Q1.
A wire of length 20 cm is bent in different ways to form rectangles. For example, we can have a rectangle with length 7 cm and width 3 cm. We can also have one of length 5.5 cm and width 4.5 cm. (Think of a few more ways of forming such rectangles.)
Answer
The wire is the perimeter, so for every such rectangle
2(length + width) = 20
length + width = 10
length + width = 10
So any pair of positive lengths adding to 10 works. A few more:
| Length (cm) | 9 | 8 | 6.5 | 6 | 5 |
|---|---|---|---|---|---|
| Width (cm) | 1 | 2 | 3.5 | 4 | 5 |
| Area (cm²) | 9 | 16 | 22.75 | 24 | 25 |
Why it happens: the wire fixes the perimeter, not the shape. Since 2l + 2w = 20, we get w = 10 − l, so choosing the length fixes the width as well. Only one number is free, and it may be any value with 0 < l < 10 — there are infinitely many such rectangles, not a handful.
Try This: the area is l(10 − l). Work it out for l = 4, 4.5, 5, 5.5, 6 and you will get 24, 24.75, 25, 24.75, 24. The area is largest, 25 cm², when the rectangle is the square of side 5 cm — and it falls away symmetrically on both sides of that.