0580 · 5.5Compound shapes and parts of shapes

Counting areas and boundaries

0:00A shape can look complicated even when every part of it is familiar. A running track combines straight edges with curves, while a storage tank might combine a cylinder with a rounded end. The challenge is deciding which parts to count. By the end, you will understand how to calculate boundaries, areas, exposed surfaces and volumes without counting anything twice. We will use familiar results for rectangles, circles, cylinders, cones and spheres.

0:24Those formulas are covered in the earlier mensuration videos. Perimeter measures the distance around a flat shape, while area measures the region inside it. Surface area measures a solid’s exposed surfaces, and volume measures the space it occupies. A compound shape is made by combining simpler shapes. For area, imagine covering the inside with tiles: every tile must be counted exactly once. You can add areas of separate pieces, or subtract a missing region from a larger shape.

0:52If your pieces overlap, adding their full areas would count the overlap twice. Perimeter follows a different rule because it measures a boundary, not the space inside. Imagine tracing the edge with your finger. An internal dividing line does not belong to the perimeter. A cut-out can create new boundary edges, so subtracting an area does not mean subtracting the same lengths from the perimeter.

Areas and perimeters of compound shapes

1:14Our first exam-style question combines a rectangle with a semicircle. The rectangle is twelve centimetres long and six centimetres wide. The semicircle’s diameter is the rectangle’s right-hand edge. Calculate the total area and perimeter, giving both answers exactly in terms of pi. Halving the diameter gives a radius of three centimetres. The rectangle’s area is 12 × 6, giving seventy-two square centimetres. The semicircle’s area is ½ × π × 3², giving 9π/2 square centimetres.

1:47Adding these regions gives seventy-two plus nine pi over two square centimetres. The straight boundary contains two edges of twelve centimetres and one edge of six centimetres, totalling thirty centimetres. Half the circle’s circumference gives a curved length of 3π centimetres. The perimeter is therefore 30 + 3π centimetres. We exclude the shared diameter because it lies inside the combined shape. A semicircle on its own would include its diameter in its perimeter.

2:16Here, that same straight edge becomes an internal join, so it disappears from the outside boundary. For a sector, inspect whether its two radii are exposed or hidden. The formula for each piece is not enough; its position in the whole shape matters. Our second example starts with a rectangle measuring twelve centimetres by six centimetres. A rectangular piece measuring four centimetres by two centimetres is removed from the top-right corner.

2:41Calculate the remaining area and its perimeter. This example shows why a smaller area does not necessarily mean a shorter boundary. The original area is 12 × 6, giving seventy-two square centimetres. The removed area is 4 × 2, giving eight square centimetres. Subtracting eight from seventy-two leaves sixty-four square centimetres. The remaining top edge is 12 − 4, giving eight centimetres. The right-hand vertical edge is 6 − 2, giving four centimetres.

3:11Going anticlockwise from the bottom left corner, the bottom and right-hand edges make 12 + 4, which is sixteen centimetres. Along the indentation, 4 + 2 gives another six centimetres. The top and left edges contribute 8 + 6, which is fourteen centimetres. 16 + 6 + 14 gives a perimeter of thirty-six centimetres. That equals the original rectangle’s perimeter, even though the area fell. Removing the corner replaces two outside lengths with equal lengths along the indentation.

Cylinder and hemisphere solids

3:42Three-dimensional problems use the same distinction between occupied space and exposed boundaries. Volumes of non-overlapping parts can be added, while a hole’s volume is subtracted. Surface area requires a separate inspection of every exposed face. A cut can remove material but reveal a new surface, and a hollow object can have exposed inner walls. Our third question uses a closed solid formed from a cylinder topped by a hemisphere. A hemisphere is half a sphere, and both parts have radius three centimetres.

4:11The cylinder’s height is five centimetres. Calculate the volume and total external surface area exactly in terms of pi. The cylinder’s volume is π × 3² × 5, giving 45π cubic centimetres. Half the sphere’s volume gives 18π cubic centimetres for the hemisphere. Adding the two volumes gives 63π cubic centimetres. Neither part overlaps the other, so all their occupied space counts. The cylinder’s curved surface area is 2π × 3 × 5, giving 30π square centimetres.

4:47The hemisphere’s curved area is ½ × 4π × 3², giving 18π square centimetres. The bottom circle contributes 9π square centimetres. Adding these exposed surfaces gives 57π square centimetres. The circle where the pieces meet contributes nothing to the external surface area. Adding their separate total surface areas would count that hidden circle twice. In this example, the curved surfaces and bottom are exposed. Always establish whether a solid is closed or open before including its circular ends.

Frustum volume and surface area

5:20A frustum is the part of a cone left after its top is cut off parallel to the base. Treat it as a large cone minus a smaller cone. Our final question has bottom radius six centimetres, top radius three centimetres and perpendicular height four centimetres. Its slant height is five centimetres, and both circular ends are included. Calculate its volume and total surface area exactly in terms of pi.

5:45The removed cone is similar to the original because the cut is parallel to the base. Its radius is half the original radius, so its height is also half the original height. The frustum’s height is the difference between those two cone heights. Call the large cone’s height H, so similarity makes the small cone’s height H/2. The frustum’s height is the difference, so H minus H over two equals four.

6:10That leaves H/2 equal to four centimetres. The large height is therefore eight centimetres, and the small height is 8 ÷ 2, giving four centimetres. As a check, 4/8 = 3/6, so both cones have the same ratio of height to radius. The large cone’s volume is ⅓ × π × 6² × 8, giving 96π cubic centimetres. The small cone’s volume is ⅓ × π × 3² × 4, giving 12π cubic centimetres.

6:42Subtraction leaves 84π cubic centimetres. Using four as the large cone’s height would describe the wrong cone. Similarity also makes the small cone’s slant height half the large cone’s slant height, which we call L. Their difference is five centimetres, so L ÷ 2 = 5. The large slant height is ten centimetres. Dividing ten by two gives the small slant height of five centimetres. As a check, the square root of six squared plus eight squared is √100, which is ten.

7:14Subtracting the cone curved areas gives 60π − 15π, leaving 45π square centimetres. The bottom and top circles contribute 36π and 9π, totalling 45π square centimetres. Adding the curved area and circular ends gives 90π square centimetres. These circles must be included because cutting exposes the top and the question includes both ends. On Paper Two, you must perform suitable calculations without a calculator, including fractions and exact answers involving pi.

7:43On Paper Four, use the calculator’s pi button when a numerical answer is required. Unless instructed otherwise, give answers that are not exact to three significant figures, keeping full values during working. For example, 84π becomes approximately two hundred and sixty-four cubic centimetres. Common mistakes include counting internal joins, omitting exposed cut surfaces and confusing perpendicular height with slant height. Volume uses perpendicular height, while a cone’s curved surface area uses slant height.

8:15Keep length, area and volume units distinct, and convert incompatible measurements before calculating. The central method is always the same: count the required space or boundary exactly once.