Soalan 10 (10 markah):
Penyelesaian secara lukisan berskala tidak diterima.
Rajah menunjukkan sisi empat PQRS pada suatu satah mengufuk.


VQSP ialah sebuah piramid dengan keadaan PQ = 12 m dan V adalah 5 m tegak di atas P.
Cari
(a) ∠QSR,
(b) panjang, dalam m, bagi QS,
(c) luas, dalam m², bagi satah condong QVS.


Penyelesaian: 
(a)
$\begin{array}{l}\frac{\sin \angle QSR}{20.5}=\frac{\sin {64}^o}{22}\\ \sin \angle QSR=\frac{\sin {64}^o}{22}\times 20.5\\ \sin \angle QSR=0.8375\\ \angle QSR={56}^{o}52\text{'}\end{array}$


(b)
$\begin{array}{l}\angle QRS={180}^o-{64}^o-{56}^{o}52\text{'}\\ ={59}^{o}8\text{'}\\ \frac{QS}{\sin {59}^{o}8\text{'}}=\frac{22}{\sin {64}^o}\\ QS=\frac{22}{\sin {64}^o}\times \sin {59}^{o}8\text{'}\\ QS=21.01\text{ m}\end{array}$


(c)

$\begin{array}{l}Q{V}^2=P{Q}^2+V{P}^2\\ QV=\sqrt{{12}^2+5^2}\\ QV=13\text{ m}\\ \\ S{V}^2=P{S}^2+V{P}^2\\ SV=\sqrt{{10}^2+5^2}\\ SV=\sqrt{125}\text{ m}\\ \\ QS=21.01\text{ m}\\ \\ {21.01}^2={13}^2+{\left(\sqrt{125}\right)}^2-2\left(13\right)\left(\sqrt{125}\right)kos\theta \\ 26\left(\sqrt{125}\right)kos\theta =169+125-441.42\\ kos\theta =\frac{169+125-441.42}{26\left(\sqrt{125}\right)}\\ kos\theta =-0.5071\\ \theta ={120}^{o}28\text{'}\\ \\ \text{Luas }\Delta QVS\\ =\frac{1}{2}\times 13\times \sqrt{125}\times \sin {120}^{o}28\text{'}\\ =62.64{\text{ m}}^2\end{array}$

Soalan 9 (10 markah):
Penyelesaian secara lukisan berskala tidak diterima.
Rajah 8 menunjukkan prisma lutsinar dengan tapak PQRS berbentuk segi empat tepat. Permukaan condong PQUT ialah segi empat sama dengan sisi 12 cm dan permukaan condong RSTU ialah segi empat tepat. PTS ialah keratan rentas seragam bagi prisma itu. QST ialah satah berwarna hijau di dalam prisma itu.
Diberi bahawa ∠PST = 37^o dan ∠TPS = 45^o.
Cari
(a) panjang, dalam cm, bagi ST,
(b) luas, dalam cm², satah berwarna hijau.
(c) panjang terdekat, dalam cm, dari titik T ke garis lurus QS.


Penyelesaian:
(a)



$\begin{array}{l}\frac{ST}{\sin {45}^o}=\frac{12}{\sin {37}^o}\\ ST=\frac{12}{\sin {37}^o}\times \sin {45}^o\\ ST=14.1\text{ cm}\end{array}$


(b)
$\begin{array}{l}Q{T}^2=Q{P}^2+P{T}^2\\ Q{T}^2={12}^2+{12}^2\\ QT=\sqrt{{12}^2+{12}^2}=16.97\text{ cm}\\ \\ \angle PTS={180}^o-{45}^o-{37}^o={98}^o\\ \\ \frac{PS}{\sin {98}^o}=\frac{12}{\sin {37}^o}\\ PS=\frac{12}{\sin {37}^o}\times \sin {98}^o\\ PS=19.75\text{ cm}\\ \\ Q{S}^2=Q{P}^2+P{S}^2\\ QS=\sqrt{Q{P}^2+P{S}^2}\\ QS=\sqrt{{12}^2+{19.75}^2}\\ QS=23.11\text{ cm}\end{array}$



$\begin{array}{l}Q{S}^2=Q{T}^2+S{T}^2-2\left(QT\right)\left(ST\right)kos\angle QTS\\ {23.11}^2={16.97}^2+{14.1}^2-2\left(16.97\right)\left(14.1\right)kos\angle QTS\\ kos\angle QTS=\frac{{16.97}^2+{14.1}^2-{23.11}^2}{2\left(16.97\right)\left(14.1\right)}\\ kos\angle QTS=-\frac{47.28}{478.55}\\ \angle QTS={95.67}^o\\ \\ \text{Oleh itu, luas satah berwarna hijau }QTS\\ =\frac{1}{2}\left(16.97\right)\left(14.1\right)\sin {95.67}^o\\ =119.05{\text{ cm}}^2\end{array}$


(c)



$\begin{array}{l}\text{Katakan panjang terdekat dari titik }T\\ \text{ke garis }QS\text{ ialah }h.\\ \\ \text{Luas }\Delta QTS=119.05\\ \frac{1}{2}\left(h\right)\left(23.11\right)=119.05\\ h=10.3\text{ cm}\end{array}$

Soalan 8:
Rajah menunjukkan sisi empat kitaran PQRS.


(a) Hitung
(i) panjang, dalam cm, bagi PR,
(ii) ∠PRQ.
(b) Cari
(i) luas, dalam cm², bagi ∆ PRS.
(ii) jarak terdekat, dalam cm, dari titik S ke PR.


Penyelesaian:
(a)(i)
$\begin{array}{l}P{R}^2=7^2+8^2-2\left(7\right)\left(8\right)kos{80}^o\\ P{R}^2=113-19.4486\\ PR=\sqrt{93.5514}\\ PR=9.6722\text{ cm}\end{array}$


(a)(ii)
$\begin{array}{l}\text{Dalam sisi empat kitaran}\\ \angle PQR+\angle PSR=180\\ \angle PQR+80=180\\ \angle PQR={100}^o\\ \\ \frac{\sin \angle QPR}{3}=\frac{\sin 100}{9.6722}\\ \sin \angle QPR=0.3055\\ \angle QPR={17}^{o}47\text{'}\\ \\ \angle PRQ={180}^o-{100}^o-{17}^{o}47\text{'}\\ ={62}^{o}13\text{'}\end{array}$


(b)(i)
$\begin{array}{l}\text{Luas }△PRS\\ =\frac{1}{2}\times 7\times 8\sin {80}^o\\ =27.5746{\text{ cm}}^2\end{array}$


(b)(ii)

$\begin{array}{l}\text{Luas }△PRS=27.5746\\ \frac{1}{2}\times 9.6722\times h=27.5746\\ h=\frac{27.5746\times 2}{9.6722}\\ =5.7018\text{ cm}\\ \text{Jarak terdekat}=5.7018\text{ cm}\end{array}$

Soalan 7:
Rajah di bawah menunjukkan sebuah sisi empat ABCD dengan keadaan sisi AB dan sisi CD adalah selari. ∠BAC ialah sudut cakah.

Diberi bahawa AB = 14 cm, BC = 27 cm, ∠ACB = 30^o dan AB : DC = 7 : 3.
Hitung
(a) ∠BAC.
(b) panjang, dalam cm, bagi pepenjuru BD.
(c) luas, dalam cm², bagi sisi empat ABCD.

Penyelesaian:
(a)
$\begin{array}{l}\frac{\sin \angle BAC}{27}=\frac{\sin {30}^o}{14}\\ \sin \angle BAC=\frac{\sin {30}^o}{14}\times 27\\ \sin \angle BAC=0.9643\\ \angle BAC={74.64}^o\\ \angle BAC\left(\text{cakah}\right)={180}^o-{74.64}^o\\ ={105.36}^o\end{array}$


(b)
$\begin{array}{l}AB\text{ selari dengan }DC\\ \angle BAC=\angle ACD\\ \angle BCD={105.36}^o+{30}^o\\ ={135.36}^o\\ \frac{DC}{AB}=\frac{3}{7}\\ DC=\frac{3}{7}\times 14\text{ cm}\\ =6\text{ cm}\\ \\ B{D}^2={27}^2+6^2-2\left(27\right)\left(6\right)kos\angle BCD\\ B{D}^2=765-324\text{kos }{135.36}^o\\ B{D}^2=995.54\\ BD=31.55\text{ cm}\end{array}$


(c)
$\begin{array}{l}\angle ABC={180}^o-{30}^o-{105.36}^o\\ ={44.64}^o\\ \\ A{C}^2={27}^2+{14}^2-2\left(27\right)\left(14\right)kos\angle ABC\\ A{C}^2=925-756kos{44.64}^o\\ A{C}^2=387.08\\ AC=19.67\text{ cm}\\ \\ \text{Luas }△ABC\\ =\frac{1}{2}\left(14\right)\left(27\right)\sin {44.64}^o\\ =132.80{\text{ cm}}^2\\ \\ \text{Luas }△ACD\\ =\frac{1}{2}\left(19.67\right)\left(6\right)\sin {105.36}^o\\ =56.90{\text{ cm}}^2\\ \\ \text{Luas sisi empat }ABCD\\ =132.80+56.90\\ =189.7{\text{ cm}}^2\end{array}$

Soalan 6:
Rajah di bawah menunjukkan sebuah sisi empat PQRS.



(a) Cari
(i) panjang, dalam cm, bagi QS.
(ii) ∠QRS.
(iii) luas, dalam cm², bagi sisi empat PQRS.
(b)(i) Lakar sebuah segi tiga S’Q’R’ yang mempunyai bentuk berbeza daripada segi tiga SQR dengan keadaan S’R’ = SR, S’Q’ = SQ dan ∠S’Q’R’ = ∠SQR.
(ii) Seterusnya, nyatakan ∠S’R’Q’.


Penyelesaian:
(a)(i)
$\begin{array}{l}\angle P=180-76-34=70\\ \frac{QS}{\sin 70}=\frac{8}{\sin 34}\\ QS=\frac{8\times \sin 70}{\sin 34}\\ =13.44\text{ cm}\end{array}$

(a)(ii)
$\begin{array}{l}{13.44}^2=6^2+9^2-2\left(6\right)\left(9\right)kos\angle QRS\\ 108kos\angle QRS=6^2+9^2-{13.44}^2\\ kos\angle QRS=\frac{6^2+9^2-{13.44}^2}{108}\\ \angle QRS=ko{s}^{-1}\left(-0.5892\right)\\ ={126}^{o}6\text{'}\end{array}$

(a)(iii)
$\begin{array}{l}\text{Luas }PQRS\\ =\text{Luas }PQS+\text{Luas }QRS\\ =\left(\frac{1}{2}\times 8\times 13.44\times \sin 76\right)+\left(\frac{1}{2}\times 6\times 9\times \sin {126}^{o}6\text{'}\right)\\ =52.16+21.82\\ =73.98{\text{ cm}}^2\end{array}$

(b)(i)



(b)(ii)
$\begin{array}{l}\angle S\text{'}R\text{'}Q\text{'}=\angle S\text{'}RR\text{'}\\ =180-{126}^{o}6\text{'}\\ ={53}^{o}54\text{'}\end{array}$

Soalan 5:
Rajah di bawah menunjukkan trapezium ABCD.

(a) Hitung
(i) ∠BAC.
(ii) panjang, dalam cm, bagi AD.
(b) Garis lurus AB dipanjangkan ke B’ dengan keadaan BC = B’C.
(i) Lakar trapezium AB’CD.
(ii) Hitung luas, dalam, cm², bagi ∆BB’C.  


Penyelesaian:
(a)(i)
$\begin{array}{l}{5}^2=4^2+7^2-2\left(4\right)\left(7\right)\text{kos}\angle BAC\\ 25=16+49-56\text{kos}\angle BAC\\ 56\text{kos}\angle BAC=40\\ \text{kos}\angle BAC=\frac{40}{56}\\ \angle BAC={\text{kos}}^{-1}\frac{40}{56}\\ ={44}^{o}25\text{'}\end{array}$


(a)(ii)
$\begin{array}{l}\frac{AD}{\sin \angle DCA}=\frac{7}{\sin {115}^o}\\ \frac{AD}{\sin {44}^{o}25\text{'}}=\frac{7}{\sin {115}^o}\leftarrow \left(\angle DCA=\angle BAC\right)\\ AD=\frac{7}{\sin {115}^o}\times \sin {44}^{o}25\text{'}\\ AD=5.406\text{ cm}\end{array}$


(b)(i)




(b)(ii)
$\begin{array}{l}\frac{\sin \angle ABC}{7}=\frac{\sin {44}^{o}25\text{'}}{5}\\ \sin \angle ABC=\frac{\sin {44}^{o}25\text{'}}{5}\times 7\\ ={78}^{o}28\text{'}\\ \angle ABC={180}^o-{78}^{o}28\text{'}\\ \angle ABC={101}^{o}32\text{'}\left(\text{sukuan kedua}\right)\\ \\ \angle CBB\text{'}={180}^o-{101}^{o}32\text{'}={78}^{o}28\text{'}\\ \angle BCB\text{'}={180}^o-{78}^{o}28\text{'}-{78}^{o}28\text{'}={23}^{o}4\text{'}\\ \text{Luas bagi }△BB\text{'}C=\frac{1}{2}\times 5\times 5\times {23}^{o}4\text{'}\\ =4.898{\text{ cm}}^2\end{array}$

$\begin{array}{l}\frac{CD}{\sin {79.13}^{\circ }}=\frac{14}{\sin {45}^{\circ }}\\ CD=\frac{14\times \sin {79.13}^{\circ }}{\sin {45}^{\circ }}=19.44\text{cm}\end{array}$

∠ACB = 180^o – (50^o + 70^o) = 60^o
$\begin{array}{l}\frac{AB}{\sin {60}^o}=\frac{4}{\sin {50}^o}\\ AB=\frac{4\times \sin {60}^o}{\sin {50}^o}\\ AB=4.522\text{cm}\end{array}$

$\begin{array}{l}\frac{28}{\sin {54}^o}=\frac{26}{\sin \angle ACB}\\ \sin \angle ACB=\frac{26\times \sin {54}^o}{28}\\ \sin \angle ACB=0.7512\\ \angle ACB={48.7}^o\end{array}$