Turn on thread page Beta
    • Thread Starter
    Offline

    13
    ReputationRep:
    How would I divide the area of a circle in the ratio 3:1 with one straight cut (the smaller piece will have an area equal to a quarter of the area of the circle)?

    I am trying to find the distance from the edge of the circle to the cut which I have called x. (See attached for my diagram).

    My plan was to work out the area of the sector OPQ and then subtract the area of the triangle. By equating this to \frac {\pi r^2}{4} I would then get the length x.

    I have worked out the area of the triangle OPQ using Pythagoras and \frac {1}{2} \times base \times height to be:

    (r-x) \sqrt {r^2-(r-x)^2}

    For the area of the sector OPQ I used:

    \frac {1}{2} r^2 \theta

    Therefore:

    \frac {\pi r^2}{4} = \frac {1}{2} r^2 \theta - (r-x) \sqrt {r^2-(r-x)^2}

    I tried rearranging for x but didn't really get anywhere. Also I don't know \theta so was never going to reach an answer for x anyway.

    EDIT: P and Q are meant to be where the cut meets the circle and \theta is meant to be the angle at O.
    Attached Images
     
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by SherlockHolmes)
    How would I divide the area of a circle in the ratio 3:1 with one straight cut (the smaller piece will have an area equal to a quarter of the area of the circle)?

    I am trying to find the distance from the edge of the circle to the cut which I have called x. (See attached for my diagram).

    My plan was to work out the area of the sector OPQ and then subtract the area of the triangle. By equating this to \frac {\pi r^2}{4} I would then get the length x.

    I have worked out the area of the triangle OPQ using Pythagoras and \frac {1}{2} \times base \times height to be:

    (r-x) \sqrt {r^2-(r-x)^2}

    For the area of the sector OPQ I used:

    \frac {1}{2} r^2 \theta

    Therefore:

    \frac {\pi r^2}{4} = \frac {1}{2} r^2 \theta - (r-x) \sqrt {r^2-(r-x)^2}

    I tried rearranging for x but didn't really get anywhere. Also I don't know \theta so was never going to reach an answer for x anyway.

    EDIT: P and Q are meant to be where the cut meets the circle and \theta is meant to be the angle at O.
    Without loss of generality you can take the radius =1. Then use \frac{1}{2}r^2\sin\theta
    for the area of the triangle.
    The resulting equation in theta could then be solved numerically by various methods.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by brianeverit)
    Without loss of generality you can take the radius =1. Then use \frac{1}{2}r^2\sin\theta
    for the area of the triangle.
    The resulting equation in theta could then be solved numerically by various methods.
    Thanks for the reply. I have:

    \frac {1}{2} r^2 \theta - \frac {1}{2} r^2 \sin \theta = \frac {\pi r^2}{4}

    \frac {1}{2} \theta - \frac {1}{2} \sin \theta = \frac {\pi}{4}

    \theta - \sin \theta = \frac {\pi}{2}

    How would this be solved?
    Offline

    16
    ReputationRep:
    (Original post by SherlockHolmes)
    Thanks for the reply. I have:

    \frac {1}{2} r^2 \theta - \frac {1}{2} r^2 \sin \theta = \frac {\pi r^2}{4}

    \frac {1}{2} \theta - \frac {1}{2} \sin \theta = \frac {\pi}{4}

    \theta - \sin \theta = \frac {\pi}{2}

    How would this be solved?
    Needs to be solved numerically rather than algebraically
    Offline

    17
    ReputationRep:
    Is this a level or degree??
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by TenOfThem)
    Needs to be solved numerically rather than algebraically
    I have used the Newton-Raphson Method and got an answer of 2.31... radians. Then using some trig I have got x to be 0.586... which seems to be about right.

    Thanks for the help all.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by benwalters1996)
    Is this a level or degree??
    It's probably classed under A level knowledge. It was a question given by my teacher.
    Offline

    16
    ReputationRep:
    (Original post by SherlockHolmes)
    I have used the Newton-Raphson Method and got an answer of 2.31... radians. Then using some trig I have got x to be 0.586... which seems to be about right.

    Thanks for the help all.
    No Problem
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 16, 2013

University open days

  • Heriot-Watt University
    School of Textiles and Design Undergraduate
    Fri, 16 Nov '18
  • University of Roehampton
    All departments Undergraduate
    Sat, 17 Nov '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 17 Nov '18
Poll
Black Friday: Yay or Nay?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Equations

Best calculators for A level Maths

Tips on which model to get

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.