2003 - JAMB Mathematics Past Questions and Answers - page 3

Since PQ = QR, Δ PQR is a right angle isosceles triangle

∴ ∠PQR = 45^o

=> The bearing of P from R = 180 + 45

= 225^o

θ = 30 x 2(∠ at center twice ∠ at θ)

∴ θ = 60^o

Length of arc = (\frac{\theta}{360}\times 2\pi r\

=\frac{60}{360}\times 2 \times \frac{22}{7}\times \frac{21}{1}\

=22cm)

Locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1) is the perpendicular bisector of the straight line joining Q and R

Mid point QR = (x₂+x₁)/2 . (y₂+y₁)/2

= ²⁺⁰/₂ . ¹⁺⁰/₂

= ¹/₂

= -2

Equation of Pm = y - y₁ = m(x-x₁)

i.e y - ¹/₂ = -2(x-1)

2y - 1 = -4(x-1)

2y - 1 = -4x + 4

2y + 4x = 5

γ + S = 180(Allied angles)

S = 180 - γ

α + P = 180(Allied angles)

p = 180 - α

β = P + S

β = 180 - α + 180 - γ

β = 360 - α - γ

β + α + γ = 360

θ = 45 x 2 (∠ at center twice ∠ at θ)

θ = 90

Area of sector = ^θ/₃₆₀ x πr²

= ⁹⁰/₃₆₀ x ²²/₇ x 7 x 7

= ⁷⁷/₂cm²

Area of triangle = ¹/₂ ab sin θ

= ¹/₂ x 7 x 7 sin 90

= ¹/₂ x 7 x 7 x 1

= ⁴⁹/₂

Area of shaded region = ⁷⁷/₂ - ⁴⁹/₂

= ²⁸/₂

= 14cm²

R = P (base ∠s of isc Δ, ps = RS)

∴R = 40^o

∠PSQ = R (∠s in alternate segment)

∴∠PSQ = 40^o

Midpoint of PQ (\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\

=\frac{-1+(-3)}{2};\frac{6-4}{2}\

=\frac{-4}{2};\frac{2}{2}\

=-2,1)

Midpoint of QR = (\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\

\frac{-3+1}{2};\frac{-4-4}{2}\

\frac{-2}{2};\frac{-8}{2}\

-1,-4)

Midpoint of PQ and QR = (-2,1) and (-1,-4)

PS = SR

(\bigtriangleup)PSR is an isosceles

< P = < SPQ = 40^o

< R = < SRQ

< PSQ = < SRQ (alternate segment)

< PSQ = 40^o