a) 20cm
If you imagine the diagram on the right looking like an 'I' shape, you can see that the diagram on the left has two 'I's and a vertical rectangle. Because one 'I' is 13cm, two 'I's will be 26cm. Subtract 26cm from 33cm and we now know that the length of the small rectangle is 7cm.
Now, look at the other diagram. We now know that the length is 7cm, and the total length is 13cm, so we can say that 7cm + 2x = 13cm. We can say this because we are trying to find out the width and the length (7cm) plus 2 widths = 13cm. When we are finished, we find that the width is 3cm, by moving the equation round to become 13cm - 7cm = 2x, which becomes 6cm = 2x and finally 3cm = x.
We now know that the length is 7cm, the width is 3cm, therefore the perimeter is 20cm.
We can also do it algebraically. Imagine that x is the length of the small rectangle and y is the width. In the larger diagram, there are 4 widths and 3 lengths, so we can say 4y + 3x = 33cm. We can also see that in the smaller diagram, there are 2 widths and 1 length, therefore 2y + x = 13cm. This also makes 4x - 2x = 26cm. (Multiply the second equation by two). If we take 4x - 2x = 26cm from 4y + 3x = 33cm, we get y = 7cm. Knowing this, we can find out the width by the first method and get the perimeter as, yet again, 20cm.
b) 16cm
If you look, you can see that there are five rectangles. Since their combined area is 60cm squared, this means that the area of one of the five rectangles is 12 cm squared. Then, we see that the length of one rectangle is the same as the three widths. We can now form the equations:
xy = 12
x = 3y
- x is the length, y is the width.
Thus, we can substitute x for 3y, making 3y x y = 12. Divide each side by 3 to make y squared equal to 4. The square root of 4 is 2, so the width of the rectangle is 2. In that case, as x is 3 times y, we can multiply 2 by 3 to get the length as 6. Therefore, the perimeter will equal 2(2+6) which equals 16. We can check this is right as 2 x 6 makes 12.
c) 50mm
Very similar to the previous, except that there are four widths to one length, meaning that there are now six rectangles, and the area is 600mm squared. We jiggle round to get:
x = 4y
xy = 100mm
We substitute x for 4y, making 4y x y = 100mm. Dividing both sides by 4, we make y squared equal to 25. This makes the width 5mm. Since the length is 4 times the width, the length would be 4 x 5 which makes 20mm. According to this, we then find the perimeter being 50mm.
d) 18cm
This time, we have something slightly different. We have five widths to four lengths. Similarly, since we have nine rectangles, the area of a small rectangle is 20cm squared. We can now make these equations:
4x = 5y
xy = 20cm
This is harder to fiddle with, but is still solvable. First of all, we can fiddle around with xy = 20cm to make y = 20cm/x. Then, we multiply everything by five, to make 5y = 100cm/x. We then substitute 5y with 4x, making 4x now equal to 100cm divided by x. We multiply both sides by x, to make 4x squared equal to 100cm. Then, we divide everything by 4, which makes x squared 25. Finally, we square root x to make 5. (Clue: You have seen this equation - 4x squared = 100 before). Then, we find the x times y makes twenty. Since x, the length is 5, 5 x 4 is 20cm, then y = 4. Finally, we times the sum of the length and the width (5 and 4) by two, making the perimeter of the small rectangle 18cm.
The last question : Not as hard as you might think.
It looks impossible - they have not even given you the lengths of all the sides, yet you can still work it out. Look at this document: