We have no dimensions to work with. I'll pick some and try and comply with the conditions of the problem.
Suppose you have an object that is 14 by 22 by 27 cm. These three numbers have no common factor so they cannot be reduced any further, which is helpful for this problem.
Find the Volume
Volume
l = 27 cm
w = 14 cm
h = 22 cm
V = 27 *14 * 22
V = 8316 cm^3
Find the surface area
SA = 2*l*w + 2*l*h + 2*w*h
SA = 2*27*14 + 2*27*22 + 2*14*22
SA = 756 + 1188 + 616
SA = 2558
Just looking at these numbers The surface area is about 1/3 of the volume. I don't think this is always true.
Another way to do this is to consider a cube which might give you a more useful result.
s = L = W = H all three dimensions are equal in a cube.
The volume of a cube is s*s*s = s^3
The surface area of a cube is 2*s*s + 2*s*s + 2s*s = 6s^2


That means whatever the side length, the Surface Area to volume = 6/the side length which is kind of an interesting result.
Answer:
Ac=cd & ab=bd +common side so abc=dbc
same for ehf and ghf but difference in common right angle.
Answer:
q1=11
q3= 33
Step-by-step explanation:
The data set has 44 number of students. The first quartile is 25 % of the numbers in the data set . So
25 % of 44 = 25/100 * 44= 0.25 *44 = 11
So the first quartile lies at 11.
Similarly the third quartile lies at the 75 % of the numbers of the data set . So
75 % of 44 = 75/100 * 44= 0.75 *44 = 33
So the third quartile lies at 33.
1501 / 200
this is what i looked up an got