2006 - JAMB Mathematics Past Questions and Answers - page 7

Area of PQRS = length x breath(7 x 10)cm2 = 70
Area of semi-circle (\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2})
= 19.5cm2 = 20cm2(appr.)
Area of the figure = (70 + 20)cm2
= 90cm2
Users' Answers & Comments
< QRS = 30o(corresponding < ), < QSR = 50o
< SQR = 180o - (30 + 50)o (Sum of < S in a (\bigtriangleup))
180o - 80o = 100o
Users' Answers & Comments
Construction RQ
< PRQ = 90o(amgle substended at the circumfernce of a circle by diameter). Also in figure PQRS
< PQS + PSQ = 180(supplementary angle)
< PQS = 180 - 145 = 35
In (\bigtriangleup)PRQ x = 180 - 935 = 90)
x = 55
Users' Answers & Comments
Draw a line perpendicular to |SR| to form |PT| in (\bigtriangleup) PST, cos 60 = (\frac{|ST|}{8}(
|PT| = 8 cos 6 = 4cm
Since |TR| = |PQ|
SR = ST + TR
= (4 + 10)cm
= 14cm
Users' Answers & Comments
Reflex of < 260 = 360 - 260
= 10(angle at a point0
x = 100 - 55 = 45o
Users' Answers & Comments
school fees is 90o
Transport = 45o
Housing and food = 360 - (90 + 45) = 225
food (\frac{2}{3} \times 225 = 150)
expenses on food = (\frac{150}{360} \times 30000)
= N12500
Users' Answers & Comments((\frac{52}{160} = \frac{36}{160}))% = (\frac{88}{160})%
= 55%
Users' Answers & Comments