Sec 1 Maths: Triangles, Quadrilaterals & Polygons — practice questions & worked solutions

Interior $+$ exterior $= 180^\circ$, split in the ratio $13 : 2$ ($15$ parts):

$$\begin{aligned}\text{exterior angle} &= \frac{2}{15} \times 180^\circ\\ &= 24^\circ\end{aligned}$$

Number of sides:

$$\begin{aligned}n &= \frac{360^\circ}{24^\circ}\\ &= 15\end{aligned}$$

(i) Sum of interior angles of an $n$-sided polygon $=(n-2)\times 180^\circ$. The polygon has $6$ sides:

$$(6-2)\times 180^\circ=720^\circ$$

(ii)(a) $QR$ (the horizontal segment at height $y=3$).

(ii)(b) $TU$ (the segment with gradient $1$ and $y$-intercept $-3$).

Sum of interior angles $=(n-2)\times 180^\circ$:

$$\begin{aligned}(n-2)\times 180 &= 84+126+150+(n-3)\times 120\\ 180n-360 &= 360+120n-360\\ 180n-360 &= 120n\\ 60n &= 360\\ n &= 6\end{aligned}$$

Convert the given interior angles to exterior angles:

$$\begin{aligned}180^\circ – 86^\circ &= 94^\circ \\ 180^\circ – 115^\circ &= 65^\circ \\ 180^\circ – 153^\circ &= 27^\circ\end{aligned}$$

Let $k$ be the number of $153^\circ$ interior angles. The exterior angles of a polygon sum to $360^\circ$:

$$\begin{aligned}3(40^\circ) + 94^\circ + 65^\circ + 27^\circ k &= 360^\circ \\ 120^\circ + 159^\circ + 27^\circ k &= 360^\circ \\ 27^\circ k &= 81^\circ \\ k &= 3\end{aligned}$$

Total number of sides:

$$\begin{aligned}n &= 3 + 2 + 3 \\ n &= 8\end{aligned}$$

(a)

Angle sum of a triangle:

$$\begin{aligned}7x + (5x – 11) + (3x – 4) &= 180\\ 15x – 15 &= 180\\ 15x &= 195\\ x &= 13\end{aligned}$$

(b)(i)

Sum of interior angles $= (7 – 2) \times 180^\circ$:

$$\begin{aligned}&= 5 \times 180^\circ = 900^\circ\end{aligned}$$

(b)(ii)

Sum of exterior angles $= 360^\circ$:

$$\begin{aligned}4(50) + (x + 7) + (x + 9) + (2x + 12) &= 360\\ 200 + 4x + 28 &= 360\\ 4x &= 132\\ x &= 33\end{aligned}$$

The smallest exterior angle gives the largest interior angle. Exterior angles: $50^\circ$, $(33 + 7) = 40^\circ$, $(33 + 9) = 42^\circ$, $(2(33) + 12) = 78^\circ$. Smallest exterior $= 40^\circ$:

$$\begin{aligned}\text{Largest interior angle} &= 180^\circ – 40^\circ = 140^\circ\end{aligned}$$

In a regular polygon all sides are equal, so $BC = CD$. A triangle with two equal sides is isosceles; hence triangle $BCD$ is isosceles.

Let the interior angle of the regular polygon be $\angle ABC$. Triangle $BCD$ is isosceles with $BC = CD$, so $\angle CBD = \angle CDB$. With interior angle $\angle BCD = \angle ABC$:

$$\begin{aligned}\angle CBD&=\frac{180-\angle BCD}{2}\end{aligned}$$

Since $\angle ABD = \angle ABC – \angle CBD = 146.25^\circ$, and $\angle BCD = \angle ABC$, let the interior angle be $I$:

$$\begin{aligned}\angle CBD&=\frac{180-I}{2}=90-\frac{I}{2}\\ I-\left(90-\frac{I}{2}\right)&=146.25\\ \frac{3I}{2}-90&=146.25\\ \frac{3I}{2}&=236.25\\ I&=157.5^\circ\end{aligned}$$

For a regular $n$-sided polygon the interior angle is $\dfrac{(n-2)180}{n}$:

$$\begin{aligned}\frac{(n-2)180}{n}&=157.5\\ 180n-360&=157.5n\\ 22.5n&=360\\ n&=16\end{aligned}$$

The sum of exterior angles of any polygon is $360^\circ$. There are 6 remaining exterior angles of $55^\circ$ each:

$$\begin{aligned}(x + 30) + (2x – 6) + (5x + 14) + 6(55) &= 360 \\ 8x + 38 + 330 &= 360 \\ 8x + 368 &= 360 \\ 8x &= -8 \\ x &= -1\end{aligned}$$

This gives $(2x – 6)^\circ = (2(-1) – 6)^\circ = -8^\circ$, a negative angle. Since an exterior angle of a polygon cannot be negative, his claim is not accurate.

Interior angle is 14 times the exterior angle, and they sum to $180^\circ$:

$$\begin{aligned}\text{exterior} + 14 \times \text{exterior} &= 180^\circ\\ 15 \times \text{exterior} &= 180^\circ\\ \text{exterior angle} &= 12^\circ\end{aligned}$$

Interior angle $= 14 \times 12^\circ = 168^\circ$.

(i) Angle $CBG$

$\angle ABC = 168^\circ$ is the interior angle, and $\angle CBG$ is its supplement on the straight line $ABG$:

$$\begin{aligned}\angle CBG &= 180^\circ – 168^\circ\\ &= 12^\circ\end{aligned}$$

(ii) Angle $DFG$

Triangle $BCG$ is isosceles with $BC = CG$, so the base angles are equal:

$$\begin{aligned}\angle CBG &= \angle CGB = 12^\circ\\ \angle BCG &= 180^\circ – 12^\circ – 12^\circ = 156^\circ\end{aligned}$$

At vertex $C$ the interior angle of the polygon $\angle BCD = 168^\circ$. Angle $GCD$ (angle of the rhombus at $C$):

$$\begin{aligned}\angle GCD &= 360^\circ – \angle BCG – \angle BCD\\ &= 360^\circ – 156^\circ – 168^\circ\\ &= 36^\circ\end{aligned}$$

In rhombus $CDFG$, opposite angles are equal, so:

$$\begin{aligned}\angle DFG &= \angle DCG = 36^\circ\end{aligned}$$

(a)(i) Angle $SPQ$

$$\begin{aligned}\angle PSR &= 180^\circ – 135^\circ = 45^\circ \quad (\text{angles on a straight line}) \\ \angle SPQ &= 180^\circ – 45^\circ = 135^\circ \quad (\text{interior angles, } PQ \parallel SR)\end{aligned}$$

(Equivalently, $\angle SPQ = \angle PST = 135^\circ$ by alternate angles.)

(a)(ii) Number of sides

Interior angle of the regular polygon $= \angle PST = 135^\circ$, so exterior angle $= 45^\circ$:

$$\begin{aligned}n &= \frac{360^\circ}{45^\circ} \\ &= 8\end{aligned}$$

(b)

The five interior angles (5-sided polygon) sum to $(5 – 2) \times 180^\circ = 540^\circ$:

$$\begin{aligned}9 + 5 + 8 + 7 + 11 &= 40 \text{ parts} \\ 1 \text{ part} &= \frac{540^\circ}{40} = 13.5^\circ \\ \text{Largest} – \text{smallest} &= (11 – 5) \times 13.5^\circ \\ &= 81^\circ\end{aligned}$$

(a) At $A$, use a protractor to draw a ray making $30^\circ$ with $AB$; mark $C$ on this ray with $AC = 7.3$ cm, then join $BC$.

(b) With compass radius more than half of $AB$, draw arcs centred at $A$ and at $B$ above and below $AB$; the line through the two arc intersections is the perpendicular bisector of $AB$.

(c) With centre $A$, draw an arc cutting $AB$ and $AC$; from those two points draw equal arcs that intersect; the line from $A$ through this intersection bisects angle $BAC$.

(d) The point equidistant from lines $AC$ and $AB$ lies on the angle bisector of $\angle BAC$; the point equidistant from $A$ and $B$ lies on the perpendicular bisector of $AB$. Label their intersection as $M$.

(a)

Exterior angle of each $157^\circ$ interior angle $= 180 – 157 = 23^\circ$. Sum of all exterior angles $= 360^\circ$. Let there be $k$ exterior angles of $22^\circ$:

$$\begin{aligned}8(23)+22k &= 360 \\ 184+22k &= 360 \\ 22k &= 176 \\ k &= 8\end{aligned}$$

Number of sides $= 8 + 8 = 16$.

(b)(i) $\angle BCW$

Interior angle of regular pentagon $= \dfrac{(5-2)\times 180}{5} = 108^\circ$, so $\angle BCD = 108^\circ$. Interior angle of regular hexagon $= \dfrac{(6-2)\times 180}{6} = 120^\circ$, so $\angle DCW = 120^\circ$. Angles round the point $C$:

$$\begin{aligned}\angle BCW &= 360-108-120 \\ &= 132^\circ\end{aligned}$$

(b)(ii)

For a regular polygon the exterior angle $= 180 – 132 = 48^\circ$ must divide $360^\circ$ exactly:

$$\begin{aligned}\frac{360}{48} &= 7.5\end{aligned}$$

Since this is not a whole number, $132^\circ$ cannot be an interior angle of a regular polygon.

Construct $DE = 8.5$ cm. At $E$ draw $\angle DEF = 65^\circ$. From $D$ mark an arc of radius $DF = 9.2$ cm to cut the $65^\circ$ ray at $F$; join $DF$.

Measuring the constructed side:

$$EF = 8.6 \text{ cm}\ (\pm 0.1)$$

(a)

Each exterior angle $= 180^\circ – 144^\circ = 36^\circ$:

$$n = \frac{360^\circ}{36^\circ}$$
$$n = 10$$

(b)

The four interior angles of $152^\circ$ give exterior angles of $28^\circ$ each. Let $k$ be the number of $22^\circ$ exterior angles. Sum of exterior angles $= 360^\circ$:

$$4(28^\circ) + 6(23^\circ) + 22^\circ k = 360^\circ$$
$$112^\circ + 138^\circ + 22^\circ k = 360^\circ$$
$$22^\circ k = 110^\circ$$
$$k = 5$$

Number of sides $= 4 + 6 + 5 = 15$.

Construct $\triangle XYZ$ with $XY = 10$ cm: with centre $X$ draw an arc of radius $8.6$ cm, and with centre $Y$ draw an arc of radius $9$ cm; their intersection is $Z$.

Measuring the constructed angle:

$$\angle YXZ = 58^\circ \ (\pm 1^\circ)$$