Here ive got another question for you all to answer and help me!

The lenth of a rectangle is twice its breadth.If its perimeter is 108 cm, find the lenth and breadth of a rectangle.

Please can you explain its answer nicely!I solved this question but i got stuck in a step!so i want to see how it would be totally solved!

rosala Jul 7, 2014

#1**+10 **

Let the breadth be B. Then, since the length is twice the breadth, call the length 28. And the perimeter of any rectangle is given by P = 2(W + L) where W is the width (breadth) and L is the length. So we have:

108 = 2(B + 2B) Divide by 2

54 = B + 2B Simplify on the right

54 = 3B Divide by 3

18 = B And that's the width (breadth) And the length is twice that = 2(18) = 36

Check that 36 + 36 + 18 + 18 = 108 .....Yep !!!!

CPhill Jul 7, 2014

#1**+10 **

Best Answer

Let the breadth be B. Then, since the length is twice the breadth, call the length 28. And the perimeter of any rectangle is given by P = 2(W + L) where W is the width (breadth) and L is the length. So we have:

108 = 2(B + 2B) Divide by 2

54 = B + 2B Simplify on the right

54 = 3B Divide by 3

18 = B And that's the width (breadth) And the length is twice that = 2(18) = 36

Check that 36 + 36 + 18 + 18 = 108 .....Yep !!!!

CPhill Jul 7, 2014

#2**+5 **

OOOHHH!NOOO!

What a SILLY ,mistake i made!

i thought the perimeter would be lxb instead of 2(l+b)!i thought hard but still the answer in my mind came lxb!Oh no what did i do!why do i always get confused in the perimeter and area of a rectangle! Now i have lost two marks , thats soo sad!is there any way to remember them both corectly CPhill?

Now i'll never make such a mistake!

BTW thank you for the answer!thumbs up from me!

rosala Jul 7, 2014

#3**+5 **

Think of it this way, rosala......"perimeter" is how far you travel if you walk "around" something.......

So the "perimeter" of a rectangle is just the sum of its sides.....namely W + W + L + L = 2W + 2L = 2(W + L)

Think of "area" as a rug covering something.....in this case, a rectangle!!! And the area is just L X W...........

Does that help???

CPhill Jul 7, 2014