Question 13:
Solution by scale drawing are not accepted.
Diagram 13 shows a transparent glass block with a base ABCD on a horizontal plane. AFB is a coloured plane inside the block.
It is given that the plane ABE is an isosceles triangle with an area of 9√2 cm2 and ∠BAE = 45o.
(a) Show that AB = 6 cm. [2 marks]
(b) Given that BC = 12 cm, AF = 10 cm, CF = 7 cm and ∠FCB = 90o, calculate
(i) ∠AFB,
(ii) the area, in cm2, of the coloured plane by using the Heron’s formula.
[5 marks]
(c) It is given that AB = A‘B‘, AF = A‘F‘ and ∠AFB = A’F’B’.
(i) Sketch ΔA‘B‘F‘ which has a different shape from ΔABF.
(ii) Hence, calculate the ΔA‘B‘F‘.
[3 marks]
Answer:
(a)
$$ \begin{aligned} & \text { Area of } A B E=9 \sqrt{2} \mathrm{~cm}^2\\ &\begin{aligned} \frac{1}{2}(x)(x) \sin 45^{\circ} & =9 \sqrt{2} \mathrm{~cm}^2 \\ \frac{1}{2 \sqrt{2}} x^2 & =9 \sqrt{2} \mathrm{~cm}^2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} x^2 & =(9 \sqrt{2})(2 \sqrt{2}) \mathrm{cm} \\ x^2 & =36 \mathrm{~cm} \\ x & =6 \mathrm{~cm} \end{aligned} $$
$$ \therefore A B=6 \mathrm{~cm} \text { (shown) } $$
(b)(i) $$ \begin{aligned} F B & =\sqrt{12^2+7^2} \\ & =13.89 \mathrm{~cm} \end{aligned} $$
$$ \begin{aligned} A B^2 & =A F^2+B F^2-2(A F)(B F) \mathrm{cos} \angle A F B \\ 6^2 & =10^2+(13.89)^2-2(10)(13.89) \mathrm{cos} \angle A F B \\ 36 & =292.9321-277.8 \mathrm{cos} \angle A F B \\ \operatorname{cos} \angle A F B & =\frac{292.9321-36}{277.8} \\ \angle A F B & =\mathrm{cos}^{-1} (0.9249) \\ \angle A F B & =22.35^{\circ} \end{aligned} $$
(b)(ii) $$ \begin{aligned} s & =\frac{10+13.89+6}{2} \\ & =14.945^2 \end{aligned} $$
$$ \begin{aligned} \text { Area } & =\sqrt{14.945(14.945-6)(14.945-10)(14.945-13.89)} \\ & =\sqrt{697.4210} \\ & =26.4087 \mathrm{~cm}^2 \end{aligned} $$
(c)(i)
(c)(ii)
$$ \begin{aligned} \frac{\sin \angle A B F}{10} & =\frac{\sin 22.35^{\circ}}{6} \\ \sin \angle A B F & =\frac{\sin 22.35^{\circ}}{6} \times 10 \\ \angle A B F & =\sin ^{-1}(0.6338) \\ \angle A B F & =39.33^{\circ} \end{aligned} $$
$$ \therefore \begin{aligned} \angle A^{\prime} B^{\prime} F^{\prime} & =180^{\circ}-39.33^{\circ} \\ & =140.67^{\circ} \end{aligned} $$
Solution by scale drawing are not accepted.
Diagram 13 shows a transparent glass block with a base ABCD on a horizontal plane. AFB is a coloured plane inside the block.
It is given that the plane ABE is an isosceles triangle with an area of 9√2 cm2 and ∠BAE = 45o.
(a) Show that AB = 6 cm. [2 marks]
(b) Given that BC = 12 cm, AF = 10 cm, CF = 7 cm and ∠FCB = 90o, calculate
(i) ∠AFB,
(ii) the area, in cm2, of the coloured plane by using the Heron’s formula.
[5 marks]
(c) It is given that AB = A‘B‘, AF = A‘F‘ and ∠AFB = A’F’B’.
(i) Sketch ΔA‘B‘F‘ which has a different shape from ΔABF.
(ii) Hence, calculate the ΔA‘B‘F‘.
[3 marks]
Answer:
(a)
$$ \begin{aligned} & \text { Area of } A B E=9 \sqrt{2} \mathrm{~cm}^2\\ &\begin{aligned} \frac{1}{2}(x)(x) \sin 45^{\circ} & =9 \sqrt{2} \mathrm{~cm}^2 \\ \frac{1}{2 \sqrt{2}} x^2 & =9 \sqrt{2} \mathrm{~cm}^2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} x^2 & =(9 \sqrt{2})(2 \sqrt{2}) \mathrm{cm} \\ x^2 & =36 \mathrm{~cm} \\ x & =6 \mathrm{~cm} \end{aligned} $$
$$ \therefore A B=6 \mathrm{~cm} \text { (shown) } $$
(b)(i) $$ \begin{aligned} F B & =\sqrt{12^2+7^2} \\ & =13.89 \mathrm{~cm} \end{aligned} $$
$$ \begin{aligned} A B^2 & =A F^2+B F^2-2(A F)(B F) \mathrm{cos} \angle A F B \\ 6^2 & =10^2+(13.89)^2-2(10)(13.89) \mathrm{cos} \angle A F B \\ 36 & =292.9321-277.8 \mathrm{cos} \angle A F B \\ \operatorname{cos} \angle A F B & =\frac{292.9321-36}{277.8} \\ \angle A F B & =\mathrm{cos}^{-1} (0.9249) \\ \angle A F B & =22.35^{\circ} \end{aligned} $$
(b)(ii) $$ \begin{aligned} s & =\frac{10+13.89+6}{2} \\ & =14.945^2 \end{aligned} $$
$$ \begin{aligned} \text { Area } & =\sqrt{14.945(14.945-6)(14.945-10)(14.945-13.89)} \\ & =\sqrt{697.4210} \\ & =26.4087 \mathrm{~cm}^2 \end{aligned} $$
(c)(i)
(c)(ii)
$$ \begin{aligned} \frac{\sin \angle A B F}{10} & =\frac{\sin 22.35^{\circ}}{6} \\ \sin \angle A B F & =\frac{\sin 22.35^{\circ}}{6} \times 10 \\ \angle A B F & =\sin ^{-1}(0.6338) \\ \angle A B F & =39.33^{\circ} \end{aligned} $$
$$ \therefore \begin{aligned} \angle A^{\prime} B^{\prime} F^{\prime} & =180^{\circ}-39.33^{\circ} \\ & =140.67^{\circ} \end{aligned} $$
Question 14:
Table 14 shows information for three ingredients used to make a product.
(a) The price of ingredient R decreased by 7% from the year 2022 to the year 2024.
(i) State the value of x.
(ii) Hence, find the price of ingredient R in the year 2022 if the price of ingredient R in the year 2024 is RM18.60.
[3 marks]
(b)(i) It is given that the ratio of the cost of ingredient P to ingredient Q is 2 : 5 while the ratio of the cost of ingredient Q to ingredient R is 3 : 4.
Calculate and interpret the composite index for the cost of ingredients of the product in the year 2024 with 2022 as the base year.
(ii) The cost of all ingredients is expected to increase by 15% from the year 2024 to the year 2026. The manufacturer wants to obtain a profit of at least 50% more than the cost of the ingredients of the product in the year 2026.
If the cost of the ingredients of the product in the year 2022 is RM40, determine the range of the selling price of the products in the year 2026.
[7 marks]
Answer:
(a)(i) $$ \begin{aligned} x & =100-7 \\ & =93 \end{aligned} $$
(a)(ii) $$ \begin{aligned} R: I_{2024 / 2022} & =93 \\ \frac{18.60}{P_{22}} \times 100 & =93 \\ P_{22} & =\frac{18.60}{93} \times 100 \\ P_{22} & =\text { RM } 20 \end{aligned} $$
(b)(i)
$$ \begin{aligned} \bar{I}_{2024 / 2022} & =\frac{(130)(6)+(120)(15)+(93)(20)}{6+15+20} \\ & =108.29 \end{aligned} $$
The cost on ingredients of the product increases 8.29% from year 2022 to year 2024.
(b)(ii)
$$ \begin{aligned} I_{2024 / 2022} & =108.29 \\ \frac{P_{2024}}{40} \times 100 & =108.29 \\ P_{2024} & =\frac{108.29}{100} \times 40 \\ P_{2024} & =43.316 \end{aligned} $$
$$ \begin{aligned} I_{2026 / 2024} & =115 \\ \frac{P_{2026}}{43.316} \times 100 & =115 \\ P_{2026} & =\frac{115}{100} \times 43.316 \\ P_{2026} & =\text { RM49.81 } \end{aligned} $$
$$ \begin{aligned} & \text { Selling price } \geqslant \text { RM49.81 }+\left(\frac{50}{100} \times \text { RM49.81 }\right) \\ & \text { Selling price } \geqslant \text { RM74.72 } \end{aligned} $$
Table 14 shows information for three ingredients used to make a product.
(a) The price of ingredient R decreased by 7% from the year 2022 to the year 2024.
(i) State the value of x.
(ii) Hence, find the price of ingredient R in the year 2022 if the price of ingredient R in the year 2024 is RM18.60.
[3 marks]
(b)(i) It is given that the ratio of the cost of ingredient P to ingredient Q is 2 : 5 while the ratio of the cost of ingredient Q to ingredient R is 3 : 4.
Calculate and interpret the composite index for the cost of ingredients of the product in the year 2024 with 2022 as the base year.
(ii) The cost of all ingredients is expected to increase by 15% from the year 2024 to the year 2026. The manufacturer wants to obtain a profit of at least 50% more than the cost of the ingredients of the product in the year 2026.
If the cost of the ingredients of the product in the year 2022 is RM40, determine the range of the selling price of the products in the year 2026.
[7 marks]
Answer:
(a)(i) $$ \begin{aligned} x & =100-7 \\ & =93 \end{aligned} $$
(a)(ii) $$ \begin{aligned} R: I_{2024 / 2022} & =93 \\ \frac{18.60}{P_{22}} \times 100 & =93 \\ P_{22} & =\frac{18.60}{93} \times 100 \\ P_{22} & =\text { RM } 20 \end{aligned} $$
(b)(i)
$$ \begin{aligned} \bar{I}_{2024 / 2022} & =\frac{(130)(6)+(120)(15)+(93)(20)}{6+15+20} \\ & =108.29 \end{aligned} $$
The cost on ingredients of the product increases 8.29% from year 2022 to year 2024.
(b)(ii)
$$ \begin{aligned} I_{2024 / 2022} & =108.29 \\ \frac{P_{2024}}{40} \times 100 & =108.29 \\ P_{2024} & =\frac{108.29}{100} \times 40 \\ P_{2024} & =43.316 \end{aligned} $$
$$ \begin{aligned} I_{2026 / 2024} & =115 \\ \frac{P_{2026}}{43.316} \times 100 & =115 \\ P_{2026} & =\frac{115}{100} \times 43.316 \\ P_{2026} & =\text { RM49.81 } \end{aligned} $$
$$ \begin{aligned} & \text { Selling price } \geqslant \text { RM49.81 }+\left(\frac{50}{100} \times \text { RM49.81 }\right) \\ & \text { Selling price } \geqslant \text { RM74.72 } \end{aligned} $$