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Mathematics 11 Online
OpenStudy (anonymous):

Area of a rectangle is 336 m^2. Ratio of Width and Length is 3:7. Find Perimeter

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

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

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

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

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

OpenStudy (anonymous):

so perimeter is 80

OpenStudy (cwrw238):

right

OpenStudy (anonymous):

yay thanks

OpenStudy (cwrw238):

yw

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