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# End Of Chapter Exercises

## End of chapter exercises

Exercise 8.10

Calculate $$SV$$

$\begin{array}{rll} V\hat{T}U & = W\hat{V}T & \quad \text{(alt. } \angle \text{s}, WV \parallel TU) \\ \therefore TU & = VU = 35 & \quad \text{(isosceles } \triangle \text{)} \\ \frac{SW}{WT} & = \frac{SV}{VU} & \quad \text{(proportion Theorem)} \\ \therefore SV & = \frac{SW.VU}{WT} & \\ & = \frac{(10)(35)}{20} & \\ & = \text{17,5}\text{ units} & \\ \end{array}$

$$\frac{CB}{YB}=\frac{3}{2}$$. Find $$\frac{DS}{SZ}$$.

$\begin{array}{rll} DABC & \text{ is a parallelogram } & (DA \parallel CB \text{ and } DC \parallel AB)\\ DS &= SB & (\text{diagonals bisect})\\ & & \\ \dfrac{SZ}{ZB} &= \dfrac{CY}{YB} = \frac{3}{2} & (CS \parallel YZ)\\ & & \\ \dfrac{SZ}{SB} &= \dfrac{CY}{CB} = \frac{3}{5} & (CS \parallel YZ)\\ & & \\ \therefore SZ &= \frac{3}{5} SB & \\ & & \\ \therefore \dfrac{DS}{SZ} &= \dfrac{DS}{\frac{3}{5} SB} & (DS = SB) \\ & & \\ &= \frac{5}{3} & \end{array}$

Using the following figure and lengths, find $$IJ$$ and $$KJ$$ (correct to one decimal place).

$$HI= \text{20}\text{ m},KL= \text{14}\text{ m}, JL=\text{18}\text{ m}$$ and $$HJ=\text{32}\text{ m}$$.

$\begin{array}{rll} \dfrac{IJ}{LJ} & = \dfrac{HI}{KL} & \quad \text{(proportion Theorem)}\\ & & \\ IJ & = \dfrac{HI}{KL}(LJ) & \\ & & \\ & = \dfrac{20}{14}(18) & \\ & & \\ & = \dfrac{180}{7} & \\ & & \\ & = \text{25,7}\text{ m} & \end{array}$

$\begin{array}{rll} KJ & = \dfrac{LJ}{IJ}(HJ) & \\ & & \\ & = \dfrac{18}{\text{25,7}}(32) & \\ & & \\ & = \text{22,4}\text{ m} & \\ \end{array}$

Find $$FH$$ in the following figure.

$\begin{array}{rll} G\hat{F}H & = \hat{D} & \text{(corresp. } \angle \text{s}, GF \parallel ED) \\ G\hat{F}E & = F\hat{E}D & \text{(alt. } \angle \text{s}, GF \parallel ED) \\ \therefore F\hat{E}D & = \hat{D} & \\ \therefore EF &= FD = \text{45}\text{ cm} & \\ \frac{HF}{FD } &= \frac{21}{42} & \\ &= \frac{1}{2} & \\ \therefore HF & = \frac{1}{2}(45) & \\ & = \text{23,5}\text{ cm} & \end{array}$

In $$\triangle GHI$$, $$GH\parallel LJ$$, $$GJ\parallel LK$$ and $$\frac{JK}{KI}=\frac{5}{3}$$. Determine $$\frac{HJ}{KI}$$.

$\begin{array}{rll} L\hat{I}J & = G\hat{I}H &\\ J\hat{L}I & = H\hat{G}I & \text{(corresp. }\angle \text{s}, HG \parallel JL) \\ \therefore \triangle LIJ & \enspace ||| \enspace \triangle GIH & \text{(Equiangular }\triangle \text{s)} \end{array}$ $\begin{array}{rll} \frac{HJ}{JI}& =\frac{GL}{LI} & \left(\triangle LIJ \enspace ||| \enspace \triangle GIH\right)\\ \text{and }\frac{GL}{LI}& =\frac{JK}{KI} & \left(\triangle LIK \enspace ||| \enspace \triangle GIJ\right)\\ & =\frac{5}{3} \\ \therefore \frac{HJ}{JI}& =\frac{5}{3} \end{array}$ \begin{align*} \frac{HJ}{KI} & = \frac{HJ}{JI} \times \frac{JI}{KI} \end{align*} \begin{align*} JI & = JK+KI \\ & = \frac{5}{3}KI+KI \\ & = \frac{8}{3}KI \\ \frac{JI}{KI} & = \frac{8}{3} \\ & \\ \frac{HJ}{KI} & = \frac{HJ}{JI} \times \frac{JI}{KI} \\ & = \frac{5}{3}\times \frac{8}{3} \\ & = \frac{40}{9} \end{align*}

$$BF=\text{25}\text{ m}$$, $$AB=\text{13}\text{ m}$$, $$AD=\text{9}\text{ m}$$, $$DF=\text{18}\text{ m}$$.

Calculate the lengths of $$BC$$, $$CF$$, $$CD$$, $$CE$$ and $$EF$$, and find the ratio $$\frac{DE}{AC}$$.

$\begin{array}{rll} \frac{BC}{BF} & = \frac{AD}{AF} = \frac{9}{27} = \frac{1}{3} & (CD \parallel BA) \\ \therefore BC & = \frac{1}{3} \times 25 & \\ &= \text{8,3}\text{ m} & \\ & & \\ CF &= BF - BC & \\ &= 25 - \text{8,3} & \\ &= \text{16,7}\text{ m} & \\ & & \\ \frac{CD}{AB} &= \frac{DF}{AF} & (CD \parallel BA) \\ CD &= \frac{DF}{AF} \times AB & \\ &= \frac{18}{27} \times 13 & \\ &= \text{8,7}\text{ m} & \\ & & \\ \frac{CE}{CF} &= \frac{AD}{AF} & (DE \parallel AC) \\ CE &= \frac{AD}{AF} \times CF & \\ &= \frac{9}{27} \times \text{16,7} & \\ &= \text{5,6}\text{ m} & \\ & & \\ EF &= BF - (BC + CE) & \\ &= 25 - (\text{8,3} - \text{5,6}) & \\ &= \text{11,1}\text{ m} & \end{array}$

In $$\triangle XYZ$$, $$X\hat{Y}Z =\text{90}°$$ and $$YT \perp XZ$$. If $$XY = \text{14}\text{ cm}$$ and $$XT = \text{4}\text{ cm}$$, determine $$XZ$$ and $$YZ$$ (correct to two decimal places).

Use the theorem of Pythagoras to determine $$YT$$:

$\begin{array}{rll} \text{In } \triangle XTY, \quad YT^{2} &= XY^{2} - XT^{2} & (\text{Pythagoras}) \\ &= 14^{2} - 4^{2} & \\ &= 196 - 16 & \\ \therefore YT &= \sqrt{180} & \\ &= \sqrt{36 \times 5} & \\ &= 6 \sqrt{5} ~\text{cm} & \end{array}$

Use proportionality to determine $$XZ$$ and $$YZ$$:

$\begin{array}{rll} X\hat{Y}Z &= \text{90}° & \text{(given)} \\ YT &\perp XZ & \text{(given)} \\ \therefore \enspace \triangle XYT \enspace & ||| \enspace \triangle YZT \enspace ||| \enspace \triangle XZY & (\text{right-angled } \triangle \text{s}) \\ & & \\ \therefore \dfrac{YT}{TZ} &= \dfrac{XT}{YT} & (\triangle YZT \enspace ||| \enspace \triangle XYT ) \\ & & \\ \therefore YT^{2} &= TZ \cdot XT & \\ & & \\ \left( 6 \sqrt{5} \right)^{2} &= TZ \cdot 4 & \\ & & \\ \therefore TZ &= 45 & \\ & & \\ \text{And } XZ &= XT + TZ & \\ &= 4 + 45 & \\ &= \text{49}\text{ cm} & \\ & & \\ \text{In } \triangle XYZ, \quad YZ^{2} &= XZ^{2} - XY^{2} & (\text{Pythagoras}) \\ &= 49^{2} - 14^{2} & \\ \therefore YZ &= \sqrt{2205} & \\ &= \text{46,96}\text{ cm} & \end{array}$

Given the following figure with the following lengths, find $$AE$$, $$EC$$ and $$BE$$.

$$BC=15$$ cm, $$AB=4$$ cm, $$CD=18$$ cm, and $$ED=9$$ cm.

$\begin{array}{rll} B\hat{A}E & = C\hat{D}E & \text{(alt. } \angle \text{s}, AB \parallel CD) \\ A\hat{B}E & = D\hat{C}E & \text{(alt. } \angle \text{s}, AB \parallel CD) \\ A\hat{E}B & = D\hat{E}C & \text{(vert. opp. } \angle \text{s)}\\ \therefore \triangle AEB & \enspace ||| \enspace \triangle DEC & \text{(AAA)}\\ \therefore \frac{AE}{DE} & = \frac{AB}{DC} = \frac{4}{18} = \frac{2}{9} & (\triangle AEB \enspace ||| \enspace \triangle DEC) \\ & & \\ AE & = \dfrac{2}{9} DE & \\ & & \\ & = \dfrac{2}{9} (9) & \\ & & \\ & = \text{2}\text{ cm} & \\ \end{array}$

$\begin{array}{rll} \frac{EC}{BC} & = \dfrac{ED}{AD} = \dfrac{9}{11} & (AB \parallel CD) \\ & & \\ EC & = \dfrac{ED}{AD}(BC) & \\ & & \\ & = \dfrac{9}{11}(15) & \\ & & \\ & = \text{12,3}\text{ cm} & \\ \end{array}$

$\begin{array}{rll} \frac{BE}{BC} & = \dfrac{AE}{AD} = \dfrac{2}{11} & (AB \parallel CD) \\ & & \\ BE & = \dfrac{AE}{AD}(BC) & \\ & & \\ & = \dfrac{2}{11}(15) & \\ & & \\ & = \text{2,7}\text{ cm} & \\ \end{array}$

$$NKLM$$ is a parallelogram with $$T$$ on $$KL$$.

$$NT$$ produced meets $$ML$$ produced at $$V$$. $$NT$$ intercepts $$MK$$ at $$X$$.

Prove that $$\dfrac{XT}{NX} = \dfrac{XK}{MX}$$.

In $$\triangle TXK$$ and $$\triangle NXM$$:

$\begin{array}{rll} X\hat{T}K&= X\hat{N}M & (\text{alt. } \angle \text{s, } NK \parallel MV) \\ N\hat{X}M&= T\hat{X}K & (\text{vert. opp. } \angle \text{s}) \\ \therefore \triangle TXK & \enspace ||| \enspace \triangle NXM & (\text{AAA}) \\ \therefore \dfrac{TX}{NX} &= \dfrac{XK}{XM} & \end{array}$

Prove $$\triangle VXM \enspace ||| \enspace \triangle NXK$$.

In $$\triangle VXM \text{ and } \triangle NXK$$:

$\begin{array}{rll} \hat{V}&= X\hat{N}K & (\text{alt. } \angle \text{s, } NK \parallel MV) \\ M\hat{X}V&= K\hat{X}N & (\text{vert. opp. } \angle \text{s}) \\ \therefore \triangle VXM& \enspace ||| \enspace \triangle NXK & (\text{AAA}) \end{array}$

If $$XT = \text{3}\text{ cm}$$ and $$TV = \text{4}\text{ cm}$$, calculate $$NX$$.

$\begin{array}{rll} \dfrac{VX}{NX} &= \dfrac{XM}{XK} & (\triangle VXM \enspace ||| \enspace \triangle NXK \text{, proved in (b)}) \\ \text{But } \dfrac{XM}{XK} &= \dfrac{NX}{TX} & (\text{proved in (a)}) \\ \therefore \dfrac{VX}{NX} &= \dfrac{NX}{TX} & \\ NX^{2} &= VX.TX & \\ NX^{2}&= 3 \times 4 & \\ NX &= \sqrt{12} & \\ &= 2 \sqrt{3} \text{ cm} & \end{array}$

$$MN$$ is a diameter of circle $$O$$. $$MN$$ is produced to $$R$$ so that $$MN =2NR$$.

$$RS$$ is a tangent to the circle and $$ER \perp MR$$. $$MS$$ produced meets $$RE$$ at $$E$$.

Prove that:

$$SNRE$$ is a cyclic quadrilateral

$\begin{array}{rll} M\hat{S}N&= \text{90}° & (\angle \text{ in semi-circle}) \\ N\hat{R}E&=\text{90}° & (\text{given}) \\ \therefore SNRE&\text{ is a cyclic quad. } & (\text{ext. }\angle = \text{ opp. int. } \angle) \end{array}$

$$RS = RE$$

$\begin{array}{rll} N\hat{S}R&=\hat{M} = x & (\text{tangent/chord}) \\ \therefore E\hat{S}R&= \text{90}° - x & \\ \hat{E}&= \text{90}° - x & (MRE = \text{90}°, \enspace \hat{M} = x) \\ \therefore E\hat{S}R&= \hat{E} & \\ \therefore RS&= RE & (\text{isos. } \triangle) \end{array}$

$$\triangle MSN \enspace ||| \enspace \triangle MRE$$

In $$\triangle MSN$$ and $$\triangle MRE$$:

$\begin{array}{rll} \hat{M} &=\hat{M} & \\ M\hat{S}N &= \text{90}° & (\angle \text{ in semi-circle}) \\ M\hat{R}E &= \text{90}° & (\text{given}) \\ \therefore M\hat{S}N &= M\hat{R}E & \\ \therefore \triangle MSN& \enspace ||| \enspace \triangle MRE & (\text{AAA}) \end{array}$

$$\triangle RSN \enspace ||| \enspace \triangle RMS$$

In $$\triangle RSN$$ and $$\triangle RMS$$:

$\begin{array}{rll} \hat{R}&= \hat{R} & (\text{common } \angle) \\ R\hat{S}N&= \hat{M} & (\text{tangent/chord}) \\ \therefore \triangle RSN & \enspace ||| \enspace \triangle RMS & (\text{AAA}) \end{array}$

$$RE^{2} = RN.RM$$

$\begin{array}{rll} \dfrac{RS}{RN} &= \dfrac{RM}{RS} & \\ RS^{2} &= RN.RM & \\ \text{But } RS &= RE & \\ RE^{2} &= RN.RM & \end{array}$

$$AC$$ is a diameter of circle $$ADC$$. $$DB \perp AC$$.

$$AC = d, AD = c, DC = a \text{ and } DB = h$$.

Prove that $$h = \dfrac{ac}{d}$$.

$\begin{array}{rll} \triangle ADB& \enspace ||| \enspace \triangle DCB \enspace ||| \enspace \triangle ACD & (A\hat{D}C = \text{90}°, DB \perp AC) \\ \therefore \dfrac{DB}{AD}&= \dfrac{CD}{AC} & \\ \therefore \dfrac{h}{c}&= \dfrac{a}{d} & \\ \therefore h&= \dfrac{ac}{d} & \end{array}$

Hence, deduce that $$\enspace \dfrac{1}{h^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{c^{2}}$$.

$\begin{array}{rll} h^{2}&= \dfrac{a^{2}c^{2}}{d^{2}} & \\ \text{But } d^{2}&= a^{2} + c^{2} & (\text{In } \triangle ADC, \hat{D} = \text{90}°, \text{ Pythagoras}) \\ \therefore h^{2}&= \dfrac{a^{2}c^{2}}{a^{2}+c^{2}} & \\ \therefore \dfrac{1}{h^{2}}&= \dfrac{a^{2}+c^{2}}{a^{2}c^{2}} & \\ \dfrac{1}{h^{2}}&= \dfrac{a^{2}}{a^{2}c^{2}} + \dfrac{c^{2}}{a^{2}c^{2}} & \\ \dfrac{1}{h^{2}}&= \dfrac{1}{c^{2}} + \dfrac{1}{a^{2}} & \end{array}$

$$RS$$ is a diameter of the circle with centre $$O$$. Chord $$ST$$ is produced to $$W$$. Chord $$SP$$ produced meets the tangent $$RW$$ at $$V$$. $$\hat{\text{R}}_{1} = \text{50}°$$.

[NCS, Paper 3, November 2011]

Calculate the size of $$\text{W}\hat{\text{R}}\text{S}$$.

$$W\hat{R}S = \text{90}°\qquad (\text{tangent } \perp \text{ radius})$$

Find $$\hat{\text{W}}$$.

$\begin{array}{rll} R\hat{S}T & = \text{50}° & \qquad(\text{tangent chord th.})\\ \hat{W} & = \text{40}° & \qquad(\angle \text{ sum } \triangle) \end{array}$

OR

$\begin{array}{rll} \hat{T}_{1} & = \text{90}° & \qquad(\angle \text{s in semi-circle})\\ \hat{W} + \hat{R}_{1} & = \hat{T}_{1} & \qquad(\text{ ext. } \angle \triangle)\\ \hat{W} & = \text{40}° & \end{array}$

Determine the size of $$\hat{\text{P}}_1$$.

$\begin{array}{rll} \hat{R}_{2} & = \text{40}° & \qquad(\text{tangent } \perp \text{ radius})\\ \hat{P}_{1} & = \text{40}° & \qquad(\angle \text{s in same seg.}) \end{array}$

Prove that $$\hat{\text{V}}_1 = \text{P}\hat{\text{T}}\text{S}$$.

$\begin{array}{rll} \hat{P}_{1} & = \hat{W} & \qquad(= \text{40}°)\\ \text{WVPT is}& \text{ a} \text{ cyclic quadrilateral} &(\text{ext. } \angle = \text{ int. opp.})\\ \hat{V}_{1} & = P\hat{T}S & \qquad (\text{ext. } \angle \text{ cyclic quad.}) \end{array}$

OR

$\begin{array}{rll} \hat{T}_{1} & =\text{90}° & \qquad(\angle \text{s in semi-circle})\\ P\hat{T}S & = \text{90}° + \hat{T}_{2} & \\ \hat{T}_{2} & = \hat{S}_{1} & \qquad(\angle \text{s in same seg.})\\ P\hat{T}S & = \text{90}° + \hat{S}_{1} & \qquad(\text{ext. } \angle \triangle) \\ \hat{V}_{1} & = P\hat{T}S & \\ \end{array}$

OR

$\begin{array}{rll} \hat{P}_{2} & = \text{140}° & \qquad(\angle \text{s on str. line})\\ \hat{W} + \hat{P}_{2} & = \text{180}° & \\ \text{WVPT} &\text{is a cyclic quadrilateral} & \qquad(\text{opp. } \angle \text{s suppl})\\ \hat{V}_{1} & = P\hat{T}S & \qquad(\text{ext } \angle \text{ cyclic quad}) \end{array}$

OR

$\begin{array}{rll} \hat{V}_{1} & = \hat{R}_{1} + \hat{R}_{2} + \hat{S}_{1} & \qquad(\text{ext. } \angle \triangle)\\ \hat{V}_{1} & = \text{90}° + \hat{S}_{1} & \\ P\hat{T}S & = \text{90}° + \hat{T}_{2} & \\ \text{but } \hat{T}_{2} & = \hat{S}_{1} & \qquad(\angle \text{s in same seg.})\\ \hat{V}_{1} & = P\hat{T}S & \end{array}$

OR

In $$\triangle PTS$$ and $$\triangle WVS$$

$\begin{array}{rll} \hat{P}_{1} & = \hat{W} & \qquad(\text{40}°) \\ \hat{S}_{2} & \text{ is common} &\\ \hat{V}_{1} & = P\hat{T}S & \qquad (\angle \text{ sum } \triangle) \end{array}$

$$ABCD$$ is a cyclic quadrilateral and $$BC = CD.$$

$$ECF$$ is a tangent to the circle at $$C$$. $$ABE$$ and $$ADF$$ are straight lines.

Prove:

$$AC \text{ bisects } E\hat{A}F$$

$\begin{array}{rll} BC &= CD & (\text{given}) \\ \therefore B\hat{A}C &= D\hat{A}C & (\angle \text{s on equal chords}) \end{array}$

$$BD \parallel EF$$

$\begin{array}{rll} D\hat{C}F&= \hat{A_{2}} & (\text{tangent/chord}) \\ B\hat{D}C&= \hat{A_{1}} & (\angle \text{s on same chord}) \\ \hat{A_{1}} &= \hat{A_{2}} & (\text{proved in (a)}) \\ \therefore B\hat{D}C &= D\hat{C}F & \\ \therefore BD & \parallel EF & (\text{alt. } \angle \text{s are equal}) \end{array}$

$$\triangle ADC \enspace ||| \enspace \triangle CBE$$

In $$\triangle ADC$$ and $$\triangle CBE$$:

$\begin{array}{rll} \hat{A_{2}} &= D\hat{B}C & (\angle \text{s on chord } CD) \\ &= B\hat{C}E & (\text{alt. } \angle \text{s, } BD \parallel EF) \\ A\hat{D}C &= E\hat{B}C & (\text{ext. } \angle \text{ of a cyclic quad}) \\ \therefore \triangle ADC & \enspace ||| \enspace \triangle CBE & (\text{AAA}) \end{array}$

$$DC^{2} = AD.BE$$

$\begin{array}{rll} \dfrac{DC}{AD} &= \dfrac{BE}{BC} & (\triangle ADC \enspace ||| \enspace \triangle CBE) \\ \text{But } DC &= BC & (\text{given}) \\ \therefore \dfrac{DC}{AD} &= \dfrac{BE}{DC} & \\ \therefore DC^{2} &= AD.BE & \end{array}$

$$CD$$ is a tangent to circle $$ABDEF$$ at $$D$$. Chord $$AB$$ is produced to $$C$$. Chord $$BE$$ cuts chord $$AD$$ in $$H$$ and chord $$FD$$ in $$G$$. $$AC \parallel FD$$ and $$E = AB$$. Let $$D_{4} = x$$ and $$D_{1} = y$$.

[NCS, Paper 3, November 2011]

Determine THREE other angles that are each equal to $$x$$.

$\begin{array}{r@{\;}l@{\quad}l} \hat{A} & = \hat{D}_{4} = x & (\text{tangent chord th.})\\ \hat{E}_{2} & = x & (\text{tangent chord th.})\\ \hat{D}_{2} & = \hat{A} = x & (\text{alt. } \angle \text{s, } CA \parallel DF) \end{array}$

Prove that $$\triangle \text{BHD} \enspace ||| \enspace \triangle \text{FED}$$.

In $$\triangle BHD$$ and $$\triangle FED$$

1. $$\hat{B}_{2} = \hat{F} \quad(\angle \text{s in same seg.})$$
2. $$\hat{D}_{3} = \hat{D}_{1} \quad(\text{chord subtends } = \angle \text{s})$$

$$\triangle BHD \enspace ||| \enspace \triangle FED \quad (\angle \angle \angle)$$

Hence, or otherwise, prove that $$AB \cdot BD = FD \cdot BH$$.

$\begin{array}{r@{\;}l@{\quad}l} \frac{FE}{BH} & = \frac{FD}{BD} & (||| \enspace \triangle\text{s})\\ \text{But } FE & = AB & (\text{given})\\ \frac{AB}{BH} & = \frac{FD}{BD} &\\ AB \cdot BD & = FD \cdot BH & \end{array}$