Question 11:
Diagram 7 shows triangle ABC.
$$ \text { It is given that } B C=(2+x) \mathrm{cm}, A D^2=1+\sqrt{2} \text { and } A B D \text { is a straight line. } $$
$$ \text { Show that } x=\frac{1+5 \sqrt{2}}{49} \mathrm{~cm} \text {. } $$
[5 marks]
Answer:
$$ \begin{aligned} B D & =\sqrt{(2+x)^2-(2-x)^2} \\ & =\sqrt{8 x} \end{aligned} $$
$$ \sqrt{x}+\sqrt{8 x}=\sqrt{1+\sqrt{2}} $$
$$ \begin{aligned} &\text { Squaring both sides, }\\ &\begin{aligned} (\sqrt{8 x}+\sqrt{x})^2 & =(\sqrt{1+\sqrt{2}})^2 \\ 9 x+2 x \sqrt{8} & =1+\sqrt{2} \\ x(9+2 \sqrt{8}) & =1+\sqrt{2} \\ x & =\frac{1+\sqrt{2}}{9+2 \sqrt{8}} \times \frac{9-2 \sqrt{8}}{9-2 \sqrt{8}} \\ & =\frac{9+5 \sqrt{2}-8}{(9)^2-(2 \sqrt{8})^2} \\ & =\frac{1+5 \sqrt{2}}{(9)^2-(2 \sqrt{8})^2} \\ & =\frac{1+5 \sqrt{2}}{49} \end{aligned} \end{aligned} $$
Diagram 7 shows triangle ABC.
$$ \text { It is given that } B C=(2+x) \mathrm{cm}, A D^2=1+\sqrt{2} \text { and } A B D \text { is a straight line. } $$
$$ \text { Show that } x=\frac{1+5 \sqrt{2}}{49} \mathrm{~cm} \text {. } $$
[5 marks]
Answer:
$$ \begin{aligned} B D & =\sqrt{(2+x)^2-(2-x)^2} \\ & =\sqrt{8 x} \end{aligned} $$
$$ \sqrt{x}+\sqrt{8 x}=\sqrt{1+\sqrt{2}} $$
$$ \begin{aligned} &\text { Squaring both sides, }\\ &\begin{aligned} (\sqrt{8 x}+\sqrt{x})^2 & =(\sqrt{1+\sqrt{2}})^2 \\ 9 x+2 x \sqrt{8} & =1+\sqrt{2} \\ x(9+2 \sqrt{8}) & =1+\sqrt{2} \\ x & =\frac{1+\sqrt{2}}{9+2 \sqrt{8}} \times \frac{9-2 \sqrt{8}}{9-2 \sqrt{8}} \\ & =\frac{9+5 \sqrt{2}-8}{(9)^2-(2 \sqrt{8})^2} \\ & =\frac{1+5 \sqrt{2}}{(9)^2-(2 \sqrt{8})^2} \\ & =\frac{1+5 \sqrt{2}}{49} \end{aligned} \end{aligned} $$
Question 12:
Diagram 8 shows a purple coloured right angled triangle pattern on a net of a rectangular roll of paper.
It is given that the length of the base of the first triangle is a cm and the length of each subsequent base increases by d cm.
(a) Derive that the length of the base of the nth triangle is Tn = a + (n – 1)d. [3 marks]
(b) On the net, the 31st purple coloured triangle is the last triangle with an area of 72 cm2 and it is given that a = 30d.
(i) Find the value of d, in cm.
(ii) Zul wants to send a number of rolls of the paper to his friend through a courier company. Table 2 shows an information related to parcel delivery charges.
If Zul has RM15 and the mass of the paper per cm2 is 4 × 10-5 kg, calculate the maximum number of rolls of the paper that can be sent.
[6 marks]
Answer:
(a) $$ \begin{aligned} T_1 & =a, \\ T_2 & =a+d, \\ T_3 & =a+d+d \\ & =a+2 d \end{aligned} $$
$$ \begin{aligned} &\begin{aligned} & T_1=a+(1-1) d \\ & T_2=a+(2-1) d \\ & T_3=a+(3-1) d \end{aligned}\\ &\text { Jadi / So, }\\ &T_n=a+(n-1) d \end{aligned} $$
(b)(i) $$ \begin{aligned} \frac{1}{2} \times 12 \times x & =72 \\ x & =12 \end{aligned} $$
$$ \begin{aligned} T_{31} & =a+(31-1) d \\ & =12 \end{aligned} $$
$$ \begin{aligned} &\text { Given, } a=30 d\\ &\begin{aligned} 30 d+(31-1) d & =12 \\ 60 d & =12 \\ d & =\frac{1}{5} \end{aligned} \end{aligned} $$
(b)(ii) $$ \begin{aligned} a & =30\left(\frac{1}{5}\right) \\ & =6 \end{aligned} $$
$$ \begin{aligned} &\text { Mass of a roll of paper }\\ &\begin{aligned} & =S_{31} \times 12 \times 4 \times 10^{-5} \\ & =\frac{31}{2}\left[2(6)+(31-1)\left(\frac{1}{5}\right)\right] \times 12 \times 4 \times 10^{-5} \\ & =0.1339 \mathrm{~kg} \end{aligned} \end{aligned} $$
$$ \text { RM9: } \frac{2.5}{0.1339} \approx 18 \text { rolls } $$
$$ \text { RM6: RM1.20 every } 0.5 \mathrm{~kg} \rightarrow \mathrm{RM} 2.40 \text { every kg } $$
$$ \begin{aligned} & \frac{6}{2.4}=2.5 \mathrm{~kg} \\ & \frac{2.5}{0.1339} \approx 18 \text { rolls } \end{aligned} $$
$$ \begin{aligned} & 18+18=36 \text { rolls } \\ & 36 \times 0.1339 \mathrm{~kg}=4.820 \mathrm{~kg} \\ & 5-4.820 \mathrm{~kg}=0.18 \mathrm{~kg} \rightarrow 1 \text { more roll } \end{aligned} $$
$$ \text { The number of maximum rolls is }=36+1=37 $$
Diagram 8 shows a purple coloured right angled triangle pattern on a net of a rectangular roll of paper.
It is given that the length of the base of the first triangle is a cm and the length of each subsequent base increases by d cm.(a) Derive that the length of the base of the nth triangle is Tn = a + (n – 1)d. [3 marks]
(b) On the net, the 31st purple coloured triangle is the last triangle with an area of 72 cm2 and it is given that a = 30d.
(i) Find the value of d, in cm.
(ii) Zul wants to send a number of rolls of the paper to his friend through a courier company. Table 2 shows an information related to parcel delivery charges.
If Zul has RM15 and the mass of the paper per cm2 is 4 × 10-5 kg, calculate the maximum number of rolls of the paper that can be sent.
[6 marks]
Answer:
(a) $$ \begin{aligned} T_1 & =a, \\ T_2 & =a+d, \\ T_3 & =a+d+d \\ & =a+2 d \end{aligned} $$
$$ \begin{aligned} &\begin{aligned} & T_1=a+(1-1) d \\ & T_2=a+(2-1) d \\ & T_3=a+(3-1) d \end{aligned}\\ &\text { Jadi / So, }\\ &T_n=a+(n-1) d \end{aligned} $$
(b)(i) $$ \begin{aligned} \frac{1}{2} \times 12 \times x & =72 \\ x & =12 \end{aligned} $$
$$ \begin{aligned} T_{31} & =a+(31-1) d \\ & =12 \end{aligned} $$
$$ \begin{aligned} &\text { Given, } a=30 d\\ &\begin{aligned} 30 d+(31-1) d & =12 \\ 60 d & =12 \\ d & =\frac{1}{5} \end{aligned} \end{aligned} $$
(b)(ii) $$ \begin{aligned} a & =30\left(\frac{1}{5}\right) \\ & =6 \end{aligned} $$
$$ \begin{aligned} &\text { Mass of a roll of paper }\\ &\begin{aligned} & =S_{31} \times 12 \times 4 \times 10^{-5} \\ & =\frac{31}{2}\left[2(6)+(31-1)\left(\frac{1}{5}\right)\right] \times 12 \times 4 \times 10^{-5} \\ & =0.1339 \mathrm{~kg} \end{aligned} \end{aligned} $$
$$ \text { RM9: } \frac{2.5}{0.1339} \approx 18 \text { rolls } $$
$$ \text { RM6: RM1.20 every } 0.5 \mathrm{~kg} \rightarrow \mathrm{RM} 2.40 \text { every kg } $$
$$ \begin{aligned} & \frac{6}{2.4}=2.5 \mathrm{~kg} \\ & \frac{2.5}{0.1339} \approx 18 \text { rolls } \end{aligned} $$
$$ \begin{aligned} & 18+18=36 \text { rolls } \\ & 36 \times 0.1339 \mathrm{~kg}=4.820 \mathrm{~kg} \\ & 5-4.820 \mathrm{~kg}=0.18 \mathrm{~kg} \rightarrow 1 \text { more roll } \end{aligned} $$
$$ \text { The number of maximum rolls is }=36+1=37 $$