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0580 · 4.4Similarity

Cambridge IGCSE Mathematics (0580) · Topic 4, Geometry · 10:53

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In this video

  1. 0:00Similar shapes and scale factors. Defines similarity, corresponding parts, congruence and scale factors, including how to calculate enlargement and reduction factors.
  2. 2:29Calculating lengths in similar shapes. Uses corresponding rectangle sides to calculate an unknown length and explains why similarity uses multiplication rather than addition.
  3. 3:38Proving triangles are similar. Explains triangle similarity conditions and uses parallel lines to prove nested triangles similar before finding a missing length.
  4. 6:13Area scale factors. Shows why area changes by the square of the length scale factor and applies square roots to recover missing lengths.
  5. 7:46Surface area, volume and exam skills. Explains squared surface-area and cubed volume scale factors, reverse calculations, and common examination mistakes.

Transcript: Similarity

Similar shapes and scale factors

0:00Imagine making a model building twice as tall as the original. If every length doubles, its surface area becomes four times as large. Its volume grows to eight times the original, even though its height only doubled. Similarity explains these different changes, and helps us calculate missing measurements without measuring every part. By the end of this video, you will understand what makes shapes similar and how to match their corresponding sides.

0:24You will also see why areas use a squared scale factor and volumes use a cubed scale factor. We will use those connections to calculate lengths and explain why triangles are similar. The key is understanding what changes, and what stays the same. You need a little knowledge of ratios, multiplication and angles. A ratio compares quantities in a stated order, so two to three is different from three to two.

0:47We will also use the fact that angles in a triangle total one hundred and eighty degrees. Any other angle facts will be identified when they help us explain similarity. Similar shapes have the same shape, but they do not have to have the same size. Their corresponding angles are equal, and their corresponding lengths are in the same ratio. Corresponding means matching parts that occupy equivalent positions in the shapes.

1:10A similar copy can be turned over or rotated without losing its similarity. Think of resizing a photograph without stretching it unevenly. Every horizontal and vertical distance changes by the same multiplier, so the proportions stay unchanged. That multiplier is called the scale factor. If the width doubles but the height stays unchanged, the photograph becomes distorted and is no longer a similar copy. Having equal angles alone does not establish similarity for every kind of shape.

1:37All rectangles have four right angles, but a long narrow rectangle is not similar to a square. Their side lengths do not have matching proportions. Congruent shapes have the same shape and size, so they are the special case with scale factor one. To calculate a scale factor, divide a length in the new shape by its corresponding length in the original shape. State the direction before calculating, because reversing it reverses the fraction.

2:02A factor greater than one makes the shape larger, and a positive factor less than one makes it smaller. A factor of exactly one leaves the size unchanged. For example, changing a corresponding length from four to six centimetres gives a scale factor of 6 ÷ 4. That simplifies to 3/2, or one point five. Every other corresponding length must also be multiplied by one point five. Going back from the larger shape uses 2/3 instead.

Calculating lengths in similar shapes

2:29Our first exam-style example uses two similar rectangles. The smaller rectangle has width eight centimetres and height five centimetres. The larger rectangle has corresponding width twelve centimetres, and we must calculate its height. We first match width with width, rather than choosing lengths just because they are nearby. Dividing twelve by eight gives a scale factor of 3/2. The larger height is therefore 5 × 3/2. Multiplying the numerator gives 15/2.

2:58Dividing fifteen by two gives seven point five centimetres, which is the required height. This answer makes sense because the larger width is fifty per cent greater than the smaller width. Its height must increase by the same percentage. Adding four centimetres to both dimensions would not preserve the proportions. Similarity uses one common multiplier, not one common amount added. Perimeter also changes by the length scale factor because it is a sum of side lengths.

3:25If every side is multiplied by 3/2, their total receives the same multiplier. This applies to any corresponding distance, including a diagonal or a height. Those lengths must match geometrically, even when they are not edges of the shape.

Proving triangles are similar

3:38Triangles give us useful ways to establish similarity before calculating anything. If two angles in one triangle equal two angles in another, their third angles must also match. This follows because each triangle has an angle total of one hundred and eighty degrees. Matching angles fixes the shape of a triangle, although its size can still change. This is the angle angle similarity condition. Another condition is that all three pairs of corresponding sides are in the same ratio.

4:05A third condition uses two pairs of proportional sides and equal included angles. The included angle is the angle between those two sides, not an unrelated angle elsewhere in the triangle. Do not assume that two proportional side pairs alone prove similarity. Without suitable information about an angle or the remaining sides, the triangles could have different shapes. Likewise, one pair of equal angles is not enough. The evidence must establish the whole shape, rather than just one feature that happens to match.

4:35In our second example, a line joins points S and T inside triangle PQR. Point S lies on the side from P to Q, and point T is on the side from P to R. The line from S to T is parallel to the base from Q to R. We must show that the triangles are similar, then calculate the full base length. Angle SPT equals angle QPR because they are the same angle at P.

4:59Angle PST equals angle PQR because corresponding angles between parallel lines are equal. Triangle PST is therefore similar to triangle PQR by the angle angle condition. These statements provide geometric reasons, rather than relying on how the diagram looks. Writing the triangle names in matching order helps us identify corresponding sides. Vertex P corresponds to itself, while S pairs with Q and T goes with R. The side from P to S therefore matches the whole side from P to Q.

5:30The internal line from S to T matches the full base from Q to R. First, add six and three to obtain the full length from P to Q, which is nine centimetres. The scale factor from the smaller triangle to the larger one is 9 ÷ 6, giving 3/2. Multiply ten by 3/2 to calculate the corresponding base. The length from Q to R is fifteen centimetres. The important detail is that six centimetres corresponds to nine centimetres, not to the remaining three centimetres.

6:01The two similar triangles both start at P, so we need their complete sides. In nested diagrams, mark the triangles you are comparing before choosing numbers. This prevents a correct calculation being applied to the wrong lengths.

Area scale factors

6:13Area measures a two-dimensional region, so it responds to changes in two directions. A unit square enlarged by a length factor of two becomes two units wide and two units high. It now contains four unit squares rather than one. Doubling each length therefore multiplies area by four, not by two. The same reasoning works with any positive scale factor. Both the base and perpendicular height of a similar triangle receive that multiplier.

6:38Their product therefore receives the scale factor squared, while the factor of one half remains unchanged. For similar shapes, the area scale factor is always the length scale factor squared. A length scale factor of three gives an area scale factor of nine. Halving every length leaves only 1/4 of the original area, because ½² is 1/4. When area information is given first, take a square root to recover the length factor.

7:04Our third example gives similar shapes with areas of thirty-two and seventy-two square centimetres, and a smaller corresponding length of ten centimetres. Dividing seventy-two by thirty-two gives an area factor of 9/4. Taking its square root gives the length factor 3/2. Multiplying ten by 3/2 gives the larger corresponding length of fifteen centimetres. Using 9/4 directly on the length would give the wrong answer because that factor describes area, not distance.

7:32The square root removes one of the two dimensions involved in the area change. Notice also that area is measured in square centimetres, but the final length uses centimetres. Units help identify which relationship the question needs.

Surface area, volume and exam skills

7:46For similar solids, every corresponding linear measurement changes by the same scale factor. Each surface has two dimensions, so total surface area changes by the scale factor squared. Volume involves three dimensions, so its scale factor is the length scale factor cubed. These relationships require similar solids, not merely objects with one matching measurement. Imagine doubling the width, height and depth of a cube. Each layer holds four times as many unit cubes, and there are twice as many layers.

8:15The total number of unit cubes is therefore eight times as large. This explains why volume uses the cube of the length factor rather than its square. Our fourth example compares similar cuboids with corresponding widths of four and six centimetres. Dividing six by four gives the length scale factor 3/2. Squaring this gives a surface area factor of 9/4. Multiplying the smaller surface area of eighty by 9/4 gives one hundred and eighty square centimetres.

8:42For volume, cube 3/2 to obtain 27/8. The larger volume is 48 × 27/8. Simplifying 48 ÷ 8 to six makes the multiplication easier. 6 × 27 gives one hundred and sixty-two cubic centimetres. Volume information can also work backwards to a length. Take the cube root of a volume ratio to recover the corresponding length ratio. For instance, a volume ratio of one hundred and twenty-five to sixty-four gives a length ratio of five to four.

9:14The corresponding surface area ratio would then be twenty-five to sixteen. If the smaller solid in that example has height eight centimetres, multiply eight by 5/4. The corresponding larger height is ten centimetres. To compare surface areas, square each part of the length ratio to get twenty-five to sixteen. We have moved from volume to length first, then from length to area. In examinations, commands such as calculate require a numerical result with clear working.

9:43Show that two triangles are similar requires matching geometric facts and reasons. Typical reasons are a common angle, vertically opposite angles, and alternate or corresponding angles between parallel lines. A simple explanation should identify the relationship that justifies your conclusion. For example, doubling all corresponding lengths multiplies volume by eight because the volume scale factor is 2³. These methods must work without a calculator on Paper two, including simple fractions, squares and cube roots.

10:12Paper four allows a calculator, but it still requires the correct relationship and corresponding measurements. Keep exact fractions or full calculator values during the working, rather than rounding early. For answers that are not exact, normally give three significant figures unless the question requests another accuracy. Watch especially for reversed ratios, mismatched sides and using an area factor directly on a length. State the direction, identify corresponding parts and decide whether you are comparing lengths, areas or volumes.

10:41Similarity means equal corresponding angles and proportional corresponding lengths. Its central pattern is one scale factor for length, its square for area, and its cube for volume.