I need to check the answers- help (1 Viewer)

[lyounamu]

I don't have an access to the answer from my book - I mean, they are like "prove" questions.


Here we go:

Two circles with centres X and Y intersect at two points A and B.
(a) Draw a neat sketch joining XA, XB, YA, YB, XY, AB. Let P be the point where XY meets AB.

(b) Prove that the triangles AXY and BXY are congruent.

(c) Prove that AP = BP.

(d) Given that XA is also a tangent to the circle with centre, Y, prove that XAYB is a cyclic quadrilateral.

Well, I did "prove" them but I am not sure if I am entirely correct. Please provide your working out below.

 

[axlenatore]

I don't have an access to the answer from my book - I mean, they are like "prove" questions.


Here we go:

Two circles with centres X and Y intersect at two points A and B.
(a) Draw a neat sketch joining XA, XB, YA, YB, XY, AB. Let P be the point where XY meets AB.

(b) Prove that the triangles AXY and BXY are congruent.

(c) Prove that AP = BP.

(d) Given that XA is also a tangent to the circle with centre, Y, prove that XAYB is a cyclic quadrilateral.

Well, I did "prove" them but I am not sure if I am entirely correct. Please provide your working out below.

Consider Triangles AXY and BXY
XY is common
AX = BX (radii to edge on same circle)
AY = BY (radii to edge on same circle)
There Triangle AXY is congruent to triangle BXY (SSS)

Consider triangles XAP and BXP
XP is common
XB = XA (radius of same circle)
angle BXP = angle AXP (proved in (ii) as they are corresponding angles in congruent triangles)
Therefore triangle AXP is congruent to triangle BXP (SAS)
Therefore corresponding sides are equal in length
Therefore AP = BP

As for the last one, im not sure i know that the tangent will 90 a degree angle with the radius and if you can prove that the other side of the quad is 90 then you can prove its a cyclicquad

Im probably wrong but i didnt want to give up the chance of making a diagram

EDIT: Past hsc question or just one from a text book?

 

[lyounamu]

Consider Triangles AXY and BXY
XY is common
AX = BX (radii to edge on same circle)
AY = BY (radii to edge on same circle)
There Triangle AXY is congruent to triangle BXY (SSS)

Consider triangles XAP and BXP
XP is common
XB = XA (radius of same circle)
angle BXP = angle AXP (proved in (ii) as they are corresponding angles in congruent triangles)
Therefore triangle AXP is congruent to triangle BXP (SAS)
Therefore corresponding sides are equal in length
Therefore AP = BP

As for the last one, im not sure i know that the tangent will 90 a degree angle with the radius and if you can prove that the other side of the quad is 90 then you can prove its a cyclicquad

Im probably wrong but i didnt want to give up the chance of making a diagram

EDIT: Past hsc question or just one from a text book?

Yep, yep. That's what I got. The question that I really wanted to ask was the last one though.

For the last question, since ax = bx, bx is also a tangent (you know tangents at a point are equal). So I said that they both subtend 90 degrees at each end and add up to 180 degrees. It would be acceptable, right?

Thanks for the awesome diagram as well. I love it.

By the way, yeah, this is a HSC question but that stupid solution book didn't bother to give solution. :angry:

 

[axlenatore]

Yep, yep. That's what I got. The question that I really wanted to ask was the last one though.

For the last question, since ax = bx, bx is also a tangent (you know tangents at a point are equal). So I said that they both subtend 90 degrees at each end and add up to 180 degrees. It would be acceptable, right?

Thanks for the awesome diagram as well. I love it.

By the way, yeah, this is a HSC question but that stupid solution book didn't bother to give solution. :angry:

If you tell me what year and question it is, ill look it up, i have a book of past papers with worked answers, ill photograph the answer if you want

 

[lyounamu]

If you tell me what year and question it is, ill look it up, i have a book of past papers with worked answers, ill photograph the answer if you want

The book that you have shouldn't have it because this is 1986 HSC question. But just in case, here we go:

Q2 ii) d)

Thanks.

 

[axlenatore]

Sorry only my two unit book goes back that far, but your answer sounds about right, someone might check it later,

 

[bored of sc]

(d) Since XA is a tangent to circle centre y < XAY = 90^o (angle between tangent and radius is a right angle).

Since triangle AXY is congruent to triangle BXY < XAY = < XBY = 90^o (equal matching <'s of congruent triangles AXY and BXY).

< XAY + < XBY = 90^o + 90^o = 180^o

Therefore < XAY is supplemantary to < XBY.

Thus, XAYB is a cyclic quadrilateral (opposite angles XAY and XBY are supplemantary).

 

[lyounamu]

That's a bit iffy. See my proof above.

How?

Your solution is correct but I don't see how mine isn't.

xa and bx are obviously of same length as proved earlier and they are both tangents due to that. (tangents at a point are equal and conversly, the lines that are same must be tangents to a circle).

Therefore, since one makes 90 degrees at one end the other must make another 90 degrees at one other end meaning that they add to 180 degrees.

 

[friction]

1986 ur kidding.

 

[lyounamu]

1986 ur kidding.

lol?

 

[bored of sc]

How?

Your solution is correct but I don't see how mine isn't.

xa and bx are obviously of same length as proved earlier and they are both tangents due to that. (tangents at a point are equal and conversly, the lines that are same must be tangents to a circle).

Therefore, since one makes 90 degrees at one end the other must make another 90 degrees at one other end meaning that they add to 180 degrees.

Yeah, I just realised that I am definitely wrong with what I said. Sorry about that.