Circle Geometry xp (1 Viewer)

xwrathbringerx · May 23, 2009

Hi

ABCD is a cyclic quadrilateral and FAE is a tangent at A. Angle DAE = 50 degrees. Calculate the size of

(a) angle BAF

(b) angle BCD given also that BD is parallel to FE.

Could someone please help me?

Oh and does anyone have the worked solutions to the chapter on plane geometry of the New Senior Mathematics 3 U Course by Fitzpatrick (the green small book), esp. 20 (c) and 20 (d)???

Thanx

omniscience · May 23, 2009

hei gais how many sides does triangle have?

i am reallly struggling with my 6 unit maths

Timothy.Siu · May 23, 2009

BD is parallel to FE is a condition given to you.

its not only for the second part, but is for the first part also

xwrathbringerx · May 23, 2009

stupid me =='

How do i do this one...

Two circles touch internally at A. The tangent at P on the smaller circle cuts the larger circle at Q and R. Prove that AP bisects angle RAQ.

gurmies · May 23, 2009

Draw tangent at A & draw PS so that S lies on RA
< RAT = < SPA (angle in the alternate segment)
< RAT = < AQR (angle in the alternate segment)
.'. < SPA = < AQR
< QPA = < PSA (angle in the alternate segment)
.'. ∆QPA ||| ∆SPA (equiangular)
.'. < QAP = < SAP (corresponding angles of similar triangles), & thus PA bisects < QAR.

xwrathbringerx · May 24, 2009

Hey

I was wondering if anyone has the solutions to these last questions of 20 (c) in the green Fitzpatrick book. I'm struggling to solve them since I'm weak at circle geometry. Please, any help?

1. Two circles, centres O and P, touch externally at A. The direct common tangent touches the circles at X and Y respectively. The common tangent at A meets the direct common tangent at Z. Prove that angle OZP = 90 degrees.

<O

11. Three circular discs whose radii are 10 cm, 20 cm and 30 cm respectively touch each other. Calculate:
(a) the length of the perimeter
(b) the area enclosed between the discs.

<O

12. Three circular discs, each of radius length a units, touch each other. Calculate in terms of a
(a) the area enclosed between the discs
(b) the radius of the largest disc which can fit into the space between the three discs.

Thanx a lot!

kurt.physics · May 24, 2009

xwrathbringerx said:
<o><o><o><o><o><o><o><o><o><o>
12. Three circular discs, each of radius length a units, touch each other. Calculate in terms of a
(a) the area enclosed between the discs
(b) the radius of the largest disc which can fit into the space between the three discs.

Thanx a lot!

(a)

$Join the three centres of the circles to form an equilateral triangle with side length '2a'.$

$Using the formula for the area of a triangle:$

$A = \frac{1}{2}\cdot b \cdot h$

$= \frac{1}{2} \cdot 2a \cdot 2a$

$= 2a^2$

$There are two methods to proceed from here, one uses year 12 trigonometry, the other is more logical$

$Note that each angle subtended by the three arcs (ie in the triangle), are \,\,$60^{\circ}\,\,$ (because the triangle is equalateral)$

$Now, if you imagine taking each of the three sectors and putting them together such that the 3 points where the$\,\, 60^{\circ}\,\,$ angle is are joined together, then you would find is what you now have is a semi circle with radius 'a'.$

$$The area of a semi circle is:$

$\frac{\pi r^2}{2}$

$$with$\,\,r = a$

$A = \frac{\pi a^2}{2}$

$If we subtract this from the area of the triangle that we found earlier, than that will leave us with the area enclosed between the three circles$

$=2a^2 - \frac{\pi a^2}{2}$

$=\frac{4a^2 - \pi a^2}{2}$

$=\frac{a^2(4- \pi)}{2}$

$= \frac{1}{2}a^2 \left(4- \pi \right)$

(b)

$Draw the circle enclosed by the three larger circles$

$Join each centre of the larger circles to the centre of the smaller centre. Note that these three lines are all the same lenght$

$Let the smaller circles radius be 'r'$

$Note that there are 3 equal triangles formed by our newly drawn lines$

$Furthermore, note that their biggest angles all meet at the small circles centre$

$With this information$

$3 \times\,\,$(the angle at the centre of small circle)$\,\, = 360^{\circ}$

$$(this is because they are all equal and they all meet at a point, and we know that the sum of the angles at a point is$\,\,360\,\,)$

$\therefore\,\,$the angles at the centre of the small circle are$\,\,120^{\circ}$

$Furthermore, as each of the lines formed from the joining of the centres of the larger circles to the centre of the smaller circle are all equal, then there are 3 equilateral triangles$

$\therefore\,\,$there the base angle of each one of the three triangles are$\,\,30^{\circ}$

$Consider the sine rule:$

$\frac{a}{\sin a} = \frac{b}{\sin b}$

$Consider just one of the equalateral triangles$

$\frac{a+r}{\sin 30} = \frac{2a}{\sin 120}$

$=(a+r)\times \sin 120 = 2a \times sin 30$

$\sin 120 = \sin{(180-120)}$

$= \sin 60$

$From the exact values triangle$

$\sin 60 = \frac{\sqrt{3}}{2}$

$\sin 30 = \frac{1}{2}$
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$\therefore (a+r) \times \frac{\sqrt{3}}{2} = 2a \times \frac{1}{2}$

$a + r = \frac{a}{\frac{\sqrt{3}}{2}}$

$a+r = \frac{2a}{\sqrt{3}}$

$a + r = \frac{2a\sqrt{3}}{3}$

$\therefore r = \frac{2a\sqrt{3}}{3} - a$

$= \frac{2a\sqrt{3} - 3a}{3}$

$= \frac{a(2\sqrt{3} - 3)}{3}$
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kurt.physics · May 24, 2009

Here is the diagram for the above question:

kurt.physics · May 24, 2009

xwrathbringerx said:
Hey<o><o><o><o><o><o><o><o><o>
11. Three circular discs whose radii are 10 cm, 20 cm and 30 cm respectively touch each other. Calculate:
(a) the length of the perimeter
(b) the area enclosed between the discs.<o>
Thanx a lot!

$Consider the fact that the line connecting two centres of a circle goes through the point of contact. If we join all three of the centres to form a triangle, we find an interesting fact: the sides of the triangle are 30, 40 and 50. This is a pythagoran triple, it means that this triangle is right angled. Consider the angles in the triangle, and also the fact that the perimeter (length) of an arc (of a circle) with angle$\,\,\theta\,\,$is$

$C = \frac{ \theta }{360^{\circ}} \times 2 \pi r$

$Can you figure it out now?$
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xwrathbringerx · May 24, 2009

I'm actually still struggling with question 1. How on earth do i do it?

kurt.physics · May 24, 2009

xwrathbringerx said:
1. Two circles, centres O and P, touch externally at A. The direct common tangent touches the circles at X and Y respectively. The common tangent at A meets the direct common tangent at Z. Prove that angle OZP = 90 degrees.
<o><o><o><o><o><o><o><o><o><o>

Let the circle on the left be circle A and the circle on the right be circle B.

Furthermore, construct the common tangent and the direct common tangent.

The 2 radii make right angles with the direct common tangent, also the other two radii make right angles with the common tangent.

In circle A:

We have the two right angles,

we have two equal sides

and the hypotenuse is common

By RHS,

the two triangles are congrugent

The same argument holds for the two triangles in circle B

So, if you see, there are two equal blue dots (because of the congruence) and similarly, there are 2 red dots.

2b + 2r = 180 (angles on a straight line)

.: b + r = 90

.: Angle OZP= 90

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xwrathbringerx · May 24, 2009

Thanx a lot!

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