4U Revising Game (1 Viewer)

[Kirjava]

Sorry but your assuming that angle MPN is a rightangle.

Aha, my mistake. The diagram I drew was rather misleading.

Perhaps this is the more general result -- put forward with due tentativeness after the last mistake (i'd scan it up if i could).

angle MPN = 360 - (90 + 90 +2tan-1(b/a)) let tan-1(b/a) = @
= 180 - 2@

Area = 1/2absin(c) = 1/2*MP*NP*sin(180 - 2@)
=1/2*MP*NP*sin(2@)

tan@ = b/a (draw a triangle here, whatever helps)
so sin@ = b/sqr(a^2 + b^2)
cos@ = a/sqr(a^2 + b^2)

Area = 1/2*MP*NP*2sin@cos@
=(a^2 * b^2 / a^2 + b^2)*(ab/a^2 + b^2) [I'm assuming the result in part ii]
=a^3*b^3/(a^2 + b^2)^2

Please check this, otherwise i'd be glad to move on from a page 1 question, heh.

 

[namburger]

P(x) is a monic polynomial of degree 4 with interger coefficients and constant term 4. One zero is root 2, another zero is rational and the sum of the zeros is positive. Factorise P(x) over real numbers.

(x-root2)(x+root2)(x-2)(x+1)

 

[namburger]

Aha, my mistake. The diagram I drew was rather misleading.

Perhaps this is the more general result -- put forward with due tentativeness after the last mistake (i'd scan it up if i could).

angle MPN = 360 - (90 + 90 +2tan-1(b/a)) let tan-1(b/a) = @
= 180 - 2@

Area = 1/2absin(c) = 1/2*MP*NP*sin(180 - 2@)
=1/2*MP*NP*sin(2@)

tan@ = b/a (draw a triangle here, whatever helps)
so sin@ = b/sqr(a^2 + b^2)
cos@ = a/sqr(a^2 + b^2)

Area = 1/2*MP*NP*2sin@cos@
=(a^2 * b^2 / a^2 + b^2)*(ab/a^2 + b^2) [I'm assuming the result in part ii]
=a^3*b^3/(a^2 + b^2)^2

Please check this, otherwise i'd be glad to move on from a page 1 question, heh.

Congrats

 

[tommykins]

Your one of the shit I spilled.

sorry if i hurt your feelings, no harm done =)

So you admit to having piece of shit in your mouth?

Okay man.

 

[ronnknee]

conics2008 is the answer to the triangle and circle question pi?

 

[conics2008]

I dont know, how did you find pi, let see your working out ???

 

[ronnknee]

Where'd you get the question?
And as if post a question you don't know the solution to here...

 

[tommykins]

Q:?

A triangle has sides 3,4 & 5. A circle has been inscribed.

Find the area of the circle

Hrm, I know a property of the triangle that if there are 3 lines bisecting the angles of the triangle and meeting the other side, they meet at the same point within the triangle. This point is the centre of the circle inscribed within the triangle.

I could approach this to find the centre of the circle within the triangle (Note centre of triangle =/= centre of circle) but that seems too algebraically wrong.

 

[conics2008]

you could draw this up on an x and y plane. and then find the equation of the line with length 5 and then find the perpd distance between the line and point (0,0)..

this is a year 10 question. answer will be posted on monday...

 

[tommykins]

you could draw this up on an x and y plane. and then find the equation of the line with length 5 and then find the perpd distance between the line and point (0,0)..

this is a year 10 question. answer will be posted on monday...

A line has an infinite length.

 

[conics2008]

tommy the x and y crap right. draw up a triangle with length 3 and 4 and the hyp 5.. you can assume its a rightangle triangle..

 

[ronnknee]

I had the idea but I manipulated the y = 4/3 x incorrectly
And then I ended up getting x₁ as 3 and y₁ as -6 T_T

 

[tommykins]

tommy the x and y crap right. draw up a triangle with length 3 and 4 and the hyp 5.. you can assume its a rightangle triangle..

I've already done that.

Actually, let me draw up a diagram to see if I'm even intepretting this question right.

http://img58.imageshack.us/img58/2387/37072473uk3.jpg

PS. The circle is touching all 3 sides of the triangle.

Since it's a year 10 question - I'll think of something simpler.

If the triangle is concyclic - the right angle makes the hypotenuse the diameter of the circle. (property of circle)

http://img139.imageshack.us/img139/4953/27528020th7.jpg

r of circle is therefore 2.5

area of circle = 2.5²pi -- year 10 level questino although it leaves me really disappointed if that is the answer.

 

[ngogiathuan]

Let AB = 4, AC=3, BC=5 and O be the centre of the circle inscribed inside.
area of AOB = (1/2) AB * r (r is radius of the circle)
area of AOC = (1/2) BC * r
area of BOC = (1/2) AC * r
so area of triangle ABC = (1/2) (AB+BC+CA) r = 6r
However, area of triangle ABC also equal (1/2) AB AC = 1/2 *3 * 4 = 6
therefore 6r=6 so r =1
area of this circle = pi r^2 = pi

To draw this circle exactly, u draw 3 bisectors of 3 angles BAC, ABC and ACB. Where they intersect is the centre of the circle inscribed.

 

[conics2008]

tommy yeah you got the diagram right xD ill just read the other guys working out

 

[tommykins]

Let AB = 4, AC=3, BC=5 and O be the centre of the circle inscribed inside.
area of AOB = (1/2) AB * r (r is radius of the circle)
area of AOC = (1/2) BC * r
area of BOC = (1/2) AC * r
so area of triangle ABC = (1/2) (AB+BC+CA) r = 6r
However, area of triangle ABC also equal (1/2) AB AC = 1/2 *3 * 4 = 6
therefore 6r=6 so r =1
area of this circle = pi r^2 = pi

To draw this circle exactly, u draw 3 bisectors of 3 angles BAC, ABC and ACB. Where they intersect is the centre of the circle inscribed.

Congrats.

 

[ronnknee]

I did it the Year 11 way - ie. finding the perpendicular distance (radius) and equating it to the other radii in terms of x and y. From there, find the centre by solving the two equations simultaneously and find x and y to get a value for the radius

http://community.boredofstudies.org/showpost.php?p=3477848&postcount=90

 

[conics2008]

welldone =)


nicely done ronnknee and the other guy xD

 

[undalay]

Hereeeeee

edit: oh damn i got beat to it : P

 

[conics2008]

does any one have some integration question, i just finished that chapter...

can some one post some please....