Results 1 to 11 of 11

Math Help - geometry help

  1. #1
    Newbie Zyger's Avatar
    Joined
    May 2008
    Posts
    8

    Exclamation geometry help

    I need help solving these math problems, I have no idea how to do them:


    the circle one I need the are aof the shaded figure
    the p-gram one, i need the area
    the pentagon, I need the apothem, area, and side length
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The area of the shaded regions in the circle.

    The sides of the triangle with hypoteneuse 6 inches are the same, They are the radius of the circle:

    6=\sqrt{a^{2}+a^{2}}

    a=3\sqrt{2}

    The subtended angle is 90 degrees.

    Area of circular segment:

    \frac{1}{2}(3\sqrt{2})^{2}(\frac{\pi}{2}-sin(\frac{\pi}{2}))

    That's one of them. Multiply by 2.

    For the rhombus, find the area of one of the right triangles and multiply by 4.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by Zyger View Post
    I need help solving these math problems, I have no idea how to do them:


    the circle one I need the are aof the shaded figure
    the p-gram one, i need the area
    the pentagon, I need the apothem, area, and side length
    For the circle problem, you need to compute the radius first, which is 3\sqrt{2}. Then, the area of the semicircle is \frac{1}{2}\pi r^2. We subtract out the area of the triangle to obtain the area of the shaded region: \frac{1}{2}\pi (3\sqrt{2})^2 - (3\sqrt{2})^2 = 9\pi - 18.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie Zyger's Avatar
    Joined
    May 2008
    Posts
    8
    may please have some more help on the others?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Zyger View Post
    I need help solving these math problems, I have no idea how to do them:


    the circle one I need the are aof the shaded figure
    the p-gram one, i need the area
    the pentagon, I need the apothem, area, and side length
    A useful formula to know is that

    For a regular n-gon of sidelength b

    A_{n-gon}=\frac{1}{4}nb^2\cot\bigg(\frac{\pi}{n}\bigg)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie Zyger's Avatar
    Joined
    May 2008
    Posts
    8
    Quote Originally Posted by Zyger View Post
    I need help solving these math problems, I have no idea how to do them:


    the circle one I need the are aof the shaded figure
    the p-gram one, i need the area
    the pentagon, I need the apothem, area, and side length


    All I need now is the one with the pentagon
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Zyger View Post
    All I need now is the one with the pentagon
    A general desire for the underyling is never hurtful

    usign my formula \frac{5}{4}\cdot{b^2}\cdot\cot\bigg(\frac{\pi}{5}\  bigg)=\frac{10(\sqrt{5}+1)\sqrt{2}x^2}{\sqrt{5-\sqrt{5}}}\approx{27.52b^2}

    So now apply it

    plug in your b value

    and use the fact that

    A=\frac{1}{2}p\cdot{a}

    to find the apothem
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie Zyger's Avatar
    Joined
    May 2008
    Posts
    8
    Quote Originally Posted by Mathstud28 View Post
    A general desire for the underyling is never hurtful

    usign my formula \frac{5}{4}\cdot{b^2}\cdot\cot\bigg(\frac{\pi}{5}\  bigg)=\frac{10(\sqrt{5}+1)\sqrt{2}x^2}{\sqrt{5-\sqrt{5}}}\approx{27.52b^2}

    So now apply it

    plug in your b value

    and use the fact that

    A=\frac{1}{2}p\cdot{a}

    to find the apothem
    I am so sorry, but that math looks like a foreign language to me. Can you go a little slower with it please?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Zyger View Post
    I am so sorry, but that math looks like a foreign language to me. Can you go a little slower with it please?
    Of course this is a site to learn

    Ok so basically we have that A=\frac{1}{2}a\cdot{p}

    where a is apothem and p is permieter

    Now since we know that the exterior angle of a n-gon is \frac{360}{n} we know due to the linear pair postulate that the interior angle of an n-gon is

    180-\frac{360}{n}

    so for a pentagon(5-gon)

    the interior angle would be

    180-\frac{360}{5}=108

    So now we know that the angle at a vertex is 108 we know that the apothem bisects it making

    the angle that the apothem cuts off 44...so now we have a right triangle with angles ,90,44,46

    So we need to calculate the apothem or in this case we are given it but we need to find sidelenght

    so we use trig \sin(46)=\frac{x}{15}\Rightarrow{x=\sin(46)\cdot{1  5}\approx{10.875}}

    and since that gives us half of our side we see that the sidelenghts are

    21.75

    now we go to our formula

    A=\frac{1}{2}a\cdot{p}


    Now since P_{pentagon}=5n where n is the sidelength we see that A=5\cdot{21.75}=107.8

    so now we see

    A=\frac{1}{2}\cdot{107.8}\cdot{15}=808.75
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie Zyger's Avatar
    Joined
    May 2008
    Posts
    8
    SORRY POR MY POST, i HAVE TO EDIT NOW
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie Zyger's Avatar
    Joined
    May 2008
    Posts
    8
    Quote Originally Posted by Mathstud28 View Post
    Of course this is a site to learn

    Ok so basically we have that A=\frac{1}{2}a\cdot{p}

    where a is apothem and p is permieter

    Now since we know that the exterior angle of a n-gon is \frac{360}{n} we know due to the linear pair postulate that the interior angle of an n-gon is

    180-\frac{360}{n}

    so for a pentagon(5-gon)

    the interior angle would be

    180-\frac{360}{5}=108

    So now we know that the angle at a vertex is 108 we know that the apothem bisects it making

    the angle that the apothem cuts off 44...so now we have a right triangle with angles ,90,44,46

    So we need to calculate the apothem or in this case we are given it but we need to find sidelenght

    so we use trig \sin(46)=\frac{x}{15}\Rightarrow{x=\sin(46)\cdot{1  5}\approx{10.875}}

    and since that gives us half of our side we see that the sidelenghts are

    21.75

    now we go to our formula

    A=\frac{1}{2}a\cdot{p}


    Now since P_{pentagon}=5n where n is the sidelength we see that A=5\cdot{21.75}=107.8

    so now we see

    A=\frac{1}{2}\cdot{107.8}\cdot{15}=808.75

    Thank you sooooooooooooooooooooooooooooooooo much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Finite Geometry: Young's Geometry
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: September 15th 2010, 07:20 AM
  2. Geometry (1)
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 7th 2009, 06:44 PM
  3. geometry (2)
    Posted in the Geometry Forum
    Replies: 4
    Last Post: November 7th 2009, 06:41 PM
  4. geometry (3)
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 7th 2009, 07:22 AM
  5. Modern Geometry: Taxicab Geometry Problems
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 30th 2009, 07:32 PM

Search Tags


/mathhelpforum @mathhelpforum