Areas and boundaries of circular composites
Choose retained parts for area and trace complete boundaries without counting shared internal edges.
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Decide what the question measures
For area, identify the region that remains. Add non-overlapping parts that belong to it, or subtract a cut-out from a whole whose area you know. For perimeter, trace the complete solid outline of that retained region. A shared edge inside the region is not part of its perimeter.
These are different decisions. An area expression combines square units; a perimeter expression combines lengths. Do not add the complete perimeters of the parts before checking where their edges meet.
In the figures below, shading marks the retained region. Unshaded cut-outs have been removed. Dashed lines show original component boundaries or joins for reference; follow the solid boundaries when finding the retained perimeter. Use the stated measurements, not measurements taken from the picture.
A joined diameter becomes an inside edge
Rectangle A has sides 14 cm and 10 cm. Semicircle B is joined outside it along the full 14 cm side, so B has radius 7 cm. Use π = 22/7.
The join is inside the shaded region
A is the rectangular component and B is the semicircular component. The shared diameter is dashed.
View figure data
| Given | Information |
|---|---|
| A | Rectangle, given figure; side lengths: 14, 10, 14, 10 cm. |
| B | Semicircle, given figure; radius 7 cm. |
| Boundary convention | The shaded region is retained; unshaded regions do not belong to it. Trace every solid boundary, including any enclosed hole edge. Dashed construction guides, if shown, are not extra boundary edges. |
| Pi instruction | Use pi = 22/7. Do not round the result. |
| Drawing convention | Use the stated physical measurements rather than measuring the picture. Drawing coordinates describe layout only. |
Check your understanding 1 Write two different expressions
Find the retained area and perimeter. Identify the shared edge and explain why adding the rectangle’s and semicircle’s complete perimeters would give the wrong answer.
Show answer
The area is 14 × 10 + 1/2 × 22/7 × 7² = 140 + 77 = 217 cm². The curved edge is half the circle’s circumference: 1/2 × 2 × 22/7 × 7 = 22 cm. The outside perimeter is 14 + 10 + 10 + 22 = 56 cm.
The complete component perimeters total 48 + 36 = 84 cm, but that counts the internal 14 cm join once for each part. Removing both copies gives 84-2 × 14 = 56 cm. Tracing the outside directly avoids counting the join in the first place.
A removed area can leave a curved boundary
A quarter circle B of radius 14 cm is cut from a square A whose side is 14 cm. B’s centre is at the original square corner, and its two straight radii lie along two original sides. Use π = 22/7.
Keep the shaded region outside the cut-out
B is the unshaded quarter-circle cut-out. A labels the retained part of the original square.
View figure data
| Given | Information |
|---|---|
| A | Square, given figure; side lengths: 14, 14, 14, 14 cm. |
| B | Quarter circle, given figure; radius 14 cm. |
| Boundary convention | The shaded region is retained; unshaded regions do not belong to it. Trace every solid boundary, including any enclosed hole edge. Dashed construction guides, if shown, are not extra boundary edges. |
| Pi instruction | Use pi = 22/7. Do not round the result. |
| Drawing convention | Use the stated physical measurements rather than measuring the picture. Drawing coordinates describe layout only. |
Check your understanding 2 Subtract areas, then start a fresh boundary trace
Find the shaded area and its complete perimeter. Explain why neither the original square perimeter nor the quarter-circle perimeter answers the boundary question.
Show answer
The square’s area is 14² = 196 cm². The cut-out area is 1/4 × 22/7 × 14² = 154 cm², so the shaded area is 196-154 = 42 cm².
The retained boundary has two straight sides of 14 cm and a quarter-circle arc. That arc is 1/4 × 2 × 22/7 × 14 = 22 cm long. The complete perimeter is 14 + 14 + 22 = 50 cm. The two dashed original sides are no longer boundary edges of the shaded region.
A triangle’s height is not another boundary length
Triangle A has a base of 6 cm and perpendicular height of 8 cm. Its side lengths are 6 cm, 8 cm and 10 cm; these are given, so no extra side calculation is required. A semicircle B is joined outside the triangle along the full 6 cm base. Keep π exact.
Choose the dimensions for each measure
The 6 cm triangle base is also the semicircle's full diameter. The two exposed triangle sides are 8 cm and 10 cm.
View figure data
| Given | Information |
|---|---|
| A | Triangle, given figure; a base is 6 cm and its perpendicular height is 8 cm. Side lengths: 6, 10, 8 cm. |
| B | Semicircle, given figure; radius 3 cm. |
| Boundary convention | The shaded region is retained; unshaded regions do not belong to it. Trace every solid boundary, including any enclosed hole edge. Dashed construction guides, if shown, are not extra boundary edges. |
| Pi instruction | Keep pi exact; type pi or π. Do not substitute a decimal approximation. |
| Drawing convention | Use the stated physical measurements rather than measuring the picture. Drawing coordinates describe layout only. |
Check your understanding 3 Keep the common diameter out of the perimeter
Give the retained area and complete perimeter in terms of π. Explain where the 6 cm measurement enters each calculation and which straight sides remain exposed.
Show answer
The triangle’s area is 1/2 × 6 × 8 = 24 cm². The semicircle radius is 6/2 = 3 cm, so its area is (1/2)π × 3² = (9/2)π cm². The total area is 24 + (9/2)π cm².
For perimeter, the exposed straight sides are 8 cm and 10 cm. The curved edge is half the full circumference, 1/2 × 2π × 3 = 3π cm. Thus the perimeter is 18 + 3π cm. The shared 6 cm base helps determine the radius but is not an extra outside edge.
An enclosed cut-out has its own boundary
A square A of side 28 cm has an enclosed circular cut-out made from two semicircles B and C, each of radius 7 cm. The two semicircles share a diameter and together remove one full circle. This question asks for the total boundary length of the shaded region, including the hole, not only the outside square. Use π = 22/7.
Count both solid boundary paths
A labels the retained square region. B and C label the two unshaded semicircular cut-outs.
View figure data
| Given | Information |
|---|---|
| A | Square, given figure; side lengths: 28, 28, 28, 28 cm. |
| B | Semicircle, given figure; radius 7 cm. |
| C | Semicircle, given figure; radius 7 cm. |
| Boundary convention | The shaded region is retained; unshaded regions do not belong to it. Trace every solid boundary, including any enclosed hole edge. Dashed construction guides, if shown, are not extra boundary edges. |
| Pi instruction | Use pi = 22/7. Do not round the result. |
| Drawing convention | Use the stated physical measurements rather than measuring the picture. Drawing coordinates describe layout only. |
Check your understanding 4 Separate removed area from added boundary
Find the shaded area and the requested total boundary length. Explain why removing this inner region reduces area but adds a boundary to the outside square’s perimeter.
Show answer
The square area is 28² = 784 cm². The two semicircles together remove π × 7² = 154 cm², leaving 784-154 = 630 cm².
The outside square boundary is 4 × 28 = 112 cm. The inner circle boundary is 2 × 22/7 × 7 = 44 cm. Both bound the shaded region, so the requested total is 112 + 44 = 156 cm. The diameter between B and C belongs inside the removed region and is not counted.
Overlapping parts need a counting correction
Square C has side 14 cm. Quarter circles A and B each have radius 14 cm, with their centres at opposite corners of the square. Only their common overlap is shaded. Use π = 22/7.
Two quarter circles cover the square and overlap
A and B label the quarter circles that meet in the shaded lens. C identifies the whole reference square, including its shaded and unshaded parts.
View figure data
| Given | Information |
|---|---|
| A | Quarter circle, given figure; radius 14 cm. |
| B | Quarter circle, given figure; radius 14 cm. |
| C | Square, given figure; side lengths: 14, 14, 14, 14 cm. |
| Boundary convention | The shaded region is retained; unshaded regions do not belong to it. Trace every solid boundary, including any enclosed hole edge. Dashed construction guides, if shown, are not extra boundary edges. |
| Pi instruction | Use pi = 22/7. Do not round the result. |
| Drawing convention | Use the stated physical measurements rather than measuring the picture. Drawing coordinates describe layout only. |
Imagine counting the area of A, then the area of B. Every unshaded part of the square is counted once, but the shaded overlap is counted twice. The square is completely covered by the two quarter circles. Subtracting the square’s area removes one count everywhere: the unshaded parts are left with zero counts and the overlap with one.
Check your understanding 5 Find the overlap without counting it twice
Find the shaded area and its perimeter. Explain why subtracting the square’s area once keeps the overlap, even though the overlap lies inside that square. Then identify the two arcs that form the solid boundary.
Show answer
Each quarter-circle area is 1/4 × 22/7 × 14² = 154 cm². The square area is 14² = 196 cm². The overlap area is 154 + 154-196 = 112 cm². Subtracting the square once corrects the double count of the overlap; it does not remove two copies of it.
The lens boundary consists of two quarter-circle arcs. Each is 1/4 × 2 × 22/7 × 14 = 22 cm long, so its perimeter is 22 + 22 = 44 cm. None of the square’s straight sides bounds the shaded region.
Another valid boundary calculation here is 50 + 50-56 = 44 cm: add the two full quarter-circle perimeters, then remove the square’s four straight sides. This works because those four sides are exactly the unwanted edges. Check the actual edges before transferring an area subtraction rule to perimeter.
A two-ended badge
A badge is made from a rectangle A measuring 14 cm by 10 cm, with a semicircle joined outside each of its opposite 14 cm sides. B and C are these two semicircles. Use π = 22/7.
A badge with two curved ends
B and C lie outside opposite 14 cm sides of rectangle A. The whole badge is shaded.
View figure data
| Given | Information |
|---|---|
| A | Rectangle, given figure; side lengths: 14, 10, 14, 10 cm. |
| B | Semicircle, given figure; radius 7 cm. |
| C | Semicircle, given figure; radius 7 cm. |
| Boundary convention | The shaded region is retained; unshaded regions do not belong to it. Trace every solid boundary, including any enclosed hole edge. Dashed construction guides, if shown, are not extra boundary edges. |
| Pi instruction | Use pi = 22/7. Do not round the result. |
| Drawing convention | Use the stated physical measurements rather than measuring the picture. Drawing coordinates describe layout only. |
Check your understanding 6 Find both measures
Find the badge’s area and perimeter. Show your expressions and explain which given lengths belong in each calculation. You can work on paper before comparing your reasoning with the answer.
Show answer
Each semicircle has diameter 14 cm and radius 7 cm. Their combined area is one full circle’s area, so the badge area is 14 × 10 + 22/7 × 7² = 140 + 154 = 294 cm².
The two 14 cm joins are internal. Only the rectangle’s two 10 cm sides and the two semicircular arcs form the outside boundary. The two arcs together have one circle’s circumference, giving 10 + 10 + 2 × 22/7 × 7 = 20 + 44 = 64 cm.
The same two semicircles combine into a full circular area for one calculation and a full circumference for the other. Neither calculation includes the shared straight diameters as extra boundary lengths.
Check the model before the arithmetic
For area, ask whether every retained part is counted once and every stated cut-out is removed once. For perimeter, name the boundary you are tracing and decide whether any enclosed hole is included in the question. Keep lengths separate from square units, distinguish radius from diameter, and apply the stated π and final-rounding instruction. A calculation based on the wrong region cannot be repaired just by doing its arithmetic carefully.
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