Help for some Maths!!! (2 Viewers)

[lookoutastroboy]

hey guys, lookoutastroboy here,

Mind if you help a potentially good yr 9 student out here with some of his maths homework? Here are the questions, and post your answers (particularly, our solutions so I can understand how you got the answer) under each question when you have completed each one.

1. A swimming pool has the shape of a trapezoidal prism as shwon in the diagram. Find:

http://community.boredofstudies.org/attachment.php?attachmentid=16527&stc=1&d=1211598742

Problem: Find how far the water level will be from the top of the pool if it is three-quarters full. (Answer to the nearest centimetre.)

2. A marquee is in the shape of a pentagonal prism. Use the dimensions shown to calculate the surface area. (There is no floor.)

http://community.boredofstudies.org/attachment.php?attachmentid=16528&stc=1&d=1211597865

3. A barn is made from aluminium. Calculate the area of metal used in its construction. (Assume it has a total window area of 8.8 m sq and a length of 7.2m.)

http://community.boredofstudies.org/attachment.php?attachmentid=16529&stc=1&d=1211598306

Thank you in advance for solving these maths problems.

 

[lyounamu]

hey guys, lookoutastroboy here,

Mind if you help a potentially good yr 9 student out here with some of his maths homework? Here are the questions, and post your answers (particularly, our solutions so I can understand how you got the answer) under each question when you have completed each one.

1. A swimming pool has the shape of a trapezoidal prism as shwon in the diagram. Find:

http://community.boredofstudies.org/attachment.php?attachmentid=16527&stc=1&d=1211598742

Problem: Find how far the water level will be from the top of the pool if it is three-quarters full. (Answer to the nearest centimetre.)

2. A marquee is in the shape of a pentagonal prism. Use the dimensions shown to calculate the surface area. (There is no floor.)

http://community.boredofstudies.org/attachment.php?attachmentid=16528&stc=1&d=1211597865

3. A barn is made from aluminium. Calculate the area of metal used in its construction. (Assume it has a total window area of 8.8 m sq and a length of 7.2m.)

http://community.boredofstudies.org/attachment.php?attachmentid=16529&stc=1&d=1211598306

Thank you in advance for solving these maths problems.

1)41/80 metre = 51 cm

2) 363.079425... m^2 = 363.08 m^2 (to the nearest 2nd decimals).

I am just tring to see if I got them right. I will post solutions soon.

I cannot understand the third diagram. It looks too ambiguous.

 

[lookoutastroboy]

hey guys,

anyone who has looked at this thread, this is my last cry for help on these questions. If you can do any of them, that'd be FANTASTIC!!!












thanks in advance and hope to see you soon, lookoutastroboy.

(p.s. lyounamu, if you read this, please post up those solutions a.s.a.p
thanks mate!!!)

 

[bored of sc]

1) 51m

volume 3D shape = volume triangular prism + volume rectangular prism
= [(2.1 * 50)/2 * 15] + [1 * 50 *15]
= 787.5 + 750
= 1537.5 m^3

3/4 full = 3/4 * 1537.5
= 1153.125 m^3 water

1153.125 - 787.5 = amount of water left in rectangular prism part of pool (the 1m by 50m by 15m part)
= 365.625m^3

therefore the height can be measured by the equation x * 50 * 15 = 365.625
x = 0.4875m
= 48.75cm

therefore distance from the top of the pool = 100cm - 48.75cm
= 51.25cm
= 51cm to nearest centremetre

hope that is right

 

[lyounamu]

hey guys,

anyone who has looked at this thread, this is my last cry for help on these questions. If you can do any of them, that'd be FANTASTIC!!!












thanks in advance and hope to see you soon, lookoutastroboy.

(p.s. lyounamu, if you read this, please post up those solutions a.s.a.p
thanks mate!!!)

1) Volume of the pool when it is 3/4 full = 3/4 x ((3.1+1) x 50)/2 x 15 = 1153.125 m^2

Now, I will calculate the bottom half of the pool (triangular prism) = (2.1 x 50)/2 x 15 = 787.5 m^2

Now, the top half (rectangular prism) must be equal to whole volume - the triangular prism which is 365.625 m^2

Therefore, 50 x 15 x X (the height) = 365.625
X = 39/80 metres.

However, the distance between the top of the pool and the water is 1 - X
which is 1 - 39/80 = 41/80 metres.

Therefore, the answer is 51.25 = 51 cm.

2) Using pythagoras' theorem to find the unknown length x,
5.5^2 + 0.6^2 = x^2
x = square root of 3061/100
11x2.4x2 +(11x0.6x2)/2 + 16x2.4x2 + 16x square root of 3061/100 x 2 = 313.244175... = 313.24 m^2

3) I don't understand the diagram. It looks too ambiguous. Sorry!

 

[bored of sc]

2) 313.24m^2 (to 2.d.p) Not too sure as the shape is confusing - I assumed it was in the house-shape.

first things first - perpendicular height of roof-triangle = 3m - 2.4m = 0.6m
height of equal sides of roof-isosceles triangle = x
x^2 = 0.6^2 + 5.5^2 (pythagoras and halving the base of triangle)
x = 5.5326m

surface area = 11 * 2.4 * 2 (area of 2 (front and back) width (or base) * height of rectangular part) + 16 * 2.4 * 2 (area of other 2 (left and right sides) base length * height) + (0.6*11)/2 * 2 (front and back triangles) + 5.5326 * 16 * 2 (2 side roof/rectangular parts)

= 313.2432 m^2
= 313.24 m^2 (to 2.d.p)

 

[bored of sc]

2) Using pythagoras' theorem to find the unknown length x,
5.5^2 + 3^2 = x^2
x = square root of 157/4.

i think that 3 should be 0.6 but don't trust me! justification: triangular perpendicular height = 3 - 2.4 (height of entire marquee - height of rectangular prism part)

 

[lyounamu]

i think that 3 should be 0.6 but don't trust me! justification: triangular perpendicular height = 3 - 2.4 (height of entire marquee - height of rectangular prism part)

Wait, I thought the roof was 3m!

Well, I will try using 0.6 as it looks like it makes more sense.

 

[bored of sc]

3) Pretty confusing as Lyounamu said - I'll have a dig when I get back this arvo...

 

[lyounamu]

3) Pretty confusing as Lyounamu said - I'll have a dig when I get back this arvo...

It looks like as if it is 3m (in terms of length given) even though it looks more like 0.6m (in terms of diagram) as you said.

 

[bored of sc]

Wait, I thought the roof was 3m!

Well, I will try using 0.6 as it looks like it makes more sense.

Oh, well that changes everything.

P.S Question 3 is super hard to understand - there are measurements going everywhere!

 

[bored of sc]

Conclusion to be made - diagrams to hard to accurately understand all the measurements...

 

[lookoutastroboy]

sorry guys,

i wish i could somehow make the diagrams a bit better. Just consider that my textbook does not allow reproduction of questions other than in school for any purpose. What i mean by reproduction is that the questions cannot be produced in the same way (i altered diagrams to be the same in my own way, though, so that is allowable). For the 3rd question, that is basically what the diagram looks like except i've had a hard attempt at putting the measurements in but it should be correct. I'll post the answer very soon.







thanks in advance, lookoutastroboy.

 

[lookoutastroboy]

bored of sc and lyounamu,

first two questions are correct!!! thank you so much guys and I hope you can finish the 3rd one as well.











Thanking you a lot in advance, lookoutastroboy.

 

[lookoutastroboy]

hey guys,

mistake, sorry. The answer to the marquee question is (closest metre squared) 363m squared approximately, not approx. 313m squared (thanks for trying anyway=)) The answer for the 3rd (ambiguous question) is 204.5 metres suqraed (to 1 d.p.). I'll post a better diagram of it in the best way possible.











Thanking you in advance, lookoutastroboy.

 

[lyounamu]

hey guys,

mistake, sorry. The answer to the marquee question is (closest metre squared) 363m squared approximately, not approx. 313m squared (thanks for trying anyway=)) The answer for the 3rd (ambiguous question) is 204.5 metres suqraed (to 1 d.p.). I'll post a better diagram of it in the best way possible.











Thanking you in advance, lookoutastroboy.

2) Using pythagoras' theorem to find the unknown length x,
5.5^2 + 3^2 = x^2
x = square root of 157/4
11x2.4x2 +(11x3x2)/2 + 16x2.4x2 + 16x square root of 157/4 x 2 = 363.0794254... = 363.08 m^2

This answer was actually my first answer. The answer above one was the edited one after Bored of Sc pointed out that the roof length is 0.6m. However, in the end, it was 3m.

 

[lookoutastroboy]

Here (Ihope) is a better diagram of the barn question (Q.3)














Good luck solving the problem =)

 

[lookoutastroboy]

hi,

thanks lyounamu for that question! your a maths legend!!













thanks a lot mate, lookoutastroboy.

 

[bored of sc]

3) Barn question = 204.92m^2 (by my calculations)

Steps used: broke the measurements up into 4 equal roof sides, 4 equal side walls and 4 equal fence sides.

area of roof sides = (1.73 * 6) / 2 *4
= 20.76m^2

area of side walls = 3.72 * 6 * 4
= 89.28m^2

area of fencing = 10.8 * 2.4 *4
= 103.68m^2

area of metal needed = sum of above areas - area of windows
= 213.72m^2 - 8.8m^2
= 204.92m^2

P.S Not too confident about that though - correct if I am wrong and also sorry about confusion caused by other question.

 

[lookoutastroboy]

hey bored of sc,

mate, you got it right!!!!





ty for the great help, you and lyounamu have given me. This'll be sure to get me top marks in my maths test 2moro.







thanking you a lot, lookoutastroboy.