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

< QRS = 30^o(corresponding < ), < QSR = 50^o

< SQR = 180^o - (30 + 50)^o (Sum of < S in a \(\bigtriangleup\))

180^o - 80^o = 100^o

Construction RQ

< PRQ = 90^o(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

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

Reflex of < 260 = 360 - 260

= 10(angle at a point0

x = 100 - 55 = 45^o

school fees is 90^o

Transport = 45^o

Housing and food = 360 - (90 + 45) = 225

food \(\frac{2}{3} \times 225 = 150\)

expenses on food = \(\frac{150}{360} \times 30000\)

= N12500

(\(\frac{52}{160} = \frac{36}{160}\))% = \(\frac{88}{160}\)%

= 55%