Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

The perimeter of a rectangle is 20 cm and its dimensions are tripled. What is the perimeter of the larger rectangle?

OpenStudy (akashdeepdeb):

So 2(l+b) = 20 Right now if dimensions are tripled the Length = 3l, breadth = 3b So Perimeter = 2(3l+3b) = 2*3(l+b) [Find it now] Understood this? :)

OpenStudy (anonymous):

12+12+3+3=30

OpenStudy (anonymous):

no i thought it was 10 y fault

OpenStudy (anonymous):

20 cm 23 cm 40 cm 60 cm

OpenStudy (akashdeepdeb):

Did you understand what I said? :)

OpenStudy (anonymous):

60

OpenStudy (akashdeepdeb):

But did you get the concept? @scarroll40

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

An outside circular ring has a circumference of 200 cm. What is the circumference of an inner ring which is 25 cm shorter in radius? Both circles have the same center.

OpenStudy (anonymous):

Okay. We know that circumference is 2(pi)r Make an equation. 2(pi)r = 200 Calculate radius.

OpenStudy (anonymous):

you would do 20*3 and you would get your answer which is 60

OpenStudy (anonymous):

(pi)r = 100 r = 100/pi What do I take as the value of pi?

OpenStudy (anonymous):

r = 100/3.14 Rest you can solve.

OpenStudy (anonymous):

i did that but dont match the multiple choice

OpenStudy (anonymous):

43.0 cm 50.0 cm 74.4 cm 96.1 cm

OpenStudy (anonymous):

?

OpenStudy (anonymous):

the closestthing go to the anser was a but i am not sure if that is corrct or not

OpenStudy (amistre64):

\[C = 2\pi ~r\] \[\frac{C}{2\pi} = r\] \[\frac{C}{2\pi}-25 = r-25\] therefore:\[C'=2\pi(\frac C{2\pi}-25)\] sounds about right to me

OpenStudy (radar):

That looks like it, and there is an answer up there that goes along with it.

OpenStudy (radar):

It might help to simplify that a bit:|dw:1377191137468:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!