SPM Additional Mathematics 2018, Paper 2 (Question 4 – 6)


Question 4:
Diagram 2 shows the plan of a rectangular garden ABCD. The garden consists of a semicircular pond ATD and grassy area ABCDT.
It is given that DC = 6y metre and BC = 7x metre, xy. The area of the rectangular garden ABCD is 168 metre2 and the perimeter of the grassy area is 60 metre. The pond with uniform depth contains 15.4 metre3 of water.
By using  π= 22 7 , find the depth, in metre, of water in the pond.

Solution:
Area of ABCD=168 ( 6y )( 7x )=168 42xy=168 xy=4……………….( 1 ) Perimeter of grassy area=60 6y+6y+7x+( 1 2 × 2 × 22 11 7 × 7 x 2 )=60 12y+18x=60 2y+3x=10……………….( 2 ) From ( 1 ):xy=4 x= 4 y ……………….( 3 ) Substitute (3) into (2): 2y+3( 4 y )=10 2 y 2 +12=10y y 2 5y+6=0 ( y2 )( y3 )=0 y2=0    y=2     or     y3=0 y=3 Substitute the values of y into (3): When y=2 x= 4 2 =2 ( ignored, xy ) When y=3 x= 4 3 Given volume of water=15.4 1 2 ( 22 7 ) ( 7x 2 ) 2 d=15.4 1 2 ( 22 7 ) [ 7( 4 3 ) 2 ] 2 d=15.4 11 7 ( 14 3 ) 2 d=15.4 308 9 d=15.4 d=0.45 m Thus, the depth of water in the pond is 0.45 m.



Question 5:
Mathematics Society of SMK Mulia organized a competition to design a logo for the society.


Diagram 3 shows the circular logo designed by Adrian. The three blue coloured regions are congruent. It is given that the perimeter of the blue coloured region is 20π cm.
[Use π = 3.142]
Find
(a) the radius, in cm, of the logo to the nearest integer,
(b) the area, in cm2, of the yellow coloured region.

Solution:
(a)
6 arcs =20π 6rθ=20π 6r[ 60 o × π 180 o 3 ]=20π 2πr=20π r=10 cm

(b)

Area of yellow coloured region =3[ area of triangle OAB ]6[ area of segment ] =3[ 1 2 absinC ]6[ 1 2 r 2 ( θsinθ ) ] =3[ 1 2 ( 10 )( 10 )sin 120 o ]6[ 1 2 ( 10 ) 2 ( θsinθ ) ] =3( 43.3013 )6[ 50( 1.0473sin1.0473 ) ] change to rad mode θ= 60 o × 3.142 180 o =1.0473 =129.90396( 9.0612 ) =129.903954.3672 =75.54  cm 2


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