# geometry

• Jun 2nd 2008, 05:41 PM
AlphaRock
geometry
These are the MOST hardest questions from my book. Nobody in my class, which is an ADVANCED math honors 11 class, can get them, which is why I'm curious if it's even possible or not.

If you can answer these questions, you're definitly smarter than my whole class combined.

I'd like to know why it's that anwer because I'm curious about how to do it.

I'm clueless as to where to start. Eek!

1. Given: QR = 12 cm
<PQR = 30 degrees
Find: Area of the shaded region

2. Given: A circle is inscribed in a sector of another circle.
<A = 60 degrees, AB = 10 sqrt(3) cm

3. A spherical tank is filled with water that is 20 ft. deep and 50 ft. wide.

What is the radius of the tank?

Note: All these are circles and a cone.
• Jun 2nd 2008, 05:52 PM
topsquark
Quote:

Originally Posted by AlphaRock
1. Given: QR = 12 cm
<PQR = 30 degrees
Find: Area of the shaded region

Look at the region in two parts, the triangle and the wedge. I presume you can get the area of the wedge. As to the triangle, you know this is an isosceles triangle with a vertex angle of 120 degrees.

-Dan
• Jun 2nd 2008, 06:19 PM
AlphaRock
Quote:

Originally Posted by topsquark
Look at the region in two parts, the triangle and the wedge. I presume you can get the area of the wedge. As to the triangle, you know this is an isosceles triangle with a vertex angle of 120 degrees.

-Dan

Unfortunately, I can't.

However, the closest I got to the answer was 36pie x 60% to get the area of the mid-P-R part of the circle. I can't get the area of the 120 - 30 - 30 one...
• Jun 2nd 2008, 06:42 PM
topsquark
Quote:

Originally Posted by AlphaRock
Unfortunately, I can't.

However, the closest I got to the answer was 36pie x 60% to get the area of the mid-P-R part of the circle. I can't get the area of the 120 - 30 - 30 one...

The central angle of the wedge is 60 degrees so it is 1/6 of the circle. The area of the circle is $36 \pi$, so the area of your wedge is $6 \pi$.

For the triangle, see this site.

-Dan
• Jun 2nd 2008, 06:43 PM
topsquark
Quote:

Originally Posted by AlphaRock
2. Given: A circle is inscribed in a sector of another circle.
<A = 60 degrees, AB = 10 sqrt(3) cm

-Dan
• Jun 2nd 2008, 06:45 PM
sean.1986
Quote:

Originally Posted by topsquark

-Dan

The arrow (Rofl)
• Jun 2nd 2008, 06:47 PM
topsquark
Quote:

Originally Posted by AlphaRock
3. A spherical tank is filled with water that is 20 ft. deep and 50 ft. wide.

What is the radius of the tank?

We know that
$25 \cdot 25 = 20 \cdot x$

So
$x = \frac{25^2}{20}$

and we know that 20 + x is a diameter....

-Dan
• Jun 2nd 2008, 10:09 PM
AlphaRock
Quote:

Originally Posted by topsquark

-Dan

Thanks for catching!

Shaded area for #2 = EVERYTHING OUTSIDE of circle (EVERYTHING NOT IN THE CIRCLE).
• Jun 2nd 2008, 10:19 PM
AlphaRock
Quote:

Originally Posted by topsquark
We know that
$25 \cdot 25 = 20 \cdot x$

So
$x = \frac{25^2}{20}$

and we know that 20 + x is a diameter....

-Dan

Thanks, Dan!

I'm curious. Why is it that 25 * 25 = 20 * x

I understand what you're saying about how let the rest of the line be x. I'm having a hard time understanding why 25 * 25 = 20 * x. Anybody know?
• Jun 3rd 2008, 03:32 AM
topsquark
Quote:

Originally Posted by AlphaRock
Thanks, Dan!

I'm curious. Why is it that 25 * 25 = 20 * x

I understand what you're saying about how let the rest of the line be x. I'm having a hard time understanding why 25 * 25 = 20 * x. Anybody know?

This page does a decent job of listing some of the chord theorems. I can't derive them off the top of my head, but maybe someone else will come along an show you how.

-Dan
• Jun 3rd 2008, 04:00 AM
earboth
Quote:

Originally Posted by topsquark
...I can't derive them off the top of my head, but maybe someone else will come along an show you how.

Actually you are dealing with a right triangle.

Let denote a, b and c the sides of the right triangle, p and q the segments of the hypotenuse and h the height of the triangle, then you know:

$p+q=c$ ....... [1]

$h^2 = a^2-p^2$ ....... [2]

$h^2 = b^2-q^2$ ....... [3]

[1] + [2] will yield:

$2h^2 = \underbrace{a^2+b^2}_{c^2}-p^2-q^2$

$2h^2 = c^2 - (\underbrace{(p^2+2pq+q^2)}_{c^2} -2pq)$

Thus you have:

$\boxed{h=p \cdot q}$
• Jun 7th 2008, 02:16 PM
AlphaRock

Is anybody smart enough to do number two (the cone-shape like question)?
• Jun 8th 2008, 06:59 AM
galactus
Here's another way to tackle the sphere problem using trig and a Great Circle around the center of the sphere.

A long chord (width of water surface=50) is given by:

$2Rsin(\frac{\theta}{2})=50$...[1]

The middle ordinate is 20 feet and given by $R(1-cos(\frac{\theta}{2}))=20$...[2]

Now, we have two equations with two unknowns and we can solve for R and theta.

Solve [1] for ${\theta}=2sin^{-1}(\frac{25}{r})$, sub into [2] and solve for R and we get $R=\frac{205}{8}=25.625$

Which agrees with TQ's method.

It's certainly easier using TQ's method, but if you're interested this is something to know.

In #2, I can't follow what that cone is supposed to be. Also, someone in advanced honors should know better than to spell ${\pi}$ as 'pie'
Come on.(Smirk)
• Jun 8th 2008, 07:31 AM
earboth
Quote:

Originally Posted by AlphaRock
...

2. Given: A circle is inscribed in a sector of another circle.
<A = 60 degrees, AB = 10 sqrt(3) cm