Author Topic: Circle Geometry Help (Read 1243 times) Tweet Share

wildareal · « **on:** July 03, 2010, 04:27:35 pm »

Hi Could someone please tell me how you would do Questions 5 and 6 of this paper on the topic of Circle Geometry. Much Appreciated.

/0 · « **Reply #1 on:** July 03, 2010, 05:15:05 pm »

5
a)
$\angle QBX = \angle CAD$ (subtended by same arc)

$\angle AXP = 180-90-\angle CAD = 90 - \angle CAD$ (since $\angle APX = 90^\circ$ )

$\angle QXB = 180-90-\angle AXP$ (angles on a straight line) $= 180-90-(90-\angle CAD) = \angle CAD = \angle QBX$

b)

$BXQ$ is isosceles, so $BQ=XQ$ .

However, $XQC$ is also isosceles, since

$\angle BXQ = 90 - \angle QXC$
$\Rightarrow \angle BQX = 2\angle QXC$
$\Rightarrow \angle XQC = 180-2\angle QXC$
$\therefore \angle QCX = 180 - \angle XQC - \angle QXC = \angle QXC$

Hence, $BQ = XQ = QC$ , and Q bisects BC.

brightsky · « **Reply #2 on:** July 03, 2010, 05:45:15 pm »

Not really vector proofs, but:

6.
a) To find $\angle FAC$ , consider $\triangle AFC$ , which is made up of sides $AF$ , $FC$ and $CA$ .

By Pythagoras' theorem, we know that $AF = FC = CA = 2\sqrt{2}$ .

Hence $\triangle AFC$ is an equilateral triangle, so $\angle FAC = 60^{\circ}$

b) Consider $\triangle AFC$ again. We know that it's a equilateral triangle with side length $2\sqrt{2}$ metres. $OF$ , hence, is the altitude of the triangle, which can be found by Pythagoras' theorem:

$AO^2 + OF^2 = AF^2$

$(\sqrt{2})^2 + OF^2 = (2\sqrt{2})^2$

$OF = \sqrt{(2\sqrt{2})^2 - (\sqrt{2})^2}$

$OF = \sqrt{8 - 2}$

$OF = \sqrt{6}$

brightsky · « **Reply #3 on:** July 03, 2010, 08:25:54 pm »

c) Extend O to J such that OJ is perpendicular to AB. We need to find the slant of the cone, that is ZJ.

Consider $\triangle JOZ$ . By Pythagoras' theorem,

$OJ^2 + OZ^2 = ZJ^2$

$1^2 + 2^2 = ZJ^2$

$ZJ = \sqrt{5}$

Hence the surface area of the cone is:

$\pi \cdot 1 \cdot \sqrt{5} + \pi \cdot 1^2$

$= \sqrt{5} \pi + \pi$

$= \pi (\sqrt{5} + 1)$ m^2

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Author Topic: Circle Geometry Help (Read 1243 times) Tweet Share

wildareal

Circle Geometry Help

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brightsky

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brightsky

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