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# Let AB=6, BC=8, and AC=10.

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Let AB=6, BC=8, and AC=10. What is the area of the circumcircle of triangle ABC minus the area of the incircle triangle ABC?

Nov 20, 2015

#1
+15

Hi  Mellie,

Can you say hello to me please and let m know how you are getting on.

Also can you let me know if you understand please.

Have you already studied this stuff and the trigonometry from the last question too.

It seems very hard for you. (Hard for almost everyone.)  ://

Let AB=6, BC=8, and AC=10. What is the area of the circumcircle of triangle ABC minus the area of the incircle triangle ABC?

I will draw the triangle on the number plane.  The vertices will be B(0,0)  C(0,8) and A(6,0)

I know it is a right angled triangle because 6,8,10 is a pythagorean triad.

I am going to find the equation of the circumcircle first.  That is the big one around the outside of the triangle.

Let the centre be Q(h,k) and the radius=R

QA=QB=QC=R

The equation of the circumcircle is

$$(x-h)^2+(y-k)^2=R^2\\$$

$$QA^2=(6-h)^2+(0-k)^2\\ QA^2=36+h^2-12h+k^2\\ R^2=h^2+k^2-12h+36\qquad (1)\\$$

$$QC^2=(0-h)^2+(8-k)^2\\ QC^2=h^2+k^2-16k+64 \qquad (2)\\$$

$$QB^2=(0-h)^2+(0-k)^2\\ R^2=h^2+k^2\qquad(3)$$

$$\mbox{Sub (3) into (1)}\\ h^2+k^2=h^2+k^2-12h+36\\ 0=-12h+36\\ 12h=36\\ h=3\\ \mbox{Sub (3) into (2)}\\ h^2+k^2=h^2+k^2-16k+64\\ 0=-16k+64\\ 16k=64\\ k=4\\ \mbox{The centre of the circumcircle is (3,4)}\\ R^2=3^2+4^2=25\\ R=5\\ Area = \pi*25$$

Now lets look at the incircle

Let the centre be Z

The distance from Z to the x axis = distance of Z to the y axis = p

So the centre of the incircle must be Z(p,p) and p must also be the radius of the incircle.

Now I need the equation of line joining A(6,0) and C(0,8)

The gradient = -8/6 = -4/3   and the y intercept is  8

So the equation of the line is

y=(-4/3)x+8

3y=-4x+24

4x+3y-24=0

Now I need the perpendicular distance from    (p,p)   to the line    4x+3y-24=0

and I know that this distance is equal to p

I will use the perpendicular distance formula.

$$p=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\\ p=\frac{|4p+3p-24|}{\sqrt{4^2+3^2}}\\ p=\frac{|7p-24|}{5}\\ 5p=|7p-24|\\ 5p=7p-24 \qquad or \qquad 5p=24-7p\\ -2p=-24 \qquad or \qquad 12p=24\\ p=12 \qquad or \qquad p=2\\ \mbox{But p must be less than 6 so p=2}\\ \mbox{So the centre of the incircle is (2,2) and the radius is 2}\\ \mbox{The area of the incircle is }\pi*4\; units\; squared$$

So the area between the circumcircle and the incircle is

$$Area=25\pi-4\pi\\ Area=21\pi\;\;units \;\;squared$$

Here is the pic.  Nov 22, 2015

#1
+15

Hi  Mellie,

Can you say hello to me please and let m know how you are getting on.

Also can you let me know if you understand please.

Have you already studied this stuff and the trigonometry from the last question too.

It seems very hard for you. (Hard for almost everyone.)  ://

Let AB=6, BC=8, and AC=10. What is the area of the circumcircle of triangle ABC minus the area of the incircle triangle ABC?

I will draw the triangle on the number plane.  The vertices will be B(0,0)  C(0,8) and A(6,0)

I know it is a right angled triangle because 6,8,10 is a pythagorean triad.

I am going to find the equation of the circumcircle first.  That is the big one around the outside of the triangle.

Let the centre be Q(h,k) and the radius=R

QA=QB=QC=R

The equation of the circumcircle is

$$(x-h)^2+(y-k)^2=R^2\\$$

$$QA^2=(6-h)^2+(0-k)^2\\ QA^2=36+h^2-12h+k^2\\ R^2=h^2+k^2-12h+36\qquad (1)\\$$

$$QC^2=(0-h)^2+(8-k)^2\\ QC^2=h^2+k^2-16k+64 \qquad (2)\\$$

$$QB^2=(0-h)^2+(0-k)^2\\ R^2=h^2+k^2\qquad(3)$$

$$\mbox{Sub (3) into (1)}\\ h^2+k^2=h^2+k^2-12h+36\\ 0=-12h+36\\ 12h=36\\ h=3\\ \mbox{Sub (3) into (2)}\\ h^2+k^2=h^2+k^2-16k+64\\ 0=-16k+64\\ 16k=64\\ k=4\\ \mbox{The centre of the circumcircle is (3,4)}\\ R^2=3^2+4^2=25\\ R=5\\ Area = \pi*25$$

Now lets look at the incircle

Let the centre be Z

The distance from Z to the x axis = distance of Z to the y axis = p

So the centre of the incircle must be Z(p,p) and p must also be the radius of the incircle.

Now I need the equation of line joining A(6,0) and C(0,8)

The gradient = -8/6 = -4/3   and the y intercept is  8

So the equation of the line is

y=(-4/3)x+8

3y=-4x+24

4x+3y-24=0

Now I need the perpendicular distance from    (p,p)   to the line    4x+3y-24=0

and I know that this distance is equal to p

I will use the perpendicular distance formula.

$$p=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\\ p=\frac{|4p+3p-24|}{\sqrt{4^2+3^2}}\\ p=\frac{|7p-24|}{5}\\ 5p=|7p-24|\\ 5p=7p-24 \qquad or \qquad 5p=24-7p\\ -2p=-24 \qquad or \qquad 12p=24\\ p=12 \qquad or \qquad p=2\\ \mbox{But p must be less than 6 so p=2}\\ \mbox{So the centre of the incircle is (2,2) and the radius is 2}\\ \mbox{The area of the incircle is }\pi*4\; units\; squared$$

So the area between the circumcircle and the incircle is

$$Area=25\pi-4\pi\\ Area=21\pi\;\;units \;\;squared$$

Here is the pic.  Melody Nov 22, 2015
#2
+5

Very nice, Melody.....but....you stole my thunder on this one.....curses on you.....!!!!    [LOL!!!]

I had forgotten  - until  I thought about it awhile ago -  that  we could have had   7p - 24 =  5p  or  that 7p - 24  = -5p    and  it's obviously the second one that's correct

I might give myself points for just the thought......plus....you didn't even share any of your BBQ with me.....!!!!!

I still gave you 5 pts.....   Nov 22, 2015