Home Practice
For learners and parents For teachers and schools
Textbooks
Full catalogue
Leaderboards
Learners Leaderboard Classes/Grades Leaderboard Schools Leaderboard
Pricing Support
Help centre Contact us
Log in

We think you are located in United States. Is this correct?

End of chapter exercises

Textbook Exercise 12.2

\(ABCD\) is a rhombus with \(AM = MO\) and \(AN = NO\). Prove \(ANOM\) is also a rhombus.

4061852e9abaf25b71bb23e25f868003.png

In \(\triangle AMO\) and \(\triangle ANO\)

\(\hat{A}_1 = \hat{A}_2\) (given rhombus \(ABCD\), diagonal \(AC\) bisects \(\hat{A}\))

\(\therefore \hat{A_1} = A\hat{O}M\) (\(\angle\)s opp equal sides)

similarly \(\hat{A}_2 = A\hat{O}N\)

\(\therefore \hat{A}_2 = A\hat{O}M\) and \(\hat{A}_1 = A\hat{O}N\)

but these are alternate interior \(\angle\)s

\(\therefore AN \parallel MO\) and \(AM \parallel NO\)

\(\therefore ANOM\) is a parallelogram

\(\therefore AM = NO\) (opp sides of \(\parallel\)m)

\(\therefore AM = MO = ON = NO\)

\(\therefore ANOM\) is a rhombus (all sides equal and two pairs of sides parallel)

\(ABCD\) is a parallelogram with diagonal \(AC\). Given that \(AF = HC\), show that:

e695d05fb757a60e1b9cabd96f56ff24.png

\(\triangle AFD \equiv \triangle CHB\)

\begin{align*} \hat{A}_1 & = \hat{C}_1 \qquad \text{(alt }\angle \text{s; }AD \parallel BC \text{)}\\ AD & = BC \qquad \text{(opp sides } \parallel \text{m)}\\ AF& = HC \qquad \text{(given)}\\ \therefore \triangle AFD &\equiv \triangle CHB \qquad \text{(SAS)} \end{align*}

\(DF \parallel HB\)

\begin{align*} \hat{F}_1 &= \hat{H}_1 \qquad (\triangle AFD \equiv \triangle CHB)\\ \therefore \hat{F}_1 +\hat{F}_2 &= 180° \qquad \text{(}\angle\text{s on str line)}\\ \text{and } \hat{H}_1 +\hat{H}_2 &= 180° \qquad \text{(}\angle \text{s on str line)}\\ \therefore \hat{F}_1 &= 180° - \hat{F}_2 \\ \text{and } \hat{H}_1 &= 180° - \hat{H}_2 \\ \therefore 180° - \hat{F}_2 & = 180° - \hat{H}_2 \\ \therefore \hat{F}_2 & = \hat{H}_2 \\ \therefore DF &\parallel HB \qquad \text{(corresp } \angle \text{s equal)} \end{align*}

\(DFBH\) is a parallelogram

\begin{align*} FD & = HB \qquad (\triangle AFD \equiv \triangle CHB)\\ \text{and }DF &\parallel HB \qquad \text{ (proved above)}\\ \therefore DFBH & \text{ is a parallelogram (one pair opp sides equal and parallel)} \end{align*}

Given parallelogram \(ABCD\) with \(AE\) bisecting \(\hat{A}\) and \(FC\) bisecting \(\hat{C}\).

b648dcfa077a4220c0ef2f0d984034c3.png

Write all interior angles in terms of \(y\).

First number the angles:

e6d5479aa897cc613c6fffdd6e09d378.png

Prove that \(AFCE\) is a parallelogram.

\begin{align*} AF &\parallel EC \qquad \text{(opp sides of} \parallel \text{m)}\\ \text {and }\hat{C_1} + \hat{E_2} & = y + (180° - y)\\ \therefore & \text{ the sum of the co-interior angles is } 180°\\ \therefore AE & \parallel FC \\ \therefore AFCE & \text{ is a parallelogram (both pairs opp. sides parallel)} \end{align*}

Given that \(WZ = ZY = YX\), \(\hat{W} = \hat{X}\) and \(WX \parallel ZY\), prove that:

a302882713cc4abd23cb521052df6a6e.png

\(XZ\) bisects \(\hat{X}\)

First label the angles:

3d763cfd48a98ce0d181b03f5a9063aa.png

\(WY = XZ\)

\begin{align*} \text{Similarly, }WY &\text{ bisects }\hat{W}\\ \therefore \hat{W_1} &= \hat{W_2}\\ \text{and }\hat{W} &= \hat{X} \text{ (given)}\\ \therefore \hat{W_1} &= \hat{W_2} = \hat{X_1} = \hat{X_2}\\ \text{ and }\hat{W_1} &= \hat{Y_1} ~(\angle \text{s opp equal sides)}\\ \text{In }\triangle WZY&\text{ and }\triangle XYZ\\ WZ&=XY \text{ (given)}\\ ZY &\text{ is a common side}\\ \hat{Z} &= \hat{Y} \text{(third }\angle \text{ in }\triangle \text{)}\\ \therefore \triangle WZY &\equiv \triangle XYZ \text{ (SAS)} \\ \therefore WY & = XZ \end{align*}

\(D\) is a point on \(BC\), in \(\triangle ABC\). \(N\) is the mid-point of \(AD\). \(O\) is the mid-point of \(AB\) and \(M\) is the mid-point of \(BD\). \(NR \parallel AC\).

273c36346c5740761973c1f4069cb5d1.png

Prove that \(OBMN\) is a parallelogram.

\(AO = OB\) (given)

\(AN = ND\) (given)

\(\therefore ON \parallel BD\) (Midpt Theorem)

\(BM = MD\) (given)

\(AN = ND\) (given)

\(\therefore MN \parallel AB\) (Midpt Theorem)

\(\therefore OBMN\) is a parallelogram (both pairs opp. sides parallel)

Prove that \(BC = 2MR\).

\(AN = NC\) (given)

\(NR \parallel AC\) (given)

\(\therefore DR = RC\) (Midpt Theorem)

\(\therefore DR = \frac{1}{2} DC\)

\(MD = \frac{1}{2}BD\) (given)

\(\therefore MD + DR = \frac{1}{2} (BD + DC)\)

\(MR = \frac{1}{2}BC\)

\(\therefore BC = 2MR\)

In \(\triangle MNP\), \(\hat{M} = 90°\), \(S\) is the mid-point of \(MN\) and \(T\) is the mid-point of \(NR\).

652b3227d8367824506122aff72f3596.png

Prove \(U\) is the mid-point of \(NP\).

\begin{align*} NS&=SM \text{ (given)}\\ NT&=TR \text{ (given)}\\ \therefore ST &\parallel MR \text{ (Midpt Theorem)} \\ \therefore U & \text{ is the mid-point of } NP \text{ (converse of Midpt Theorem)} \end{align*}

If \(ST = \text{4}\text{ cm}\) and the area of \(\triangle SNT\) is \(\text{6}\) \(\text{cm$^{2}$}\), calculate the area of \(\triangle MNR\).

\begin{align*} N\hat{S}T & = 90° \text{ (corresp }\angle \text{s; } ST \parallel MR)\\ \therefore \text{ area }\triangle SNT &=\frac{1}{2} ST \times SN \\ 6 &= \frac{1}{2}(4)SN\\ \therefore SN &= \text{3}\text{ cm}\\ \therefore MN &= \text{6}\text{ cm} \\ MR &= 2ST = \text{8}\text{ cm}\\ \text{area }\triangle MNR &= \frac{1}{2} MR \times MN\\ &=\frac{1}{2} (8)(6)\\ &= \text{24}\text{ cm$^{2}$} \end{align*}

Prove that the area of \(\triangle MNR\) will always be four times the area of \(\triangle SNT\), let \(ST = x \text{ units}\) and \(SN = y \text{ units}\).

\begin{align*} \text{Let }ST \text{ be } & x \text{ units}\\ \therefore MR \text{ will be } & 2x\\ \text{Let }SN \text{ be } & y \text{ units}\\ \therefore MN \text{ will be } & 2y\\ \text{area }\triangle SNT & = \frac{1}{2}xy \\ \text{area }\triangle MNR & = \frac{1}{2} (2x)(2y) \\ & = 2xy\\ \therefore \text{ area } \triangle MNR &= 4\left(\frac{1}{2}xy\right) \\ & = 4(\text{area }\triangle SNT) \end{align*}

Given quadrilateral \(QRST\) with sides \(QR \parallel TS\) and \(QT \parallel RS\). Also given: \(\hat{Q} = y\) and \(\hat{S} = 63^{\circ}\); \(Q\hat{T}R = 38^{\circ}\) and \(R\hat{T}S = x\). Complete the proof below to prove that \(QRST\) is a parallelogram.

2f3084927d748e5ddc89059d6f3e8eaa.png

The completed proof looks like this:

\[\begin{array}{|l | l|} \hline \text{Steps} & \text{Reasons} \\ \hline Q\hat{T}R = T\hat{R}S & \text{alt } \angle \text{s; } QT \parallel RS \\ S\hat{T}R = Q\hat{R}T & \text{alt } \angle \text{s; } QR \parallel TS \\ \text{In } \triangle QRT \text{ and } \triangle RST \text{ side } RT = RT & \text{ common side } \\ \therefore \triangle QRT \equiv \triangle STR &\text{congruent (AAS)} \\ \therefore QR = TS \text{ and } QT = RS & \text{congruent triangles} \\ \hat{Q} = \hat{S} & \text{congruent triangles} \\ \therefore QRST \text{ is a parallelogram} &\text{opp sides of quad are } = \\ \hline \end{array}\]

Calculate the value of \(y\).

\(QRST\) is a parallelogram, \(\therefore \hat{Q} = \hat{S}\).

Opposite \(\angle\)s of parallelogram are equal. \(\hat{Q} = \hat{S}\) and \(\hat{R} = \hat{T}\).

Therefore, \(y = 63^{\circ}\).

Calculate the value of \(x\).

\(\angle\)s in a \(\triangle = 180 ^{\circ}\)

\(\therefore \hat{Q} + Q\hat{R}T + Q\hat{T}R = 180 ^{\circ}\)

Now we know that \(\hat{Q} = \hat{S} = 63^{\circ}\) and that \(R\hat{T}S = 79^{\circ}\).

\(\therefore \hat{x} = 180^{\circ} - 63^{\circ} - 79^{\circ} = 79^{\circ}\).

Study the quadrilateral \(QRST\) with opposite angles \(\hat{Q} = \hat{S} = 117^{\circ}\) and angles \(\hat{R} = \hat{T} = 63^{\circ}\) carefully. Fill in the correct reasons or steps to prove that the quadrilateral \(QRST\) is a parallelogram.

0153e365e117fbd29f0691acbcef84b2.png
\[\begin{array}{|l |l|} \hline \text{Steps} & \text{Reasons} \\ \hline R\hat{Q}T = R\hat{S}T & \text{given both } \angle \text{s } = 117^\circ \\ Q\hat{R}S = Q\hat{T}S & \text{given both } \angle \text{s } = 63^\circ \\ \hat{Q} + \hat{R} + \hat{S} + \hat{T} = 360 ^\circ & \text{sum of } \angle \text{s in quad} \\ R\hat{Q}T + Q\hat{T}S = 180 ^\circ & 117^\circ + 63^\circ = 180^\circ \\ \therefore QR \parallel TS & \text{co-int } \angle \text{s; } QR \parallel TS \\ \therefore RS \parallel QT & \text{co-int } \angle \text{s; } RS \parallel QT \\ \therefore QRST \text{ is a parallelogram} & \text{opp. sides parallel} \\ \hline \end{array}\]

Study the quadrilateral \(QRST\) with \(\hat{Q} = \hat{S} = 149^{\circ}\) and \(\hat{R} = \hat{T} = 31^{\circ}\) carefully. Fill in the correct reasons or steps to prove that the quadrilateral \(QRST\) is a parallelogram.

864e69615e3b8245a7aa078f80f4abe7.png
\[\begin{array}{|l |l|} \hline \text{Steps} & \text{Reasons} \\ \hline R\hat{Q}T = R\hat{S}T & \text{given both } \angle \text{s} = 149^{\circ} \\ Q\hat{R}S = Q\hat{T}S & \text{given both } \angle \text{s} = 31^{\circ} \\ \hat{Q} + \hat{R} + \hat{S} + \hat{T} = 360 ^{\circ} & \text{sum of } \angle \text{s in quad} \\ R\hat{Q}T + Q\hat{T}S = 180 ^{\circ} & 149^{\circ} + 31^{\circ} = 180^{\circ} \\ \therefore QR \parallel TS & \text{co-int } \angle \text{s; } QR \parallel TS \\ \therefore RS \parallel QT & \text{co-int } \angle \text{s; } RS \parallel QT \\ \therefore QRST \text{ is a parallelogram} & \text{opp. sides parallel} \\ \hline \end{array}\]

In parallelogram \(QTRS\), the bisectors of the angles have been constructed, indicated with the red lines below. You are also given \(QT = SR\), \(TR = QS\), \(QT \parallel SR\), \(TR \parallel QS\), \(\hat{Q} = \hat{R}\) and \(\hat{T} = \hat{S}\).

Prove that the quadrilateral \(JKLM\) is a parallelogram.

Note the diagram is drawn to scale.

5b9364c3147234964fcbe352215aa298.png

Redraw the diagram and mark all the known information:

2ee5f8218f88ef99d1c4c5ccc33d9dd8.png

Study the diagram below; it is not necessarily drawn to scale. Two triangles in the figure are congruent: \(\triangle CDE \equiv \triangle CBF\). Additionally, \(EA = ED\). You need to prove that \(ABFE\) is a parallelogram.

8eeeabe4474dcabdef3dbd523b130a79.png

Redraw the diagram and mark all known and given information:

312469f7e2fda370239a3ddbf60a54e7.png

Given the following diagram:

db5c101c981e19588b128f911fda5baa.png

Show that \(BCDF\) is a parallelogram.

\begin{align*} DF & \parallel CB \text{ (given)} \\ DC & \parallel FB \text{ (given)} \\ \therefore BCDF & \text{ is a parallelogram (both pairs opp. sides } \parallel \text{)} \end{align*}

Show that \(ADCF\) is a parallelogram.

\begin{align*} \text{In } \triangle DEC & \text{ and } \triangle FEA \\ C\hat{A}F & = A\hat{C}D \text{ (alt } \angle\text{s; } AB \parallel DC \text{)} \\ A\hat{F}D & = C\hat{D}F \text{ (alt } \angle\text{s; } AB \parallel DC \text{)} \\ DC & = FB \text{ (opp sides parm eq)} \\ \therefore DC & = FA = FB \\ \therefore \triangle DEC & \equiv \triangle FEA \text{ (ASA)} \\ \therefore DE = EF & \text{ and } CE = EA \end{align*}

But \(AE\) and \(DF\) are diagonals of \(ADCF\), \(\therefore ADCF\) is a parallelogram (diagonals bisect each other).

Prove that \(AE = EC\).

\(AE = EC\) (proved above).

\(ABCD\) is a parallelogram. \(BEFC\) is a parallelogram. \(ADEF\) is a straight line. Prove that \(AE = DF\).

0116158dd5bb5d6d97a9ce0ede75fbf9.png
\begin{align*} BC & = EF \text{ (opp sides of } \parallel \text{m)} \\ BC & = AD \text{ (opp sides of } \parallel \text{m)} \\ \therefore EF & = ED \\ AD + DE & = AE \\ EF + DE & = DF \\ \text{ but } DE & \text{ is common} \\ \therefore AE & = DF \end{align*}

In the figure below \(AB = BF\), \(AD = DE\). \(ABCD\) is a parallelogram. Prove \(EF\) is a straight line.

272898e6e1776d61a3d380aafa9647ad.png

We note that:

\begin{align*} B\hat{A}D & = B\hat{C}D \text{ (opp } \angle\text{s } \parallel \text{m)} \\ C\hat{D}E & = B\hat{C}D \text{ (alt } \angle\text{s; } AE \parallel BC \text{)} \\ F\hat{B}C & = B\hat{C}D \text{ (alt } \angle\text{s; } AF \parallel DC \text{)} \\ \therefore C\hat{D}E & = F\hat{B}C \end{align*}

We also note that:

\begin{align*} AD & = BC \text{ (opp sides parm eq)} \\ AB & = DC \text{ (opp sides parm eq)} \end{align*}

Now we can show that \(\triangle DEC\) is congruent to \(\triangle BCF\):

\begin{align*} \text{in } \triangle DEC & \text{ and } \triangle BCF\\ C\hat{D}E & = F\hat{B}C \qquad \text{(proven above)} \\ DC = AB & = BF \qquad \text{(given)} \\ DE = AD & = BC \qquad \text{(given)} \\ \therefore \triangle DEC & \equiv \triangle BCF \text{ (SAS)} \end{align*}

Finally we can show that \(ECF\) is a straight line:

\begin{align*} \therefore B\hat{F}C & = D\hat{C}E \text{ (}\triangle DEC \equiv \triangle BCF \text{)} \\ B\hat{C}F & = D\hat{E}C \text{ (}\triangle DEC \equiv \triangle BCF \text{)} \\ \text{but } F\hat{B}C + B\hat{F}C + B\hat{C}F & = 180° \text{(sum of }\angle\text{s in } \triangle \text{)} \\ \therefore D\hat{C}E + B\hat{C}F + B\hat{C}D & = 180° \\ \therefore ECF & \text{ is a str line} \end{align*}