This note covers similar shapes, including how lengths, areas and volumes scale, and the circle theorems used to find angles in circles. Similarity lets you find unknown lengths from a model or diagram, and circle theorems give a toolkit for proving and calculating angles.
Key ideas
Similar shapes. Two shapes are similar if one is an enlargement of the other. Their corresponding angles are equal and their corresponding sides are in the same ratio. This ratio is the linear scale factor \(k\).
Similar triangles. Two triangles are similar if either of these is true:
- Their angles are equal in pairs (two equal angles is enough, because the third must then match).
- Their three pairs of sides are in the same ratio.
A line drawn parallel to one side of a triangle creates a smaller similar triangle, because of corresponding angles.
Scaling lengths, areas and volumes. If the linear scale factor is \(k\), then
- lengths are multiplied by \(k\),
- areas (including surface areas) are multiplied by \(k^2\),
- volumes are multiplied by \(k^3\).
If you are given the ratio of areas, take the square root to get the ratio of lengths. If you are given the ratio of volumes, take the cube root. The solids must be similar, which means every dimension scales by the same factor. Also, the masses of similar solids made of the same material are in the ratio of their volumes.
Congruent shapes. Congruent triangles are identical in shape and size. Two triangles are congruent if they satisfy one of the conditions SSS, SAS, ASA (or AAS) or RHS (right angle, hypotenuse, side).
Circle theorems. For each theorem, learn the statement and be able to give it as a reason.
- The angle at the centre is twice the angle at the circumference, when both stand on the same arc.
- The angle in a semicircle is a right angle.
- Angles in the same segment (standing on the same chord) are equal.
- Opposite angles of a cyclic quadrilateral (all four vertices on the circle) add up to \(180^\circ\).
- A tangent is perpendicular to the radius at the point of contact.
- Tangents from an outside point to a circle are equal in length.
- A perpendicular from the centre to a chord bisects the chord.
- Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment, which is the segment on the other side of the chord.
Using the theorems. Mark the centre, find a triangle of two radii (isosceles), and look for tangents (right angles) and cyclic quadrilaterals. Work step by step and state one theorem per step.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Linear scale factor \(k\) | Ratio of corresponding lengths in similar shapes. |
| Area scale factor \(k^2\) | Ratio of areas of similar shapes. |
| Volume scale factor \(k^3\) | Ratio of volumes of similar solids. |
| Angle at centre \(= 2\times\) angle at circumference | Both angles stand on the same arc. |
| Angle in a semicircle \(= 90^\circ\) | The chord is a diameter. |
| Cyclic quadrilateral | Opposite angles sum to \(180^\circ\). |
| Alternate segment theorem | Tangent-chord angle equals the angle in the opposite segment. |
| Tangents from a point | Equal in length; each meets its radius at \(90^\circ\). |
Worked examples
Example 1 Two similar bottles have heights 20 cm and 30 cm. The smaller has a surface area of 250 cm² and a volume of 400 cm³. Find the surface area and volume of the larger bottle. Then, for two other similar containers with volumes in the ratio \(8 : 27\), find the ratio of their heights and of their surface areas.
Linear scale factor \(k = \dfrac{30}{20} = 1.5\).
Surface area scale factor \(k^2 = 2.25\), so the larger surface area is \(250 \times 2.25 = 562.5\) cm².
Volume scale factor \(k^3 = 3.375\), so the larger volume is \(400 \times 3.375 = 1350\) cm³.
For the volume ratio \(8 : 27\), take cube roots: heights are in the ratio \(2 : 3\).
Surface areas are in the ratio \(2^2 : 3^2 = 4 : 9\).
Answer: 562.5 cm² and 1350 cm³; heights \(2:3\); surface areas \(4:9\)
Example 2 In triangle \(ABC\), \(D\) lies on \(AB\) and \(E\) lies on \(AC\) with \(DE\) parallel to \(BC\). \(AD = 4\) cm, \(DB = 6\) cm and \(DE = 5\) cm. (a) Find \(BC\). (b) The area of triangle \(ADE\) is 7 cm². Find the area of the trapezium \(DBCE\).
(a) Triangles \(ADE\) and \(ABC\) are similar: angle \(A\) is shared, and angles \(ADE\) and \(ABC\) are corresponding angles between parallel lines.
\(AB = 4 + 6 = 10\) cm, so the scale factor from \(ADE\) to \(ABC\) is \(k = \dfrac{10}{4} = 2.5\).
\(BC = 5 \times 2.5 = 12.5\) cm.
(b) Area scale factor \(= 2.5^2 = 6.25\). Area of \(ABC = 7 \times 6.25 = 43.75\) cm².
Trapezium \(DBCE = 43.75 - 7 = 36.75\) cm².
Answer: (a) 12.5 cm (b) 36.75 cm²
Example 3 (a) Points \(A, B, C, D\) lie in order on a circle with centre \(O\). Angle \(AOC = 132^\circ\), and \(B\) lies on the major arc \(AC\). Find angles \(ABC\) and \(ADC\). (b) Tangents from a point \(T\) touch a circle with centre \(O\) at \(A\) and \(B\). Angle \(ATB = 50^\circ\). Find angle \(AOB\), the angle between the tangent \(TA\) and the chord \(AB\), and state the angle \(ACB\) for \(C\) on the major arc.
(a) Angle \(ABC\) stands on the same arc as the centre angle \(AOC\), so it is half of it: \(132^\circ \div 2 = 66^\circ\) (angle at centre is twice the angle at the circumference).
\(ABCD\) is a cyclic quadrilateral, so opposite angles add up to \(180^\circ\): angle \(ADC = 180^\circ - 66^\circ = 114^\circ\).
(b) Radii \(OA\) and \(OB\) meet the tangents at \(90^\circ\). In quadrilateral \(OATB\), angle \(AOB = 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ\).
Tangents from \(T\) are equal, so triangle \(TAB\) is isosceles with \(TA = TB\). The angle between tangent \(TA\) and chord \(AB\) is \((180^\circ - 50^\circ) \div 2 = 65^\circ\).
By the alternate segment theorem, angle \(ACB = 65^\circ\) for \(C\) on the major arc. This agrees with the angle at the centre: \(130^\circ \div 2 = 65^\circ\).
Answer: (a) \(66^\circ\) and \(114^\circ\) (b) \(130^\circ\), \(65^\circ\) and \(65^\circ\)
Common errors
- Common error: using the linear scale factor for areas or volumes. Square it for area and cube it for volume.
- Common error: matching the wrong sides in similar triangles. List the corresponding vertices in order first, for example \(A \to A\), \(D \to B\), \(E \to C\).
- Common error: using the angle-at-centre theorem when the two angles do not stand on the same arc.
- Common error: saying opposite angles of any quadrilateral add up to \(180^\circ\). That is only true when the quadrilateral is cyclic.
- Common error: giving a reason such as "circle theorem" without naming it. Always state the actual theorem.
Exam tip: When a question says "give reasons" or "show that", write each step as a short line with its theorem, for example "angle \(OAT = 90^\circ\) (tangent and radius)".
Check yourself
- Two similar cones have heights 6 cm and 15 cm. The smaller has volume 80 cm³. Find the volume of the larger.
Answer
\(k = \tfrac{15}{6} = 2.5\), so the volume scale factor is \(2.5^3 = 15.625\). Volume \(= 80 \times 15.625 = 1250\) cm³. - Two similar shapes have areas 36 cm² and 100 cm². The perimeter of the smaller is 24 cm. Find the perimeter of the larger.
Answer
The area ratio is \(36 : 100\), so the linear ratio is \(6 : 10\), a scale factor of \(\tfrac{10}{6}\). Perimeter \(= 24 \times \tfrac{10}{6} = 40\) cm. - \(AB\) is a diameter of a circle and \(C\) is on the circumference. Angle \(CAB = 35^\circ\). Find angle \(ABC\).
Answer
Angle \(ACB = 90^\circ\) (angle in a semicircle). So angle \(ABC = 180^\circ - 90^\circ - 35^\circ = 55^\circ\). - \(PQRS\) is a cyclic quadrilateral. Angle \(PQR = 98^\circ\) and angle \(QRS = 71^\circ\). Find angles \(PSR\) and \(QPS\).
Answer
Opposite angles add up to \(180^\circ\). Angle \(PSR = 180^\circ - 98^\circ = 82^\circ\). Angle \(QPS = 180^\circ - 71^\circ = 109^\circ\).