Totally OT: Geometry help!!
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Totally OT: Geometry help!!
I admit I can no longer help my 15 yr old son with his homework.. he's stumped & I have NO clue...HELP!!! Is anyone a math wiz? I was NOT!
He's in Honor Geometry ... here's the problem
We have two above ground pools (top view)... imagine a rectangle with two semicircles on the ends...the rectangles' width is 1/2 of it's length & it is 38" deep.
One is 8000 gallons, the other 10000 gallons... we need the dimensions, in inches, of both...
so far he knows that 8000 gallons = 1848000 cubic inches & a 10000 gallons = 2310000 cubic inches but that's where he gets stuck...
Anyone?
Bueller Bueller?
He's in Honor Geometry ... here's the problem
We have two above ground pools (top view)... imagine a rectangle with two semicircles on the ends...the rectangles' width is 1/2 of it's length & it is 38" deep.
One is 8000 gallons, the other 10000 gallons... we need the dimensions, in inches, of both...
so far he knows that 8000 gallons = 1848000 cubic inches & a 10000 gallons = 2310000 cubic inches but that's where he gets stuck...
Anyone?
Bueller Bueller?
too much on my plate!!
Join Date: Jan 2005
Location: not kissing frogs anymore
Posts: 646
Wow honors Geometery! Thats awesome! Sorry I can't help either, I always sucked at math, barely got through it in college.
Hopefully someone will come along and be able to help.
Hopefully someone will come along and be able to help.
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Join Date: May 2003
Location: El Paso, Tx
Posts: 5,862
Math teacher here, not sure I understand what dimensions do you need? Volume is done in cubic inches/feet etc, so if that's what he needs, he's already got his answer.
Volume for the rectangle would be Lx Hx W
Volume for the semi-circles essentially a cylinder when put together would be pi (3.14)x r^2 x h where r is the radius (from center to one point on the semi-circle) or take 1/2 of the diameter and h is the depth.
If he needs the area of the figures he would need to use area of the base of one cyliner (pi)r^2 + the area of the rectangle (lxw). If he's needing one dimension, he sets the forumulas with the information he has inserted and set it equal to his dimensions he has, only for volume as that's all he has to work with.
Tell him he has to set up an equation (using the formulas) and he's solving for the missing variable.
Volume for the rectangle would be Lx Hx W
Volume for the semi-circles essentially a cylinder when put together would be pi (3.14)x r^2 x h where r is the radius (from center to one point on the semi-circle) or take 1/2 of the diameter and h is the depth.
If he needs the area of the figures he would need to use area of the base of one cyliner (pi)r^2 + the area of the rectangle (lxw). If he's needing one dimension, he sets the forumulas with the information he has inserted and set it equal to his dimensions he has, only for volume as that's all he has to work with.
Tell him he has to set up an equation (using the formulas) and he's solving for the missing variable.
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Join Date: May 2003
Location: El Paso, Tx
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Never give a math teacher a challenge! *LOL*, he's missing one variable, either radius, or length. Is it your actual pool? He'll need to measure to get one of them, I suggest length. I'm assuming dimensions based on volume.
V= 3.14 r^2h + lwh (but width is 1/2 length) so w= 1/2 l (length)
1848000 = 3.14x r^2 x 38 + l x (1/2 l )x38
1848000 = 38(3.14x r^2 x 38 + 1/2 l^2) x38 ( factor out 38 and l x 1/2 l = 1/2 l ^2)
That's as far as he can go without either the r or l for this problem.
Once he has one of the variables inserted ( l or r ) he'll solve the equation for the other, this he'll do again and then he'll have all his dimensions for length (l), width and height( he already has=38).
Now if he knows the area of the rectangle he can use that to find the length ( l variable). For example if the area is 500 and A=lxw and w= 1/2 l then he would set it up as 500 = 1/2 l ^2, divide by 1/2 and square root to get the answer for l.
Hope that helps a bit.
V= 3.14 r^2h + lwh (but width is 1/2 length) so w= 1/2 l (length)
1848000 = 3.14x r^2 x 38 + l x (1/2 l )x38
1848000 = 38(3.14x r^2 x 38 + 1/2 l^2) x38 ( factor out 38 and l x 1/2 l = 1/2 l ^2)
That's as far as he can go without either the r or l for this problem.
Once he has one of the variables inserted ( l or r ) he'll solve the equation for the other, this he'll do again and then he'll have all his dimensions for length (l), width and height( he already has=38).
Now if he knows the area of the rectangle he can use that to find the length ( l variable). For example if the area is 500 and A=lxw and w= 1/2 l then he would set it up as 500 = 1/2 l ^2, divide by 1/2 and square root to get the answer for l.
Hope that helps a bit.
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Is this really ENGLISH or is it a special language only those with math empowered minds can speak?
CHY...I am not worthy! 
Maybe the hand that rocks the cradle doesn't rule the world...Teachers do!
Volume for the rectangle would be Lx Hx W
Volume for the semi-circles essentially a cylinder when put together would be pi (3.14)x r^2 x h where r is the radius (from center to one point on the semi-circle) or take 1/2 of the diameter and h is the depth.
If he needs the area of the figures he would need to use area of the base of one cyliner (pi)r^2 + the area of the rectangle (lxw). If he's needing one dimension, he sets the forumulas with the information he has inserted and set it equal to his dimensions he has, only for volume as that's all he has to work with.
Volume for the semi-circles essentially a cylinder when put together would be pi (3.14)x r^2 x h where r is the radius (from center to one point on the semi-circle) or take 1/2 of the diameter and h is the depth.
If he needs the area of the figures he would need to use area of the base of one cyliner (pi)r^2 + the area of the rectangle (lxw). If he's needing one dimension, he sets the forumulas with the information he has inserted and set it equal to his dimensions he has, only for volume as that's all he has to work with.
Maybe the hand that rocks the cradle doesn't rule the world...Teachers do!
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Join Date: Dec 2004
Posts: 193
I spent hours online last night trying to find an answer.. until my Comcast went down again...
Chy - it is not an actual pool just a drawing of the top view... a rectangle with half circles on the ends... the ONLY information they gave us is that the rectangles' length is twice the size of the width (so two squares side by side with a half circle on each end), the depth is 38" & the volume will be 8000 gallons & 10000 gallons (two different calculations)... we need to determine the dimensions of the Pool in feet (eg. 16'x32')
It seems to me an impossible problem with out at least one other variable as you suggested...
I'm about to email the teacher.... if me, H, a web search & the three (yes three!) message boards I've posted on can't get me an answer how is a 15 yr old supposed to?! The kid's not stupid either he scored a 900 on the SATS when he was 12 (when a perfect score was 1600)
GRRR I give up!!
Chy - it is not an actual pool just a drawing of the top view... a rectangle with half circles on the ends... the ONLY information they gave us is that the rectangles' length is twice the size of the width (so two squares side by side with a half circle on each end), the depth is 38" & the volume will be 8000 gallons & 10000 gallons (two different calculations)... we need to determine the dimensions of the Pool in feet (eg. 16'x32')
It seems to me an impossible problem with out at least one other variable as you suggested...
I'm about to email the teacher.... if me, H, a web search & the three (yes three!) message boards I've posted on can't get me an answer how is a 15 yr old supposed to?! The kid's not stupid either he scored a 900 on the SATS when he was 12 (when a perfect score was 1600)
GRRR I give up!!
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Join Date: May 2003
Location: El Paso, Tx
Posts: 5,862
He should be able to finish the first example from here solving for l.
V= lwh + 3.14 r^2h
r = 1/2(1/2 L ) = 1/4 L radius is half the widith and width is 1/2 L
1848000 = 1/2 L ^2 x 38 + 3.14(1/4 L ^2) (38)
Factor out the 38, divide both sides by 38,combine like terms solve for L. Then he can find the width. Repeat for the next problem.
V= lwh + 3.14 r^2h
r = 1/2(1/2 L ) = 1/4 L radius is half the widith and width is 1/2 L
1848000 = 1/2 L ^2 x 38 + 3.14(1/4 L ^2) (38)
Factor out the 38, divide both sides by 38,combine like terms solve for L. Then he can find the width. Repeat for the next problem.
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Originally Posted by Chy
V= lwh + 3.14 r^2h
r = 1/2(1/2 L ) = 1/4 L radius is half the widith and width is 1/2 L
1848000 = 1/2 L ^2 x 38 + 3.14(1/4 L ^2) (38)
r = 1/2(1/2 L ) = 1/4 L radius is half the widith and width is 1/2 L
1848000 = 1/2 L ^2 x 38 + 3.14(1/4 L ^2) (38)
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Join Date: May 2003
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Originally Posted by drgnfly30
Now... do you know how to make a 9 1/2 yr old understand multiplication & division better?!?! LOL!
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