Results 1 to 2 of 2

Math Help - Tough problem, geometery+algerbra

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
    Joined
    Sep 2006
    Posts
    157

    Arrow Tough problem, geometery+algerbra

    this is a tough problem that i have been stuck on for a while. Please look at the attached picture to see what it is. Tough problem, geometery+algerbra-image.jpgany help would greatly be appreciated.
    Last edited by OnMyWayToBeAMathProffesor; June 7th 2007 at 07:24 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    this is a tough problem that i have been stuck on for a while. Please look at the attached picture to see what it is. Click image for larger version. 

Name:	image.jpg 
Views:	37 
Size:	35.6 KB 
ID:	3205any help would greatly be appreciated.
    Hello,

    let the side of the triangle b a

    Then the height of the triangle is:
    h = \frac{1}{2}\sqrt{3} \cdot a (Use Pythagorean theorem)

    The radius of the inner circle is 2/3 of the height: r = \frac{2}{3} \cdot \frac{1}{2}\sqrt{3} \cdot a = \frac{1}{3}\sqrt{3} \cdot a

    Area of the triangle:
    A_{triangle} = \frac{1}{2} \cdot a \cdot \frac{1}{2}\sqrt{3} \cdot a = \frac{1}{4} \cdot \sqrt{3} \cdot a^2

    Area of the inner circle:
    A_{\text{inner circle}} = \pi \cdot \left( \frac{1}{3}\sqrt{3} \cdot a \right)^2 = \frac{1}{3}\pi \cdot a^2

    Area of outer half circle:
    A_{\text{outer half circle}} = \frac{1}{2} \cdot \pi \cdot \left( \frac{1}{2} a \right)^2 = \frac{1}{8} \pi a^2

    Area of the three moons:
    A_{\text{3 moons}} = A_{triangle} + 3 \cdot A_{\text{outer half circle}} - A_{\text{inner circle}}

    A_{\text{3 moons}} = \frac{1}{4} \cdot \sqrt{3} \cdot a^2 + 3 \cdot \frac{1}{8} \pi a^2 - \frac{1}{3}\pi \cdot a^2 = \frac{1}{4} \cdot \sqrt{3} \cdot a^2 + \frac{1}{24}\pi \cdot a^2

    Obviously is:
    \frac{1}{8} \cdot A_{\text{inner circle}} = \frac{1}{8} \cdot \frac{1}{3}\pi \cdot a^2 = \frac{1}{24}\pi \cdot a^2

    and that's the area which is added to the area of the triangle.
    Attached Thumbnails Attached Thumbnails Tough problem, geometery+algerbra-dreimonde.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Algerbra Story Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 22nd 2010, 09:04 AM
  2. algerbra word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 4th 2008, 07:50 PM
  3. plz help me in Geometery definition
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 16th 2008, 12:46 AM
  4. REALLY quick geometery question
    Posted in the Geometry Forum
    Replies: 4
    Last Post: December 3rd 2006, 02:09 PM
  5. Could Really use Help in Geometery HW Please!
    Posted in the Geometry Forum
    Replies: 8
    Last Post: November 1st 2006, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum