Question 813318
Area of rectangle = 30cm2.
Perimeter of rectangle = 22 cm.

Assume that it's length is L and it's width is W.

So we can conclude that L.W = 30 and (2L+2W)=22 --> L+W=11.

It's just a simple linear problem. We have L = 11-W and just subtitute L for 

L.W=30. So the result is (11-W).W = 30.

11W - W^2 - 30 = 0 --> W^2 - 11W + 30 = 0 --> (W-6)(W-5)=0.

So there are 2 possibilities for it's width namely 5 cm and 6 cm.

So there are also 2 possibilities for it's length namely 6cm and 5cm.

The answer is (L,W) = (6,5) and (5,6).