Geometry and measures, Year 11: transformations and circle theorems

(−7, −5)

HintFind the step from the centre to A, multiply it by the scale factor, and take the new step from the centre. It will point the opposite way.

The whyFrom the centre (1, 1) to A is 4 right and 3 up. Multiplying by −2 gives 8 left and 6 down. Starting again from the centre: (1 − 8, 1 − 6) = (−7, −5). The image is on the other side of the centre and twice as far from it.

A rotation of 180° about (0, 0)

HintEvery vertex ends up the same distance from the centre, but straight through it on the far side.

The whyScale factor −1 sends each point straight through the centre to the same distance on the other side, so the shape keeps its size and is turned upside down. That is a half turn about the centre.

6

HintA length cannot be negative: the sign of the scale factor tells you about position, not size.

The whyEach length is multiplied by 3, the size of the scale factor, so 2 cm becomes 6 cm. The minus sign means the image lies on the opposite side of the centre and is upside down compared with the original.

A rotation of 180° about the origin (0, 0)

HintFollow one point, such as (2, 3), through both steps and compare where it started with where it ends.

The whyReflecting in the x-axis changes (x, y) to (x, −y); reflecting that in the y-axis gives (−x, −y). Every point ends up straight through the origin at the same distance, which is a half turn about the origin.

A translation with 6 on top and 0 underneath (6 to the right)

HintTrack the point (0, 0) through both mirrors, then check that a second point moves the same way.

The why(0, 0) reflects in x = 1 to (2, 0), and that reflects in x = 4 to (6, 0). Any point with x-coordinate x goes to 2 − x and then to 8 − (2 − x) = x + 6. Two reflections in parallel lines give a translation of twice the distance between the lines, here 2 × 3 = 6.

An invariant point

HintThe word describes something that does not vary.

The whyEvery point on a mirror line is invariant under that reflection, and the centre of a rotation is invariant under that rotation. A translation that actually moves the shape has no invariant points at all.

(1, 1); with the order swapped, (1, −5)

HintDo each route one step at a time, writing down the middle position.

The whyReflection first: (1, 2) goes to (1, −2), then up 3 to (1, 1). Translation first: (1, 2) goes up 3 to (1, 5), then reflects to (1, −5). The two routes disagree, so the order of a combination of transformations matters.

Side lengths, angles and area stay the same (the image is congruent); its position and the way it faces can change

HintNone of these three moves stretches or shrinks anything.

The whyRotations, reflections and translations all produce an image congruent to the original, so any combination of them does too. Only an enlargement changes lengths and area, and even then the angles stay the same.

35

HintDecide the angle at T first: it is where a radius meets the tangent.

The whyA tangent is perpendicular to the radius at the point of contact, so angle OTP is 90°. The angles of triangle OTP add to 180°, so angle OPT = 180 − 90 − 55 = 35°.

70° — tangents from the same point are equal in length, so triangle PAB is isosceles

HintCompare the lengths PA and PB before you think about angles.

The whyPA = PB, so the angles at A and B in triangle PAB are equal. They share 180 − 40 = 140° between them, which gives 70° each.

6

HintDraw the perpendicular from the centre to the chord and a radius to one end of the chord: a right-angled triangle appears.

The whyThe perpendicular from the centre bisects the chord, so each half is 8 cm. With the 10 cm radius as hypotenuse, the distance d obeys d² + 8² = 10², so d² = 36 and d = 6 cm.

Angle ACB

HintThe chord AB cuts the circle into two parts. Look across AB to the part that does not hold the 50° you were given.

The whyThis is the alternate segment theorem: the angle between a tangent and a chord equals the angle in the opposite (alternate) segment standing on that chord. The chord here is AB, so the equal angle is the one at C.

The alternate segment theorem

HintIts name mentions the region on the far side of the chord, using a word that means 'the other one'.

The whyA chord cuts a circle into two segments. The angle between tangent and chord on one side equals the angle standing on that chord in the segment on the other side, the 'alternate' one.

130°

HintOAPB is a four-sided shape, and you know what happens where each radius meets a tangent.

The whyAngles OAP and OBP are both 90°, because a tangent is perpendicular to the radius at the point of contact. The four angles of quadrilateral OAPB add to 360°, so angle AOB = 360 − 90 − 90 − 50 = 130°.

The angle at the centre standing on AB is the straight angle AOB, 180°, so the angle at the circumference is half of it, 90°

HintA, O and B lie in one line. How many degrees is that at O?

The whyAngle ACB stands on the arc AB that does not contain C. The angle at the centre on the same arc is AOB, and because AB is a diameter this is a straight line, 180°. Halving gives 90°: the angle in a semicircle is a special case of the centre theorem.

One angle at O between OA and OC is 2x, so the angle on its other side is 360° − 2x; angle ADC is half of this, 180° − x

HintThink about everything that meets at O going once all the way round.

The whyAngle ABC stands on arc ADC and angle ADC stands on arc ABC; together the two arcs make the whole circle. The two angles at the centre on those arcs fill the full turn at O, so they are 2x and 360° − 2x. Halving the second gives angle ADC = 180° − x, so the opposite angles add up to 180°.

RHS

HintEach triangle has a 90° corner where a radius meets a tangent, and the side the triangles share is opposite that corner.

The whyRight angle, Hypotenuse, Side: both triangles are right-angled, they share the hypotenuse OP, and OA = OB. Congruent triangles have all matching sides equal, so PA = PB: tangents from an external point are equal in length.

half

HintUse the theorem that links an angle at the circumference to the angle at the centre on the same arc.

The whyBoth angles stand on the same arc AB, and the angle at the centre on that arc is AOB. Each angle at the circumference is half of AOB, so the two must equal each other: angles in the same segment are equal.

Angle ABC + angle ADC = 180° (opposite angles of a cyclic quadrilateral) and angle ABC + angle CBE = 180° (angles on a straight line), so angle CBE = angle ADC

HintFind two different facts that each pair the corner at B, inside the shape, with a partner making 180.

The whyBoth angle ADC and angle CBE are what is left when angle ABC is taken from 180°, so they are equal. The result is often stated as: the exterior angle of a cyclic quadrilateral equals the interior opposite angle.