Q1. By how many % will the s.a of the cube increase if each side is doubled?

Q2. A circle of radius 7cm is divided into two sectors by cutting 120 degree sector from the centre.The 2 sectors are used to form 2 cones of l=7cm, find the ratio of bigger cone to smaller cone?

Q3. If a=csa, b=tsa, v=vol then express v in terms of a & b. (for a cuboid)

Q4. Find 1/[(2^2)-1]+1/[(4^2)-1]+1/[(6^2)-1]......1/[(20^2)-1]

Show the statements an solutions as well.

Thanks

2. Originally Posted by calbae
Q1. By how many % will the s.a of the cube increase if each side is doubled?

Q2. A circle of radius 7cm is divided into two sectors by cutting 120 degree sector from the centre.The 2 sectors are used to form 2 cones of l=7cm, find the ratio of bigger cone to smaller cone?

Q3. If a=csa, b=tsa, v=vol then express v in terms of a & b. (for a cuboid)

Q4. Find 1/[(2^2)-1]+1/[(4^2)-1]+1/[(6^2)-1]......1/[(20^2)-1]

Show the statements an solutions as well.

Thanks
Q4. I have two hints:

1. Your sum is equal to $\sum_{n=1}^{10} \frac{1}{(2n)^2 - 1}$.

2. $\frac{1}{(2n)^2 - 1} = \frac{1}{(2n - 1)(2n+1)} = \frac{1}{2} \, \left( \frac{1}{2n-1} - \frac{1}{2n+1}\right)$.

3. Originally Posted by calbae
Q1. By how many % will the s.a of the cube increase if each side is doubled?
Well the area of a square is $s^2$ where s is the length or height.

And a cube has 6 squares on it's surface, so
$A(s) = 6~s^2$

So if we double the size of our width, height, and length, then we still have a cube, as they are all equal to 2s

So we can plug this new value for length into our formula:
$A(2s) = 6~(2s)^2$

$A(2s) = 6~(4s^2)$

$A(2s) = 4~(6s^2)$

And note that our formula earlier was that $A(s) = 6~s^2$ So we can substitute this value in:
$A(2s) = 4~A(s)$

So we can see that the surface area of a cube with a side of 2s is 4 times larger than the surface area of a cube with a side of s.

Based on the wording in the question, I think they want you to say "4 is 3 more than 1, so the surface area increases 300%"