End of chapter exercises

We look at the side labels. In diagram A we note that the three pairs of corresponding sides are in different proportions. In diagram B we note the three pairs of corresponding sides are in proportion.

Therefore diagram B gives a pair of triangles that are similar.

Diagram A shows a pair of triangles with all pairs of corresponding angles equal (the same three angle markers are shown in both triangles). Diagram B shows a pair of triangles with different angles in each triangle. All six angles are different and there are no pairs of corresponding angles that are equal.

Therefore diagram A gives a pair of triangles that are similar.

We are given one angle that is equal. We are not given any equal sides (we do not know if \(x = c\)). To determine if two triangles are congruent we need to have three pieces of information (recall that the reasons for congruent triangles are: SSS, SAS, AAS and RHS). Therefore we cannot state whether or not the triangles are congruent.

Therefore, there is not enough information to determine if the two triangles are congruent.

The sides of both triangles are labelled with \(m\), \(n\) and \(p\). This means that there are three pairs of corresponding and equal sides.

Therefore, these two triangles are congruent (\(\triangle VWU \equiv \triangle YZX\)), reason: SSS.

\begin{align*} \frac{ED}{BA} &= \frac{18}{32} = \frac{9}{16} \\ \frac{DF}{AC} &= \frac{32}{64} = \frac{9}{16} \\ \frac{EF}{BC} &= \frac{45}{80} = \frac{9}{16} \end{align*}

All three pairs of sides are in proportion, \(\therefore \triangle ABC \enspace|||\enspace \triangle DEF\).

\begin{align*} y & = 30° \qquad \text{ (vert opp }\angle\text{s} = \text{)}\\ \hat{P} & = \hat{Q} \qquad \text{(}\angle \text{s opp equal sides)}\\ \text{and }\hat{S} & = \hat{T} \qquad \text{(}\angle \text{s opp equal sides)} \end{align*}

However \(\hat{P} + \hat{Q} + 30° = 180°\) (sum of \(\angle\)s in \(\triangle\)). Therefore \(\hat{P} + \hat{Q} = 150°\).

Similarly \(\hat{S} + \hat{T} = 150°\).

But \(\hat{P} = \hat{Q}\) so \(2\hat{P} = 150°\) and \(\hat{S} = \hat{T}\) so \(2\hat{S} = 150°\). Therefore \(\hat{P} = \hat{S}\).

Therefore \(\triangle PQR \enspace|||\enspace \triangle TRS\) (AAA).

Now we can use the fact that the sides are in proportion to find \(x\):

\begin{align*} \frac{x}{\text{4,8}} &= \frac{\text{3,5}}{\text{6,1}} \\ \therefore x &= \text{2,75} \end{align*}

Therefore \(x = \text{2,75}\) and \(y = 30°\).

We start by counting the number of sides. There are four sides in this figure and so it is either just a quadrilateral or one of the special types of quadrilateral.

Next we ask ourselves if there are any parallel lines in the figure. You can look at the figure to see if any of the lines look parallel or make a quick sketch of the image and see if any pairs of opposite lines meet at a point.

Both pairs of opposite sides are not parallel. This means that the figure can only be one of the following: trapezium, kite or quadrilateral.

Next we ask ourselves if one of the pairs of opposite sides is parallel, while the other is not. Neither of the two pairs of opposite sides is parallel so we must now look to see if both pairs of adjacent sides are equal in length. Both pairs of adjacent sides are equal in length. So this is a kite.

Therefore this is a kite.

Both pairs of opposite sides are parallel. That means that this shape can belong to one or more of these groups: square, rhombus, rectangle, and/or parallelogram.

The shape shown is a rhombus. It is certainly a quadrilateral (because it has four sides). It is also a parallelogram, because the opposite sides are parallel to each other. The rhombus is not a rectangle or a square because it does not have right angles. However, the rhombus is a kite, because it has two pairs of adjacent sides which are equal in length. And finally, it is a trapezium because it has a pair of opposite sides which are parallel.

Therefore the correct answer is: rhombus, parallelogram, kite, trapezium and quadrilateral.

\(\hat{E_2} = 90^{\circ}\) (diagonals of a rhombus bisect at right angles)

\begin{align*} \hat{F_1} + \hat{G_1} + 90^{\circ} &= 180^{\circ} \qquad \text{ (sum of } \angle \text{s in } \triangle\text{)} \\ 3x + 20^{\circ} + x + 10^{\circ} &= 90^{\circ}\\ 4x &= 60^{\circ} \\ \therefore x &= 15^{\circ} \end{align*}

The red line, \(TS\) or \(ST\), indicates the third side of the triangle. According to the mid-point theorem, the line joining the mid-points of two sides of a triangle is parallel to the third side of the triangle.

The line segments \(YZ\) and \(VX\) are parallel according to the mid-point theorem because segment \(VX\) is bisecting the line segments \(WZ\) and \(WY\).

From the mid-point theorem we know:

\begin{align*} VW & = 2 \times XZ \\ \left( 10 \right) & = 2x \\ \frac{1}{2} \left( 10 \right) & = x \\ 5 & = x \end{align*}

We are given that \(GH \parallel LJ\). The length of \(JL\) is 5 and the length of \(GH\) is 10, therefore \(JL = \frac{1}{2}GH\).

Therefore we know from the mid-point theorem that \(L\) is the mid-point of \(GK\) and \(J\) is the mid-point of \(HK\).

Therefore \(HJ = JK = 9\).

You can see in the figure that the segment \(LJ\) is not parallel to \(GH\). This means that the mid-point theorem cannot apply to this triangle. There are no other options to use either: this question cannot be solved. There is no solution.

In \(\triangle TRS\) and \(\triangle PTQ\):

\begin{align*} \hat{T} & = \hat{T} \qquad \text{(common }\angle \text{)} \\ T\hat{R}S & = \hat{P} \qquad \text{(corresp } \angle \text{s; } RS \parallel PQ \text{)} \\ T\hat{S}R & = \hat{Q} \qquad \text{(corresp } \angle \text{s; } RS \parallel PQ \text{)} \end{align*}

Therefore \(\triangle TRS \enspace ||| \enspace \triangle PTQ\) (AAA).

Therefore the sides are in proportion.

\begin{align*} \frac{TR}{TP} & = \frac{TS}{TQ} \\ \frac{a}{a + 15} &= \frac{\frac{b}{4}}{b} \\ \frac{a}{a + 15} &= \frac{1}{4} \\ a &= (a + 15) \left(\frac{1}{4}\right) \\ 4a & = a + 15\\ 3a & = 15\\ \therefore a & = 5\\\\ b & = \frac{b}{4} + 9\\ 4b & = b + 36\\ 3b & = 36 \\ \therefore b & = 12 \end{align*}

Therefore: \(a = 5\) and \(b=12\).

We are given two equilateral triangles, therefore in \(\triangle PSR\): \(PS = SR = PR\) and in \(\triangle PQR\): \(PQ = QR = PR\).

But \(PR\) is a common side and so \(PR = PS = SR = PQ = QR\).

Also in each triangle all the interior angles are equal to 60°. Therefore \(\hat{P}_{1} = \hat{R}_{2}\) and \(\hat{P}_{2} = \hat{R}_{2}\). Therefore \(PQ \parallel SR\) and \(PS \parallel QR\) (alt. int. \(\angle\)'s equal).

\(\therefore PQRS\) is a rhombus (all sides are equal in length, both pairs of opposite sides parallel).

\(DE \parallel BC\) (given).

\(e = c\) (alt \(\angle\)s; \(DE \parallel BC\)).

\(d = b\) (alt \(\angle\)s; \(DE \parallel BC\)).

\(d + a + e = 180°\) (\(\angle\)s on str line).

And we have shown that \(e = c\) and \(d = b\) therefore we can replace \(d\) with \(b\) and \(e\) with \(c\) to get:

\(a + b + c = 180°\).

Therefore the angles in a triangle add up to \(180°\).

\begin{align*} B\hat{C}D & = B\hat{E}C \qquad \text{(}\angle \text{s opp equal sides} \\ A\hat{B}E & = B\hat{E}C \text{ (alt } \angle\text{s; } AB \parallel DC \text{)} \\ \therefore A\hat{B}E & = B\hat{C}D \end{align*}