Question 12:
Solutions by scale drawing is not accepted.
Diagram 6 shows quadrilateral JKMN with an area of L cm2. It is given that KN = √p cm and ∠KMN = 110o.
(a) Find
(i) the value of p,
(ii) the value of L.
[8 marks]
(b) Sketch Δ K’M’N’ which has a different shape from Δ KMN, such that K’M’ = KM, K’N’ = KN and ∠K’N’M’ = ∠KNM.
Hence, state the value of ∠M’K’N’.
[2 marks]
Answer:
(a)(i) $$ \begin{aligned} (\sqrt{p})^2 & =10^2+12^2-2(10)(12) \operatorname{cos} 60^{\circ} \\ p & =124 \end{aligned} $$
(a)(ii) $$ \begin{aligned} \frac{M N}{\sin 40^{\circ}} & =\frac{\sqrt{124}}{\sin 110^{\circ}} \\ M N & =7.617 \end{aligned} $$
$$ \begin{aligned} &\text { Area }\\ &\begin{aligned} & =\frac{1}{2}(10)(12) \sin 60^{\circ}+\frac{1}{2}(\sqrt{124})(7.617) \sin 30^{\circ} \\ & =73.16 \end{aligned} \end{aligned} $$
(b)
$$
\angle N^{\prime} K^{\prime} M^{\prime}=80^{\circ}
$$
Solutions by scale drawing is not accepted.
Diagram 6 shows quadrilateral JKMN with an area of L cm2. It is given that KN = √p cm and ∠KMN = 110o.
(a) Find
(i) the value of p,
(ii) the value of L.
[8 marks]
(b) Sketch Δ K’M’N’ which has a different shape from Δ KMN, such that K’M’ = KM, K’N’ = KN and ∠K’N’M’ = ∠KNM.
Hence, state the value of ∠M’K’N’.
[2 marks]
Answer:
(a)(i) $$ \begin{aligned} (\sqrt{p})^2 & =10^2+12^2-2(10)(12) \operatorname{cos} 60^{\circ} \\ p & =124 \end{aligned} $$
(a)(ii) $$ \begin{aligned} \frac{M N}{\sin 40^{\circ}} & =\frac{\sqrt{124}}{\sin 110^{\circ}} \\ M N & =7.617 \end{aligned} $$
$$ \begin{aligned} &\text { Area }\\ &\begin{aligned} & =\frac{1}{2}(10)(12) \sin 60^{\circ}+\frac{1}{2}(\sqrt{124})(7.617) \sin 30^{\circ} \\ & =73.16 \end{aligned} \end{aligned} $$
(b)
$$
\angle N^{\prime} K^{\prime} M^{\prime}=80^{\circ}
$$
Question 13:
A factory produces guava juice and watermelon juice. Due to limitation on production capacity, the factory can produce at most 180 packs of juice in a day. The number of packs of watermelon juice produced exceeds the number of packs of guava juice is less than 60 packs. The demand for the number of packs of watermelon juice is at least ½ of the number of packs of guava juice.
(a) In a day, the factory produces x packs of guava juice and y packs of watermelon juice.
Write three inequalities other than $x ≥ 0 and y ≥ 0 that satisfy all the above constraints. [3 marks]
(b) Using a scale of 2 cm to 20 packs on both axes, construct and shade the region R which satisfies all the above constraints. [3 marks]
(c) Use the graph constructed in 13(b) to answer the following questions:
(i) State the maximum number of packs of guava juice produced.
(ii) Due to an increase in the price of guava on a particular day, each pack of guava juice causes a loss of RM5.50.
The profit from each pack of watermelon juice is RM40.
If the factory has produced 80 packs of watermelon juice on that day, calculate the maximum profit obtained.
[4 marks]
Answer:
(a) $$ \begin{aligned} & x+y \leqslant 180 \\ & y-x<60 \\ & y \geqslant \frac{1}{2} x \end{aligned} $$
(b)
(c)(i) 120
(c)(ii) (21 , 80)
Maximum profit
= −5.5(21) + 40(80)
= RM3 084.50
A factory produces guava juice and watermelon juice. Due to limitation on production capacity, the factory can produce at most 180 packs of juice in a day. The number of packs of watermelon juice produced exceeds the number of packs of guava juice is less than 60 packs. The demand for the number of packs of watermelon juice is at least ½ of the number of packs of guava juice.
(a) In a day, the factory produces x packs of guava juice and y packs of watermelon juice.
Write three inequalities other than $x ≥ 0 and y ≥ 0 that satisfy all the above constraints. [3 marks]
(b) Using a scale of 2 cm to 20 packs on both axes, construct and shade the region R which satisfies all the above constraints. [3 marks]
(c) Use the graph constructed in 13(b) to answer the following questions:
(i) State the maximum number of packs of guava juice produced.
(ii) Due to an increase in the price of guava on a particular day, each pack of guava juice causes a loss of RM5.50.
The profit from each pack of watermelon juice is RM40.
If the factory has produced 80 packs of watermelon juice on that day, calculate the maximum profit obtained.
[4 marks]
Answer:
(a) $$ \begin{aligned} & x+y \leqslant 180 \\ & y-x<60 \\ & y \geqslant \frac{1}{2} x \end{aligned} $$
(b)
(c)(i) 120
(c)(ii) (21 , 80)
Maximum profit
= −5.5(21) + 40(80)
= RM3 084.50