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Mathematics Year 6: Turning a point and a shape around a centre

Mathematics, Year 6 (Portugal): Turning a point, a triangle or a quadrilateral about a centre through a given angle and direction, using ruler, compass and protractor.

A centre, an angle and a direction

To turn a point P you need three things: the centre O, the angle and the direction (clockwise or anticlockwise). If P is 4 cm from O and you turn it by 40°, 80° and 120°, you get P₁, P₂ and P₃, all 4 cm from O. So they all lie on a circle with centre O and radius 4 cm: the point only travels round the centre, it never gets closer or further away.

Why the compass belongs in the job

A rotation pins a shape to its centre, like a door pinned to its hinges. Every point moves along a circle around the centre, and its distance from the centre never changes. That is why the compass helps: with its point on O and its opening equal to OP, it draws every place where the image of P could be. The angle tells you which of those places is the right one.

A case: turning triangle ABC by 60° about O

Triangle ABC has OA = 3 cm, OB = 5 cm and OC = 4 cm; measured from ray OA, B is at 20° and C at 65°. Turn it 60° anticlockwise. For each vertex: compass on O, opened to the distance to that vertex; protractor on O, 60° from that vertex; mark the image. A', B' and C' end up at 60°, 80° and 125° from OA. Join them. Check: 80° − 60° = 20° and 125° − 60° = 65°, the same as before. A quadrilateral works the same way, with four vertices.

The angle is measured at O, not at P

The reasonable mistake: putting the protractor on P, because P is the point that will move. But the angle of rotation is the angle POP', with its vertex at the centre O. Another mistake: marking P' in the right direction but at a different distance from O; then the point has left the circle and the shape gets distorted. Rule: protractor and compass on O, and at the end check that OP' = OP.

Where you see it: clock hands, wheels and geometry software

The minute hand of a clock is a rotation happening: from 12 to 3 it turns 90° (360° ÷ 4), from 12 to 4 it turns 120°, always clockwise, and its tip draws a circle. The same is true of a point on the rim of a bicycle wheel. In dynamic geometry software you choose the shape, the centre and the angle, and the program does the construction you did by hand.

Test what you learned

  1. Point P is 5 cm from the centre O. You turn P about O through many different angles. Where do all the images of P lie?

    • a) On the straight line through O and P
    • b) On a circle with centre P and radius 5 cm
    • c) On a circle with centre O and radius 5 cm
    • d) On circles of different sizes, bigger for bigger angles

    Correct answer: c) On a circle with centre O and radius 5 cm — Yes: turning never changes the distance to O, so every image is 5 cm from O. That describes a circle centred at O.

  2. OP = 4 cm. Turning P about O by 40° gives P₁; turning it by 130° in the same direction gives P₂. What is angle P₁OP₂, and how far is P₂ from O?

    • a) 170° and 4 cm
    • b) 90° and 4 cm
    • c) 130° and 4 cm
    • d) 40° and 4 cm

    Correct answer: b) 90° and 4 cm — Right: 130° − 40° = 90°, and a rotation never changes the distance to O, so P₂ is still 4 cm from O.

  3. You must turn point P by 60° about O using a protractor and a compass. Where do they go?

    • a) Both on O; the compass, opened to OP, keeps P' at the same distance
    • b) Both on P, measuring 60° from PO
    • c) Protractor on O only; the angle alone places P'
    • d) Protractor on the midpoint of PO, compass on P

    Correct answer: a) Both on O; the compass, opened to OP, keeps P' at the same distance — Right: the angle is measured at the centre O, and the compass on O keeps OP' = OP, so P' stays on the circle.

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