Fitzpatrick - Plane Geometry (1 Viewer)

xwrathbringerx · May 24, 2009

How on earth do I do these

Please, any help?

1. Ten equal circular discs can just be placed tightly within a frame in the form of an equilateral triangle. nine of them touch the frame; the tenth, in the centre, touches six of the other discs. If the internal sides of the frame are each of the length measure R, express the radius of each circular disc in terms of R and show that the outside perimeter of the figure which remains when the frame is removed is 11/12 (3 – root(3)) piR

<O

2. Six equal circular discs are placed so that their centres lie on the circumference of a given circle and each disc touches its two neighbours. If the radius of the given circle is r, find:
(a) the radius of a seventh disc which will touch each of the six
(b) the length of the outside perimeter of the given figure

<O

3. Inside a circle K of radius length measure R, three circular discs A, B and C each of radius r are placed so that each touches the other two and K. Express R in terms of r. In the space between K, A and B, another circular disc D is placed which just touches K, A and B. If its radius is s, show that (6 + root(x))s = (2 + root(3))r

<O

4. A playground is composed of three equal circular portions, each of which touches the other two externally, and the area between them. If the circles have radius r, find:
(a) the length measure of the perimeter of the playground
(b) its total area measure.
If a boy stands in the centre of the playground, what is the greatest distance another boy in the ground can be from him?

maplestory · May 24, 2009

Whoaaa!
i tried it out ....
but i admit those are too hard for meh ....
why dun u juz ask ur maths teacher?!
:wave:

kurt.physics · May 24, 2009

xwrathbringerx said:
1. Ten equal circular discs can just be placed tightly within a frame in the form of an equilateral triangle. nine of them touch the frame; the tenth, in the centre, touches six of the other discs. If the internal sides of the frame are each of the length measure R, express the radius of each circular disc in terms of R and show that the outside perimeter of the figure which remains when the frame is removed is 11/12 (3 – root(3)) piR
<o><o><o>

$Let each circle have radius r$

$Join the centres of all the circles touching the frame$

$\tan 60 = \frac{x}{r}$

$\therefore x = \tan 60 \cdot r$

$\tan 60 = \sqrt{3}$

$\therefore x = r\sqrt{3}$

$4r + 2r + 2 \cdot r\sqrt{3}$

$ie\,\,R = 6r + 2r\sqrt{3}$

$r(6 + 2\sqrt{3}) = R$

$\therefore r = \frac{R}{6 + 2\sqrt{3}}$

$We will have the perimeter of the semi circles - there will be 6 and then we have the 3 circles at the 3 corners of the triangle.$

$The perimeter of a semicircle is:$

$\pi \cdot r$

$\therefore\,\,$ the perimeter of the six is:$$

$6 \pi r$

$With the 3 in the corner:$

$In the triangle formed by the centres of the 9, each angle is 60 degrees, so the perimeter of that corner circle will be:$

$\frac{360-60}{360} \cdot 2\pi r$

$\frac{5}{3} \pi r$

$\frac{5\pi r}{3}$

$So the full perimeter is:$

$6 \pi \cdot r + 3\cdot \frac{5\pi r}{3}$

$6 \pi r + 5 \pi r$

$11 \pi r$

$$Now we substitute what r is in terms of R:$

$11\pi \cdot \frac{R}{6 + 2\sqrt{3}}$

$=11\pi \cdot \frac{R(6 - 2\sqrt{3})}{6^2 - (2\sqrt{3})^2}$

$=11\pi \cdot \frac{R(6 - 2\sqrt{3})}{24}$

$= 11\pi \cdot \frac{2R(3 - \sqrt{3})}{2\cdot 12}$

$= 11R\pi \cdot \frac{3 - \sqrt{3}}{12}$

$= \frac{11}{12}(3 - \sqrt{3})\pi R$

kurt.physics · May 24, 2009

xwrathbringerx said:
<o>
2. Six equal circular discs are placed so that their centres lie on the circumference of a given circle and each disc touches its two neighbours. If the radius of the given circle is r, find:
(a) the radius of a seventh disc which will touch each of the six
(b) the length of the outside perimeter of the given figure
<o><o>

$\text{Connect the centres of the 6 circles. Furthermore, connect the centre of each of the 6 circles to the centre of the 7th circle.}$

$\text{Note that there are 6 triangles, and furthermore, they are all isosceles.}$

$\text{Consider the point of the triangle where it touches the 7th circles centre. All these triangles are equal, and therefore:}$

$6 \times\text{(the top angle of the triangle)}\, = 360\,\text{(angles at a point)}$

$\therefore\,\text{The top angle of the triangle} = 60^{\circ}$

$\text{As all these triangles are isosceles, this new information, that is the top angle is 60 degrees, imply that these triangles are infact equilateral triangles}$

$\text{If we let the smaller circles have radius R, then:}$

$2R = r$

$\therefore R = \frac{1}{2}r$

$\text{Let the 7th circle have radius x, then:}$

$\frac{1}{2}r + x = r$

$\therefore x = \frac{1}{2}r$

$\text{The perimeter is made up of the outside perimeter of the 6 circles.}$

$\text{It can be shown through the angle sum of a regular hexagon that each of the interior angles is 120, so the reflex angle is}\, 360 - 120 = 240$

$\text{So the perimeter will be}$

$6 \cdot \frac{240^{\circ}}{360^{\circ}} \cdot 2 \pi R$

$= 6 \cdot \frac{2}{3} \cdot 2 \pi \frac{1}{2}r$

$= 6 \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot 2 \cdot \pi \cdot r$

$= 4 \cdot \pi \cdot r$

$= 4\pi r$

xwrathbringerx · May 25, 2009

Thanx kurt.physics!

I've been able to solve the fourth question but I'm still stuck on the third one. :burn: Please help?

Gibbatron · May 25, 2009

Damn, kurt.physics. You really DO love your circle geometry

xwrathbringerx · May 27, 2009

Oops the x in (6 + root(x))s = (2 + root(3))r is actually meant to be a 3 i.e. (6 + root(3))s = (2 + root(3))r

SORRY!

bubbybubbly · Oct 9, 2012

can you help with question 3
as I am not able to solve third part

Fitzpatrick - Plane Geometry (1 Viewer)

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