Post by: Ilovemathsmeth on October 07, 2009, 09:56:16 pm
Assuming that the rectangle is the room, and that the attic has a depth of 10 metres, how do I find the volume of the space above the ceiling (top of rectangle)? I ask because I get a totally different answer when I use scale factors and when I use the straight forward (area of cross section x depth) formula. Help, please!
Post by: Hooligan on October 07, 2009, 11:10:30 pm
When you used scale factors, which were the triangles you used?
When calculating the answers, which way gave you the correct answer? I am assuming it was the 'straight forward' way that gave you the right answer?
Post by: Ilovemathsmeth on October 07, 2009, 11:35:48 pm
I used triangles A B C and the triangle above the room which shares angle B. I found the height of triangle ABC which is (root 27). I found the height of the smaller triangle to be (root 27) - 3. Then I divided the smaller height by the larger to obtain k. I found the volume of the large triangle by multiplying 6 x 0.5 x root 27 x 10 and then cubed k. I multipled the cubed version of k with the volume of the large triangle to obtain 11.77 cubed metres.
I'm confused as to why my method is wrong - are the triangles dissimilar? I didn't think so, because the angles match up... =(
Post by: plato on October 08, 2009, 12:24:28 am
Apparently the answer is 27.94 cubic metres. That's approximately double to what you've got, maybe you forgot to multiply by two somewhere?
I used triangles A B C and the triangle above the room which shares angle B. I found the height of triangle ABC which is (root 27). I found the height of the smaller triangle to be (root 27) - 3. Then I divided the smaller height by the larger to obtain k. I found the volume of the large triangle by multiplying 6 x 0.5 x root 27 x 10 and then cubed k. I multipled the cubed version of k with the volume of the large triangle to obtain 11.77 cubed metres.
I'm confused as to why my method is wrong - are the triangles dissimilar? I didn't think so, because the angles match up... =(
I agree with
This means the height of the attic is
I then found the length marked x =
Then the width of the attic must be
So volume =
Then the volume of the attic comes out to be 27.846 which is not quite what you say the answer is.
Post by: QuantumJG on October 08, 2009, 01:06:01 am
Please check out the drawing: it's attached.
Assuming that the rectangle is the room, and that the attic has a depth of 10 metres, how do I find the volume of the space above the ceiling (top of rectangle)? I ask because I get a totally different answer when I use scale factors and when I use the straight forward (area of cross section x depth) formula. Help, please!
3/x = tan(pi/3) = sqrt(3) => x = 3/(sqrt(3)) = sqrt(3) => length of rectangle = 6 - 2*sqt(3)
height of triangle = 6*sin(pi/3) = 3*sqrt(3)
height of little triangle above ceiling = 3*sqrt(3) - 3
A = 0.5*3*(sqrt(3) - 1)*-2(sqrt(3) - 3) = -3(3 - 3*sqrt(3) - sqrt(3) + 3) = -3(6 - 4sqrt(3)) = 2.78m^2
V = A*d = 2.78*10 = 27.8m^3
Since the measurements are only given as integers being out by 0.14m^3 wont really matter
Post by: TonyHem on October 08, 2009, 01:07:40 am
Apparently the answer is 27.94 cubic metres. That's approximately double to what you've got, maybe you forgot to multiply by two somewhere?
I used triangles A B C and the triangle above the room which shares angle B. I found the height of triangle ABC which is (root 27). I found the height of the smaller triangle to be (root 27) - 3. Then I divided the smaller height by the larger to obtain k. I found the volume of the large triangle by multiplying 6 x 0.5 x root 27 x 10 and then cubed k. I multipled the cubed version of k with the volume of the large triangle to obtain 11.77 cubed metres.
I'm confused as to why my method is wrong - are the triangles dissimilar? I didn't think so, because the angles match up... =(
I'm not sure what you've have done, is it getting the ratio between the two triangles and then trying to times that to the volume of the whole triangle to get the smaller triangle?
I mucked around with a few things and I tried to divide the height of the smaller triangle by the height of the higher triangle then times it by the (smaller length of the triangle/ length of the larger triangle) then times that to the volume of the overall triangle and it gives the 27.846 that the other method . gives.
Post by: Hooligan on October 08, 2009, 09:29:44 pm
Apparently the answer is 27.94 cubic metres. That's approximately double to what you've got, maybe you forgot to multiply by two somewhere?
I used triangles A B C and the triangle above the room which shares angle B. I found the height of triangle ABC which is (root 27). I found the height of the smaller triangle to be (root 27) - 3. Then I divided the smaller height by the larger to obtain k. I found the volume of the large triangle by multiplying 6 x 0.5 x root 27 x 10 and then cubed k. I multipled the cubed version of k with the volume of the large triangle to obtain 11.77 cubed metres.
I'm confused as to why my method is wrong - are the triangles dissimilar? I didn't think so, because the angles match up... =(
Oops, yes, I did a mistake. xD
I did it again, and my answer is 27.846097.... (which is off the answer?)
I also tried it your way (similar triangles way), and I seem to get the same answer as you too... =S
What Kilbaha paper was this on? I may have a look into it...
Post by: Wei on October 11, 2009, 02:54:02 pm