Polygons Worked ExamplesIGCSE Maths 0580

(a) An $n$-sided polygon splits into $n - 2$ triangles, each contributing $180^{\circ}$.

(b) In a regular polygon all interior angles are equal, so divide.

(a) The exterior angles of any polygon always sum to $360^{\circ}$, however many sides it has.

(b) An interior angle and its exterior angle lie on a straight line.

Convert to the exterior angle first — that route avoids the algebra entirely.

Divide $360^{\circ}$ by the exterior angle.

Check with the interior-angle formula: $\frac{(20-2) \times 180}{20} = \frac{3240}{20} = 162$. ✓

Find the total angle sum for a pentagon.

Subtract the four known angles.

Set the formula equal to the given sum.

Divide both sides by $180$.

Add $2$.

(a) Use the exterior-angle route.

(b) For a shape to tessellate alone, its interior angle must divide exactly into $360^{\circ}$.

Two octagons give $270^{\circ}$ and three give $405^{\circ}$ — one leaves a gap, the other overlaps.

(c) A square has an interior angle of $90^{\circ}$. Add it to two octagon angles.

(a) The exterior angles of any polygon sum to $360^{\circ}$ — the hexagon does not need to be regular.

Collect the $x$ terms and the numbers, then solve.

(b) Each interior angle and its exterior angle lie on a straight line, so interior $= 180^{\circ} -$ exterior. The largest interior angle goes with the smallest exterior angle, which is $x = 45^{\circ}$ (at $B$ and at $D$).

(c) The exterior angle at $C$ is $2x = 90^{\circ}$.

An angle of $90^{\circ}$ between the sides means $BC$ is perpendicular to $CD$. Check the whole answer: the interior angles are $120^{\circ}, 135^{\circ}, 90^{\circ}, 135^{\circ}, 120^{\circ}, 120^{\circ}$.

(a) A pentagon splits into $3$ triangles, so its angle sum is $3 \times 180^{\circ}$. In a regular pentagon the five angles are equal.

(b) $AB = BC$ (sides of a regular pentagon), so triangle $ABC$ is isosceles with angle $ABC = 108^{\circ}$ at the apex.

(c) By symmetry triangle $AED$ is the same shape, so angle $EAD = 36^{\circ}$ too. The whole angle $BAE$ is $108^{\circ}$.

(d) In the isosceles triangle $AED$, angle $ADE = 36^{\circ}$ (the base angles are equal, as in part (b)).

$AD$ crosses $AC$ and $ED$. Angles $CAD$ and $ADE$ are on opposite sides of $AD$, so they are alternate angles. They are equal, so the lines are parallel.

(a) Each interior angle of a regular pentagon is found from the angle sum.

At the marked vertex, two pentagon angles and $y$ fit together round a point, so they sum to $360^{\circ}$.

(b) $y$ is an interior angle of the regular polygon $P$, so its exterior angle is $180^{\circ} - 144^{\circ} = 36^{\circ}$. The exterior angles of $P$ sum to $360^{\circ}$.

Each side of $P$ is also a side of one pentagon, so the ring has one pentagon per side of $P$: $10$ pentagons.

(c) Repeat the method with regular nonagons.

The inner polygon would be a regular polygon with interior angle $80^{\circ}$, so exterior angle $100^{\circ}$.

The number of sides must be a whole number, so no regular polygon has an interior angle of $80^{\circ}$ and the ring cannot close. (Between the equilateral triangle, $60^{\circ}$, and the square, $90^{\circ}$, there is no other regular polygon.)

The three angles fit exactly round a point, so they sum to $360^{\circ}$. Share $360$ in the ratio $5 : 12 : 13$, which has $5 + 12 + 13 = 30$ parts.

Work out each interior angle.

For each regular polygon, the exterior angle is $180^{\circ}$ minus the interior angle, and the exterior angles sum to $360^{\circ}$.

Repeat for tile $R$.

Tile $P$ has interior angle $60^{\circ}$, so exterior angle $120^{\circ}$: it is an equilateral triangle.

(a) All the sides of a regular polygon are equal, so $PQ = QR$. Triangle $PQR$ is isosceles, and its base angles, at $P$ and $R$, are equal.

(b) Use the angle sum of triangle $PQR$.

Angle $PQR$ and angle $QRS$ are both interior angles of the regular polygon, so they are equal. Angle $QRS$ is made of $x$ and the $144^{\circ}$.

Solve.

(c) Each interior angle is $144 + 12 = 156^{\circ}$, so each exterior angle is $180 - 156 = 24^{\circ}$.

(a) One pair of parallel sides makes it a trapezium. A trapezium only has a line of symmetry when its two sloping sides are equal, and then the line joins the midpoints of the parallel sides.

(b) A half turn maps it onto itself, so opposite sides are equal and parallel. Rectangles, rhombuses and squares all have lines of symmetry, so it must be the general case.

(c) Four equal sides means a square or a rhombus. A square's diagonals are equal, so this is the other one.

(d) Perpendicular diagonals happen in a square, a rhombus and a kite. In a square and a rhombus the diagonals bisect each other; in a kite only the diagonal that is the line of symmetry cuts the other one in half.