Angles, polygons and properties of shapes – Year 10 GCSE maths revision notes to download as PDF

Worked example 1

angles on a straight line \(112^\circ\) and \(x\): \(x=180-112=68^\circ\).

Isosceles triangle with apex \(40^\circ\): base angles \(\frac{180-40}{2}=70^\circ\) each.

Worked example 2

In triangle \(ABC\), \(AB=AC\) and angle \(BAC=48^\circ\). Work out angle \(ABC\) [3 marks].

The triangle is isosceles, so the base angles are equal. \(\frac{180-48}{2}=66^\circ\). Reason: base angles of an isosceles triangle are equal and angles in a triangle add up to \(180^\circ\).

Worked example 3

Work out the size of each exterior angle and each interior angle of a regular 9-sided polygon [3 marks].

Exterior \(=\frac{360}{9}=40^\circ\). Interior \(=180-40=140^\circ\).

Worked example 4

A parallelogram \(PQRS\) has angle \(P=3x\) and angle \(Q=2x+20\). Find \(x\) [3 marks].

Adjacent angles add to \(180^\circ\): \(5x+20=180\), \(x=32\). Angle \(P=96^\circ\) and angle \(Q=84^\circ\).

Worked example 5: bearings and back bearings

The bearing of B from A is \(065^\circ\). Find the bearing of A from B [2 marks].

Draw a north line at each point. The north lines are parallel, so the angle between AB and the north line at B is a co-interior angle with \(65^\circ\), which gives \(180-65=115^\circ\) on one side. The bearing is measured clockwise from north all the way round, so the bearing of A from B is \(180+65=245^\circ\). A quick rule: if the bearing is less than \(180^\circ\) add \(180^\circ\), and if it is greater than \(180^\circ\) subtract \(180^\circ\).

Worked example 6: finding the number of sides

Each interior angle of a regular polygon is \(140^\circ\). How many sides has it? [3 marks]

The exterior angle is \(180-140=40^\circ\) because interior and exterior angles lie on a straight line. The exterior angles of any polygon add up to \(360^\circ\), so the number of sides is \(360\div40=9\). Check by finding the interior angle of a regular 9-sided polygon: \((9-2)\times180\div9=1260\div9=140^\circ\). This agrees.

Worked example 7: an algebraic angle problem

In a parallelogram, two adjacent angles are \((3x+15)^\circ\) and \((2x+25)^\circ\). Find \(x\) and all four angles [4 marks].

Adjacent angles in a parallelogram are co-interior, so they add up to \(180^\circ\): \(3x+15+2x+25=180\). Collect like terms: \(5x+40=180\), so \(5x=140\) and \(x=28\). The angles are \(3\times28+15=99^\circ\) and \(2\times28+25=81^\circ\). Opposite angles are equal, so the four angles are \(99^\circ\), \(81^\circ\), \(99^\circ\), \(81^\circ\), which add up to \(360^\circ\).

Worked example 8: congruence reasoning

ABCD is a rectangle and the diagonals AC and BD cross at E. Prove that triangles ABE and DCE are congruent [4 marks].

In a rectangle, opposite sides are equal, so \(AB=DC\). The diagonals of a rectangle are equal and bisect each other, so \(AE=DE\) and \(BE=CE\). All three pairs of corresponding sides are equal, so the triangles are congruent by SSS. Always write the reason next to each statement, and name the triangles in matching order so that the corresponding vertices line up.