Results 1 to 2 of 2

Math Help - Radian Measure

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    Radian Measure

    Hi

    Two circles, each of radius length 10 cm, have their centres 16 cm apart. Calculate the area common to each circle.

    I've drawn a diagram:



    I keep on getting 126.45 cm2 (2 d.p.) for my answer but the answers says 32.69 cm2. I simply divided the shaded area into two (they become congruent minor segments and then I simply found the area of one and multiplied it by two).

    Could someone please show me how to do this question?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Let H be the intersection point of AB and O1O2.


    - First compute the measure (in radians) of \alpha=\widehat{AO_1H}

    Consider the triangle AO_1H, which is a right angle triangle.
    We know that O_1H=8 \text{ cm}
    Hence \cos \alpha=\frac{8}{10}=\frac 45

    So \alpha=\arccos \left(\frac 45 \right) \approx 0.6435 (radian)

    - Now compute the area \mathcal{A}_1 of the arc of circle AO_1B
    Note that \widehat{AO_1B}=2 \alpha

    I'm too rusty to remember if there is a formula for it, but here is the reasoning.
    The whole circle has a center angle of 2 \pi radians.
    The arc of circle we consider has a center angle of 2 \alpha radians.
    So basically, we're considering \frac{2 \alpha}{2 \pi}=\frac{\alpha}{\pi} th of the circle.

    It can be used for the area : \mathcal{A}_1=\frac \alpha \pi \cdot \underbrace{(\pi \cdot 10^2)}_{\text{total area of the circle}}=\alpha \cdot 10^2 \approx 64.35 \text{ cm}^2


    - You now have to substract the area of the triangle AO_1B to get half of the shaded area.

    Note that AO_1B has twice the area of AO_1H

    So let's consider the triangle AO_1H.
    It's a right angle triangle.
    By the Pythagorean theorem, AO_1^2=AH^2+O_1H^2
    That is 10^2=8^2+AH^2
    Hence AH=6 \text{ cm}

    So the area of this triangle is \frac 12 \cdot 6 \cdot 8=24 \text{ cm}^2
    Twice this area = area of triangle AO_1B = 48 cm².



    So the area of half the shaded area is \approx 64.35-48=16.35 \text{ cm}^2

    Hence the shaded area is \approx 32.70 \text{ cm}^2



    I hope it's clear enough
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Radian Measure
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: July 17th 2011, 05:30 PM
  2. Radian measure
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 24th 2009, 03:59 PM
  3. Radian Measure
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 24th 2009, 03:10 AM
  4. Radian Measure?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 9th 2009, 09:34 AM
  5. Radian Measure 2
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 22nd 2008, 11:38 PM

Search Tags


/mathhelpforum @mathhelpforum