SPM Additional Mathematics (Model Test Paper)


SPM Additional Mathematics (Model Test Paper)

Section B
[40 marks]
Answer any four questions from this section.

Question 11
In diagram 5, AOBDE, is a semicircle with centre O and has radius of 5cm. ABC is a right angle triangle.

It is given that AD DC =3.85 and DC=2.31cm.    
[Use π = 3.142]
Calculate
(a)    the value of θ, in radians,                                                                                      [2 marks]
(b)   the perimeter, in cm, of the segment ADE,                                                           [3 marks]
(c)    the area, in cm2, of the shaded region BCDF.                                                      [5 marks]

Answer and Solution:
(a)
Given  AD DC =3.85 and DC=2.31cm AD=3.85×2.31=8.8935 AC=8.8935+2.31=11.2035cm cosθ= AB AC = 10 11.2035 θ= 26.80 o θ= 26.80 o × π 180 o =0.4678 rad

(b)
ÐAOD = 3.142 – (0.4678 × 2)
            = 2.206 rad
Length of arc AED = 5 × 2.206
                                    = 11.03 cm
Therefore, perimeter of the segment ADE
= 11.03 + 8.8935
= 19.924 cm

(c)
BC = √ AC2AB2
      = √11.20352 – 102
      = 5.052 cm

2.206 rad× 180 o π = 126.38 o BOD=3.1422.206=0.936 rad Area of shaded region BCDF =Area of  ABCArea of  AODArea of segment OBD = 1 2 ( 10 )( 5.052 ) 1 2 ( 5 )( 5 )( sin126.38 ) 1 2 ( 5 ) 2 ( 0.936 ) =25.2610.0611.7 =3.50  cm 2