Perimeter, Area & Composite Figures
1. Core Idea
Perimeter is the distance around a figure (a length). Area is the space inside it (a length squared). Keeping the units straight — metres versus square metres — prevents most conceptual errors.
Composite (shaded-region) figures look intimidating but obey one rule:
Decompose or subtract. Break the shape into standard pieces, or take a big standard shape and remove a smaller one.
2. Must-Know Rules
| Figure | Area | Perimeter |
|---|---|---|
| Square (s) | s^2 | 4s |
| Rectangle (l, w) | lw | 2(l + w) |
| Triangle | (1/2)bh | a + b + c |
| Parallelogram | bh | 2(a + b) |
| Trapezoid | (1/2)(a+b)h | sum of the four sides |
| Rhombus | (d1 d2)/2 | 4s |
| Circle | pi r^2 | 2 pi r |
| Sector (angle t) | (t/360) pi r^2 | arc + 2r |
| Regular hexagon (s) | 3 s^2 sqrt(3)/2 | 6s |
The composite-figure procedure:
- Identify the standard shapes involved.
- Decide whether to add pieces or subtract one from another.
- Find every dimension you need — often one is hidden and must be deduced.
- Compute, then state units.
Perimeter of a composite figure is not the sum of the parts' perimeters — internal edges are not part of the boundary. Trace the outline with your finger to see what counts.
Scaling. Multiply all lengths by k: perimeter x k, area x k^2.
3. Worked Examples
Example 1 — Subtraction
A rectangular garden 20 m by 14 m has a circular pond of radius 5 m in the middle. What area is left for grass?
Rectangle area = 280.
Pond area = 25 pi (about 78.5).
Grass = 280 - 25 pi, about 201.5 square metres.
Example 2 — Decomposition
An L-shaped room is formed by a 10 by 8 rectangle with a 4 by 3 rectangle removed from one corner. Find the area.
80 - 12 = 68 square units.
Example 3 — Perimeter of a composite
A semicircle of radius 7 sits on top of a rectangle 14 wide and 10 tall (the semicircle's diameter matches the rectangle's width). Find the perimeter of the whole figure.
Trace the outline: two vertical sides (10 each), the bottom (14), and the semicircular arc — but not the top of the rectangle, because it is internal.
Arc = (1/2)(2 pi x 7) = 7 pi.
Perimeter = 10 + 10 + 14 + 7 pi = 34 + 7 pi, about 56.0.
4. GRE Traps
- Including internal edges in the perimeter. Example 3's rectangle top is not part of the boundary.
- Adding areas when you should subtract. Read whether the shaded region is inside or outside.
- Mixing units. Converting cm to m must happen before, not after, squaring — 1 square metre is 10,000 square cm, not 100.
- Using a slanted side as a height.
- Assuming the removed piece is centred when the figure doesn't say so — it usually doesn't matter for area but does for perimeter.
- Doubling dimensions doubles the area. It quadruples it.
- Forgetting that a sector's perimeter includes the two radii.
5. Speed Tricks
- Trace the boundary with a finger for perimeter questions. What you trace is what you add.
- For shaded regions, always do big minus small.
- Look for symmetry. Many composite figures are two or four copies of one simple shape.
- Fill in every dimension on the sketch before computing — hidden lengths usually come from the fact that opposite sides of the outer rectangle are equal.
- Keep pi symbolic until the final answer.
- Sanity check by estimating: the shaded region should be a plausible fraction of the whole figure.
6. Self-Check
Q1. A square of side 10 has a quarter circle of radius 10 removed from one corner. Find the remaining area.
Q2. A rectangle 12 by 5 has the same perimeter as a square. Find the square's area.
Q3. A running track consists of a rectangle 100 m by 60 m with a semicircle of diameter 60 m at each end. Find the total perimeter.
Answers
A1. 100 - (1/4)(100 pi) = 100 - 25 pi, about 21.5.
A2. Rectangle perimeter = 34, so the square's side is 8.5 and its area is 72.25.
A3. Two straights of 100 m each, plus two semicircles of radius 30 forming one full circle: 200 + 60 pi, about 388.5 m.
7. One-Line Summary for the Board
Decompose or subtract. Trace the outline for perimeter — internal edges do not count.