\pi(\frac{10+YZ}{2})^2$
The area of the unshaded region is $\pi(\frac{YZ}{2})^2$
Since the unshaded and shaded regions have the same area I can set up an equation to solve for YZ.
$\pi(\frac{YZ}{2})^2+\pi(\frac{YZ}{2})^2=\pi(\frac{ 10+YZ}{2})^2$
Simplifying that I got $YZ^2-20x-100=0$
Using the quadratic formula I get $YZ=24.14, -4.14$
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.
• Apr 29th 2009, 04:49 AM
earboth
Quote:

Originally Posted by iamanoobatmath
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
$26-13=13$
DC=13
Since that's the hypotenuse of a 30 60 90 triangle...
$2x=13$
$13/2=6.5$
$x=6.5$
Now to find BD...
$1x=6.5$
$x=6.5$
So BD is 6.5 centimetres

...

It would be much better if you could start a new thread for each question.

I've modified your drawing a little bit. (see attachment)

Let x denote the radius of the left circle and y the radius of the right circle. Then you know:

$\left|\begin{array}{l}x+y=13 \\\dfrac xy = \dfrac21\end{array}\right.$ $~\implies~ x=\dfrac{26}3~\wedge~y=\dfrac{13}3$

Since $y = |\overline{BD}| = \dfrac{13}3$
• Apr 29th 2009, 05:02 AM
earboth
Quote:

Originally Posted by iamanoobatmath
...

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.

$\arctan\frac{1}{2}=26.57°$
Doubling that gives me 53.14°. So now I have an angle for one vertex.
Since two opposite angles equal to 180°
$180-53.14=126.86$
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
$area=s^2\sin a$
Where s is the length of a side and a is any interior angle.
$area=15^2\sin53.14$
So the area works out to be 180.02 cm².

...

Let e denote the shorter diagonal and f the longer one.
Then you know $f = 2\cdot e$
You have 4 congruent right triangles with the legs $\frac12 f$ and $\frac12 e$ and the hypotenuse of 15 cm:

$\left(\frac12 f \right)^2 + \left(\frac12 e \right)^2 = 15^2$
Substitute f = 2e:
$\left(e \right)^2 + \left(\frac12 e \right)^2 = 15^2$

$\frac54 e^2 = 225~\implies~e^2=180~\implies~e=6\sqrt{5}$

Consequently $f = 12 \sqrt{5}$

The area of a rhombus is calculated by: $a=\dfrac12 \cdot e \cdot f$

With your values you get: $a = \dfrac12 \cdot 6\sqrt{5} \cdot 12 \sqrt{5} = 180$