PreAlgebra
posted by mysterychicken .
2 more problems that I'm stuck on, please help
1.A square floor has an area of 32 m^2. Find the length of a side to the nearest tenth of a meter.
2.The height of a parallelogram is half the length of its base. the parallelogram has an area of 67 cm^2. Find the height to the nearest tenth of a centimeter.
I just need to know how to solve these two stepbystep
Thanks
MC

1. A square has sides the same length so to find the area, you would have x times x
So x^2 = 32 Find the square root of 32 and go one place to the right of the decimal, round if necessary.
2. The base of the figure would be x and since the height is two times that it would be 2x. To find area, length times width, so x times 2x. So 2x^2 = 67.
divide both sides by 2 so that x^2=33.5
Now, find the square root of 33.5, so out one number to the right of the decimal point (tenths) and round if necessary.
Plug them back in to the area formula
(length times width) to check your answers.
Hope this helps! 
1.A square floor has an area of 32 m^2. Find the length of a side to the nearest tenth of a meter.
Let the side of the floor be x, then
x^{2}=32
x=sqrt(32)
Use the method suggested by Bob to calculate the square root:
5*5=25
6*6=36
A better estimate will be
5+(65)*(3225)/(3625)=5.6
compared to 5.657.
2.The height of a parallelogram is half the length of its base. the parallelogram has an area of 67 cm^2. Find the height to the nearest tenth of a centimeter.
Let h=height of parallelogram
then b=base = 2h
Area of parallelogram = 67 cm^{2}
then
h*2h=67
h^{2}=67/2=33.5
h=sqrt(33.5)
Again using interpolation to find the square root
5*5=25
6*6=36
h=5+(65)*(33.525)/(3625)=5.8
compared to 5.788 to three places. 
Thank you guys :D
MC 
I found the answers, thanks to both of you:
1. 5.7m
2. 5.8cm
:D
MC 
Here's a link to the manual method, if you are interested. It is a little more difficult to work it out here because of the formatting limitations.

Oops, the finger is faster than the brain. Here's the link:
http://www.itl.nist.gov/div897/sqg/dads/HTML/squareRoot.html 
Thank you
MC