Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-06-29 at 5.12.14 PM.png

Unsure how to answer this question without first using the fact that they are parallel lines and so to find angles that then prove they are parallel lines.

InteGrand · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33302

Unsure how to answer this question without first using the fact that they are parallel lines and so to find angles that then prove they are parallel lines.

Hint: draw the line through D parallel to AB.

appleibeats · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

That is what I did, but then if you do alternate angles on parallel lines for alpha and beta, doesn't that assume the lines are AB and D extended to C and EF and DC are parallel??

leehuan · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$\text{Let }CD\text{ be the line parallel to }AB\text{ through }D$

$\text{Then }\angle BAD=\angle ADC\text{ (alternate angles)}$

$\text{Then }\angle BAD = \angle EDA - \angle CDE$

$\text{Substituting in we get }\angle CDE = \beta \quad (= \angle DEF)$

$\text{Hence, since alternate angles equal, }CD \parallel EF$

$\text{So because }CD\text{ is parallel to both }AB\text{ and }EF, \quad AB \parallel EF$

davidgoes4wce · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
That is what I did, but then if you do alternate angles on parallel lines for alpha and beta, doesn't that assume the lines are AB and D extended to C and EF and DC are parallel??

leehuan · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:

I don't think you can just assume that GD is parallel to both upon a simple construction.

davidgoes4wce · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
I don't think you can just assume that GD is parallel to both upon a simple construction.

So how would you draw CD?

davidgoes4wce · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Could I say what you said leehuan?

"Let GD be the line parallel to AB through D"?

davidgoes4wce · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Somebody gave me a reputation rating for that diagram, so I'm confused as to whether that diagram is right or wrong.

leehuan · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Could I say what you said leehuan?

"Let GD be the line parallel to AB through D"?

davidgoes4wce said:
Somebody gave me a reputation rating for that diagram, so I'm confused as to whether that diagram is right or wrong.

The diagram is not wrong. It's what you defined that is wrong. So yes you should say what I said.

You let GD be parallel to both AB AND EF simultaneously in your proof.. This is not allowed - you can only let it be parallel to one thing at ONCE by DEFINITION. You have to prove the rest.

davidgoes4wce · Jun 29, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

OK Cheers I'll take on your word then.

appleibeats · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-06-30 at 3.43.14 PM.png

Stuck on part d)

Answers say equal alternate angles.
But I don't see how to show that. Probably something to do with the isosceles triangles established earlier.

InteGrand · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33303

Stuck on part d)

Answers say equal alternate angles.
But I don't see how to show that. Probably something to do with the isosceles triangles established earlier.

$\noindent Note $\angle XAB = \angle XBA = \alpha$, say (you probably showed this when you showed the triangle $ABX$ is isosceles). Then $\angle AXB = 180^\circ -2\alpha \equiv \beta$ say (angle sum of $\triangle ABX$ is $180^\circ$). So $\angle CXD = \beta$ (vertically opposite angles). Since $\triangle CDX$ is isosceles with base angles $C$ and $D$, it follows from angle sum of a triangle that $\angle XCD = \angle XDC =\frac{180^\circ -\beta}{2} = \alpha$. Hence $AB \parallel DC$ (alternate angles $DBA$ and $BDC$ are equal, as both are equal to $\alpha$).$

appleibeats · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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Dont get how to show part d) AB is perpendicular to OP

I know AM = BM from matching sides in congruent triangles

Answers says just says matching sides and angles , triangle AMO congruent to triangle BMO

But does that mean its perpendicular??

InteGrand · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
View attachment 33304

Dont get how to show part d) AB is perpendicular to OP

I know AM = BM from matching sides in congruent triangles

Answers says just says matching sides and angles , triangle AMO congruent to triangle BMO

But does that mean its perpendicular??

Yes, it means they're perpendicular. This is because angles AMO and BMO are equal since triangle AMO is congruent to triangle BMO. Since they are adjacent angles on a straight line, each is 90 degrees.

appleibeats · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-06-30 at 5.10.39 PM.png

Dont know how to answer part d)

Trying to show there is one pair of parallel lines i.e. DC parallel to AB

Answers says cointerior angles are supplementary.

davidgoes4wce · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$\noindent Note $\angle XAB = \angle XBA = \alpha$, say (you probably showed this when you showed the triangle $ABX$ is isosceles). Then $\angle AXB = 180^\circ -2\alpha \equiv \beta$ say (angle sum of $\triangle ABX$ is $180^\circ$). So $\angle CXD = \beta$ (vertically opposite angles). Since $\triangle CDX$ is isosceles with base angles $C$ and $D$, it follows from angle sum of a triangle that $\angle XCD = \angle XDC =\frac{180^\circ -\beta}{2} = \alpha$. Hence $AB \parallel DC$ (alternate angles $DBA$ and $BDC$ are equal, as both are equal to $\alpha$).$

Why did you introduce the letter X into part (d)? I didn't see anywhere where X was mentioned in the questioned. If you did introduce it, where would it be on the diagram?

InteGrand · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Why did you introduce the letter X into part (d)? I didn't see anywhere where X was mentioned in the questioned. If you did introduce it, where would it be on the diagram?

I think you're thinking of a different question of appleibeats. I was answering this one, which has an X in it:

appleibeats said:
View attachment 33303

Stuck on part d)

Answers say equal alternate angles.
But I don't see how to show that. Probably something to do with the isosceles triangles established earlier.

.

davidgoes4wce · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

OK Fair enough I was looking at a different question anyone I had a go at the Cambridge Exercise 8C Q 22

davidgoes4wce · Jun 30, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$Cambridge Year 12 Exercise 8C Q 22 (a)$

$\triangle $ ABO and $ \triangle $ ABP are isosceles. Two equal radii form two sides of each triangle. $$

$$ (b) Show $ \triangle AOP \equiv \triangle BOP$

$AO=BO (equal radii)$

$AP=BP (equal radii)$

$AB Is Common$

$\therefore \triangle AOP \equiv \triangle $ BOP (SSS Test) $$

$(c) Show $ \triangle AMO \equiv \triangle $ BMO$

$OM Is Common$

$OA=OB (equal radii)$

$\angle AMO \equiv \angle BMO = 90^\circ$

$\therefore \triangle AMO \equiv \triangle $ BMP (SAS Test) $$

$$ (d) Hence show that AM=BM and AB $ \perp OP$

$\therefore $ AM=BM (Matching Sides in Congruent Triangles) $$

$\therefore \angle AMO=\angle $ BMO (Matching Angles in Congruent Triangles) $$

$\therefore AB \perp OP$

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