Mathematics
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OpenStudy (anonymous):
Area of a rectangle is 336 m^2. Ratio of Width and Length is 3:7.
Find Perimeter
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OpenStudy (anonymous):
I know that Area = W x L and Perimeter = 2 ( W+ L ) and it has something to do with cross multiply possibly.
OpenStudy (cwrw238):
sorry I couldn't post because the chat rooms obscured the post box.
I'll have to split the posts
Let x by measure of length + width
OpenStudy (anonymous):
So perimeter is 2x
OpenStudy (cwrw238):
yes
OpenStudy (anonymous):
\[\frac{ 336 }{ x } = \frac{ 3 }{ 7 }\]
is all i got in my head right now
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OpenStudy (anonymous):
which doesnt make sense
OpenStudy (cwrw238):
no it doesn't best to do it this way
as width:length = 3:7 we can write width = 0.3 x and length = 0.7x
do you follow that?
OpenStudy (anonymous):
i dunno why it's .3 and .7
OpenStudy (cwrw238):
3 :7 means 3/10 for width and 7/10 for length ( 3+7 = 10)
OpenStudy (anonymous):
alright how do we connect that to the area
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OpenStudy (cwrw238):
area = w*l = 0.3x * 0.7x = 336
now solve for x
OpenStudy (cwrw238):
0.21x = 336
x = ?
OpenStudy (anonymous):
my teacher says we can just do 3x and 7x
OpenStudy (anonymous):
so that would be 21x
OpenStudy (anonymous):
so x would be 16
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OpenStudy (anonymous):
for .21x = 336 that woukld be like 1600
OpenStudy (cwrw238):
no that would make the perimeter = 32 and the area would not come back to 336
OpenStudy (anonymous):
wait x times x is x square right
OpenStudy (cwrw238):
right
OpenStudy (cwrw238):
sorry - my mistake
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OpenStudy (cwrw238):
so x^2 = 1600
and x = 40
OpenStudy (anonymous):
then width is .3 x so it would be .3 times 40 and same with the length
OpenStudy (cwrw238):
width would be 12 and length would be 0.7*40 = 28
and 12 8 28 = 336
so it works back ok
OpenStudy (cwrw238):
* 12 * 28
OpenStudy (cwrw238):
so the perimeter = 2x = 2*40
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OpenStudy (anonymous):
so perimeter is 80
OpenStudy (cwrw238):
right
OpenStudy (anonymous):
yay thanks
OpenStudy (cwrw238):
yw