G
Guest
Guest
I can't seem to post real long messages so here's my questions as a txt file:
http://home.iprimus.com.au/samwozza/maths.txt
http://home.iprimus.com.au/samwozza/maths.txt
The triangles are missing something (like a right angle indication or equal marks) but for the semi circles, the area of a circle is pi*(d/2)² so the semi circle areas are (pi/8)*a²,(pi/8)*b² and (pi/8)*c². You know that a²+b²=c².
For the second, if you swing the semi circle of PQ down it becomes the same as the first question.
Equate the total areas made up of the two lunes + the PQ semi circle to the area of the other two semi circles + the area of the triangle.
PS Corrected multipliers for semi circle areas.
For the second, if you swing the semi circle of PQ down it becomes the same as the first question.
Equate the total areas made up of the two lunes + the PQ semi circle to the area of the other two semi circles + the area of the triangle.
PS Corrected multipliers for semi circle areas.
Hello, Beasticly!
Denis is absolutely correct . . . I'll give it a try.
Note that: \(\displaystyle \,PQ^2\,+\,QR^2\:=\
R^2\;\;\Rightarrow\;\;PQ^2\,+\,QR^2\,-\,PR^2\:=\:0\) *
The areas can be equated like this:
\(\displaystyle [\text{Lunes}]\;=\;\[\Delta PQR]\,+\,[\text{Semicircle on }PQ]\,+\,[\text{Semicircle on }QR]\,- \,[\text{Semicircle on }PR]\)
. . . . . . . . .
. . . . . . . . .\(\displaystyle =\;\Delta PQR\,+\,\frac{\pi}{8}\left(PQ^2\,+\,QR^2\,-\,PR^2)\) *
. . . . . . . . .
Denis is absolutely correct . . . I'll give it a try.
Note that: \(\displaystyle \,PQ^2\,+\,QR^2\:=\
R^2\;\;\Rightarrow\;\;PQ^2\,+\,QR^2\,-\,PR^2\:=\:0\) *The areas can be equated like this:
\(\displaystyle [\text{Lunes}]\;=\;\[\Delta PQR]\,+\,[\text{Semicircle on }PQ]\,+\,[\text{Semicircle on }QR]\,- \,[\text{Semicircle on }PR]\)
. . . . . . . . .
. . . . . . . . .\(\displaystyle =\;\Delta PQR\,+\,\frac{\pi}{8}\left(PQ^2\,+\,QR^2\,-\,PR^2)\) *
. . . . . . . . .
G
Guest
Guest
Hey, I just tried the semi circle one, does this look right?
Area of semi circle C = Area of semi circle A + Area of semi circle B.
= (π x line A^2) /2 /2 + (π x line B^2) /2 /2
= (π x line A^2) /4 + (π x line B^2) /4
= (π x line A^2 + π x line B^2) /8
= 2 π ( line A^2 + line B^2)
Where π = Pi.
Area of semi circle C = Area of semi circle A + Area of semi circle B.
= (π x line A^2) /2 /2 + (π x line B^2) /2 /2
= (π x line A^2) /4 + (π x line B^2) /4
= (π x line A^2 + π x line B^2) /8
= 2 π ( line A^2 + line B^2)
Where π = Pi.
It isn't
(? x line A^2) /2 /2 + (? x line B^2) /2 /2
Is is
(? * line (A/2)^2)/2 + (? * line (B/2)^2)/2
= (line A^2 + line B^2) * (?/8)
(Don't use x for multiplication. It can be confused with the variable x.)
I don't see how you got to
2 ? ( line A^2 + line B^2)
Look at that step again.
(? x line A^2) /2 /2 + (? x line B^2) /2 /2
Is is
(? * line (A/2)^2)/2 + (? * line (B/2)^2)/2
= (line A^2 + line B^2) * (?/8)
(Don't use x for multiplication. It can be confused with the variable x.)
I don't see how you got to
2 ? ( line A^2 + line B^2)
Look at that step again.
G
Guest
Guest
Whoops, yeah, I screwed that up a bit.
I get that above bit, if '?' is Pi, but I don't get the next bit:
Why did you put pi over 8, and howcome 'A' and 'B' aren't over 2 anymore?
So if it's a stupid question, but math confuses me.
By the way, thanks for the help so far everyone!
I get that above bit, if '?' is Pi, but I don't get the next bit:
Why did you put pi over 8, and howcome 'A' and 'B' aren't over 2 anymore?
So if it's a stupid question, but math confuses me.
By the way, thanks for the help so far everyone!
In order to use A²+B²=C² we want A²+B² separated so the pi is factored out and combined with the 8 to get one term that is used as a multiplier for the squared terms.
(A/2)²*(1/2)=(A²/4)(1/2)=A^2/8
The same reason causes the factoring of the 8. Now we have
k(A²+B²) (where k=pi/8) = k*C²
which is what was to be proven.
(A/2)²*(1/2)=(A²/4)(1/2)=A^2/8
The same reason causes the factoring of the 8. Now we have
k(A²+B²) (where k=pi/8) = k*C²
which is what was to be proven.
G
Guest
Guest
I think I've done the lunes task. The teacher said we'd get better marks if we did it in algebra, but I have no idea how to do this. For example, in my first equation finding the area of triangle PQR I would put (A x B)/2 but then how would I express the answer to that algebraicly?
In this task I had to find how the area of the lunes (shaded areas) related to the shape of the triangle, PQR. Firstly, I replaced each side of the triangle with numbers as seen below:
Q - R = 3
P - Q = 4
P - R = 5
Triangle PQR = base x height/2
= (3 x 4)/2
= 6
Semi on PQ = (πr2)/2
= (π (4/2)2)/2
= (π22) /2
= 12.562/ 2
= 6.28
Semi on QR = (πr2)/2
= (π (3/2)2)/2
= (π1.52) /2
= 7.06500 / 2
= 3.5325
Big Semi (P – R radius) = (πr2)/2
= (π (5/2)2)/2
= (π2.52) /2
= 9.8125
Big semi (P -R radius) – Triangle PQR = 30.81125 – 6
= 24.81125
Semi on PQ + Semi on QR = 6.28 + 3.5325
= 9.8125
Now to find the area of the 2 lunes put together we must use the following equation:
Two semis – (area in big semi – triangle)
= 9.8125 – ( 9.8125 – 6 )
= 6
As you can see, the area of the lunes is exactly the same as the area of the triangle.
Is this alright? And can anyone tell me how to do it algebraicly?
Thanks.
In this task I had to find how the area of the lunes (shaded areas) related to the shape of the triangle, PQR. Firstly, I replaced each side of the triangle with numbers as seen below:
Q - R = 3
P - Q = 4
P - R = 5
Triangle PQR = base x height/2
= (3 x 4)/2
= 6
Semi on PQ = (πr2)/2
= (π (4/2)2)/2
= (π22) /2
= 12.562/ 2
= 6.28
Semi on QR = (πr2)/2
= (π (3/2)2)/2
= (π1.52) /2
= 7.06500 / 2
= 3.5325
Big Semi (P – R radius) = (πr2)/2
= (π (5/2)2)/2
= (π2.52) /2
= 9.8125
Big semi (P -R radius) – Triangle PQR = 30.81125 – 6
= 24.81125
Semi on PQ + Semi on QR = 6.28 + 3.5325
= 9.8125
Now to find the area of the 2 lunes put together we must use the following equation:
Two semis – (area in big semi – triangle)
= 9.8125 – ( 9.8125 – 6 )
= 6
As you can see, the area of the lunes is exactly the same as the area of the triangle.
Is this alright? And can anyone tell me how to do it algebraicly?
Thanks.