Use Pythagoras (see diagram)
Bob
]]>thanks...
]]>Diagram below.
I think this angle will be AIB on my diagram.
If you call the length of a side x, then you can use Pythag to get AG and then half it to get AI (and BI)
Triangle AIB is isosceles so the midpoint of AB (M) will enable you to make a right angled triangle, AMI.
Should be able to use trig to get half the angle you want.
Bob
]]>thank you.
]]>That depends on exactly what you want. You could start by looking at the MathsIsFun teaching pages:
I also find Wikipedia can be quite helpful.
http://en.wikipedia.org/wiki/Main_Page
Bob
]]>Thank you..
]]>I'm back. Do you have some answers?
Bob
]]>Have a look at
http://www.mathsisfun.com/geometry/cylinder.html
So if you add together the top, bottom and curved surface you get:
I'm getting 4021 m^2 for the tins
Q3.
To see the true angle between face VAB and the base ABCD you have to look along the line of intersection, AB
The first picture below shows the points, and the second shows the view along AB. You cannot see A because it is exactly behind B
If the size of an edge of the cube is 2S then the distance along the base to the point directly under V is S.
The angle you want can be found by tangents.
Bob
ps. I'm out now to a show rehearsal so I won't be able to reply until about 5pm GMT.
]]>i try many ways to solve it but still i fall the same answer as you did.
]]>The volume (storage) is correct. But write
600 and 1200 are also correct for the four walls. But add the floor area and ceiling area too.
Q2.
The manufacturer of tin cans received an order of 100,000 cylindrical tin cans of height 12cm and diameter 8cm how many square meters of tin cans?
The radius of the top is 4cm.
So to get the surface area of one can, first the top and bottom:
plus the curved surface area (imagine the can is unrolled to make a rectangle 12 by circumference )
Add these answers to get the total surface area.
Then x by 100 000 for all the cans.
Now you have the answer in square cm.
A square metre is 100cm by 100cm = 10 000 square cm.
So to convert your answer to square metres, divide by 10 000.
Q3. Have a look at the picture below (click to enlarge). Is that the correct model for this question ?
Bob
]]> V= LWH
= 20*10*30
= 6000 m total storage space of the dungeon
is this correct. thaks
]]>Welcome to the forum.
The internet has lots of sites that will do your homework for you and charge you.
The MIF forum doesn't charge, but we don't just do someone's homework either! Sorry.
But I'll lead you through these one by one, if you are willing to do some work too.
Q1
A dungeon is constructed with a rectangular floor 10m by 20m. The walls are vertical and 30m high. If there are ^no windows, find:
a. total area of the ceiling walls and floor
b. the storage space of the dungeon
So there are six areas altogether. eg. floor area = 10 x 20 = ceiling area.
Long walls are 20 x 30 and I'll leave you to figure out the short walls.
So calculate the six areas and add them up.
The storage space is the same as the volume so calculate length x width x height.
When you've got two answers, post them back for checking and I'll go on to Q2.
Bob
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