Could someone help me with this not-very-difficult Math Problem

[DAZ1234]

http://papers.xtremepapers.com/CIE/...Calculator Version) (4024)/4024_s10_qp_21.pdf

Q9 b in this paper; Could someone tell me what I'm supposed to apply here? I suspect it has to do with similarity. Oh and now that you're at it, could you also tell me how to do the Q11 b (iii) of the same paper?


Thanks in advance. You people are awesome!

 

[DAZ1234]

Someone? ANYONE???!!!

 

[FLRNAB]

Q11 b) (i)
K has coordinates (p,q)

The transformation matrix represents the reflection in line y = -x. In this transformation the coordinates get interchanged and their signs also get inverted.

So L would be (-q,-p)

b) (ii)

Rotation 90 center (0,0) cw has transformation matrix:
0 1
-1 0

Multiply the point K (in column vector form) with matrix above:

b) (iii)
L(-q,-p) becomes M(q,-p)
It represents reflection in Y-Axis as Y-coordinate has not changed. Matrix for reflection in Y-Axis is
-1 0
0 1

Here is a guide for transformation matrices:
https://www.xtremepapers.com/community/threads/my-guide-for-matrices-of-transformation.14874/

Reflection rules:
1) y=x
Coordinates interchanged

2) y= -x
Coordinates interchanged and signs inverted

3) x-axis
Sign of y-coordinates inverted

4) y-axis
Sign of x-coordinate inverted

 

[FLRNAB]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s10_qp_21.pdf

Q9 b in this paper; Could someone tell me what I'm supposed to apply here? I suspect it has to do with similarity. Oh and now that you're at it, could you also tell me how to do the Q11 b (iii) of the same paper?


Thanks in advance. You people are awesome!

ii) Area of sector = (Angle i.e 90 divided by 360) Multiply by pie multiply by radius square i.e 8 x 8
Area of triangle = 1/2 Multiple by base i.e MN multiply by height (x)

Area of shaded region = Area of sector - Area of triangle

iii)

Sorry for all the different forms of answers.

 

[DAZ1234]

Than

Q9) b) i) It's an equilateral triangle. So Triangle OQN = Triangle OQM. All sides of equilateral triangle must be same so QM = QN = X. So QM + QN = 2x or MN = 2x

ii) Area of sector = (Angle i.e 90 divided by 360) Multiply by pie multiply by radius square i.e 8 x 8
Area of triangle = 1/2 Multiple by base i.e MN multiply by height (x)

Area of shaded region = Area of sector - Area of triangle

iii)

Sorry for all the different forms of answers.

Thankyou SO MUCH! but can you tell me why its an equilateral triangle? thats the root of my problem :|

 

[DAZ1234]

I am sorry it's not equilateral. It's RHS property:
RHS (Right-angle-Hypotenuse-Side): If two right-angled triangles have their hypotenuses equal in length, and a pair of shorter sides are equal in length, then the triangles are congruent.

Okay, but isn't OQ supposed to be equal to ON to make MN= 2x? ( OQ and QN are not the correponding sides in the congruent triangles - hence not equal, right?)

 

[DAZ1234]

Got it! It's cx the triangle is ISOSCELES (90, 45, 45 angles) and the OQ and QN are sides opposite the equal 45 angles. Hence Equal. RIGHT?

 

[FLRNAB]

Got it! It's cx the triangle is ISOSCELES (90, 45, 45 angles) and the OQ and QN are sides opposite the equal 45 angles. Hence Equal. RIGHT?

Yeah.
You can prove it by using Tan 45 = x/QN

 

[DAZ1234]

Yeah.
You can prove it by using Tan 45 = x/QN

Hey another Q please?

http://papers.xtremepapers.com/CIE/...Calculator Version) (4024)/4024_w11_qp_12.pdf

Q28 b(ii): I just wanna know what to do when two points are collinear in vectors?

 

[FLRNAB]

Hey another Q please?

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w11_qp_12.pdf

Q28 b(ii): I just wanna know what to do when two points are collinear in vectors?

They lie on the same line. So F,E,D lie on the same line. Distance FE must be x times Distance FD aka FD is a scalar multiple of FE.
Here is a good resource

Working:

I got the value of x by trial and error.

 

[DAZ1234]

another Q please!! How do u rotate about a point (NOT the origin). As in;
http://papers.xtremepapers.com/CIE/...(Calculator Version) (4024)/4024_s06_qp_1.pdf

Q24 b (ii) here. I mean, is there a matrix thingie we can apply to these situations? Off course we're not allowed a tracing paper, but am i really gonna have to use a protractor and all that for a 2 mark question? Or can we just look at the question and make out the answer?

 

[FLRNAB]

another Q please!! How do u rotate about a point (NOT the origin). As in;
http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s06_qp_1.pdf

Q24 b (ii) here. I mean, is there a matrix thingie we can apply to these situations? Off course we're not allowed a tracing paper, but am i really gonna have to use a protractor and all that for a 2 mark question? Or can we just look at the question and make out the answer?

You can try using the matrix:
8 1
-1 3

I haven't check it though. Also it does not take that much time if you have practice.

Also rotating about a point is same as rotating about origin. Just put your compass on that point instead of the origin.

 

[DAZ1234]

but what do u do with the compass? can you please tell me how u rotate?

Also, do u know the transformation co-ordinate matrices or whatever they're called? e.g
(x,y) to (x,-y) is a reflection at the x axis.
Are there any of these for rotation and shear/stretch? (i have the reflection ones). Sorry for bothering you again and again.

 

[FLRNAB]

but what do u do with the compass? can you please tell me how u rotate?

Also, do u know the transformation co-ordinate matrices or whatever they're called? e.g
(x,y) to (x,-y) is a reflection at the x axis.
Are there any of these for rotation and shear/stretch? (i have the reflection ones). Sorry for bothering you again and again.

You are not bothering me. Here is a simulation about how to do it:
http://lgfl.skoool.co.uk/content/keystage3/maths/pc/learningsteps/RSTLC/LO_Template.swf