# SPM Additional Mathematics (Model Test Paper)

SPM Additional Mathematics (Model Test Paper)

Section C
[20 marks]
Answer any two questions from this section.

Question 12
Diagram 6 shows a quadrilateral KLMN.

Calculate
(a)    ÐKML,                                                                                                                  [2 marks]
(b)   the length, in cm, of KM,                                                                                      [3 marks]
(c)    area, in cm2, of triangle KMN,                                                                              [3 marks]
(d)   a triangle K’L’M’ has the same measurements as those given for triangle KLM, that is K’L’= 12.4 cm, L’M’= 9.5 cm and ÐL’K’M’ = 43.2o, which is different in shape to triangle KLM.
(i)     Sketch the triangle K’L’M’,
(ii)   State the size of ÐK’M’L’.                                                                             [2 marks]

(a)
$\begin{array}{l}\frac{\mathrm{sin}\angle KML}{12.4}=\frac{\mathrm{sin}{43.2}^{o}}{9.5}\\ \mathrm{sin}\angle KML=\frac{\mathrm{sin}{43.2}^{o}}{9.5}×12.4\\ \mathrm{sin}\angle KML=0.8935\\ \angle KML={63.32}^{o}\end{array}$

(b)
ÐKLM = 180o – 43.2o – 63.32o = 73.48o
KM2 = 9.52 + 12.42 – 2(9.5)(12.4) cos 73.48o
KM2 = 244.01 – 66.99
KM2 = 177.02
KM = 13.30 cm

(c)
KM2 = MN2+ KN2– 2(MN)(KN) cos ÐKNM
13.302 = 5.42 + 9.92 – 2(5.4)(9.9) cos ÐKNM
176.89 = 127.17 – 106.92 cos ÐKNM
$\begin{array}{l}\mathrm{cos}\angle KNM=\frac{127.17-176.89}{106.92}\\ \angle KNM={117.71}^{o}\end{array}$

Area of triangle KMN

(d)(i)

(d)(ii)
ÐK’M’L’ = 180o – 63.32o = 116.68o