JAMB Mathematics Past Questions & Answers - Page 2

|AP| = |PB| = \(x\) (The perpendicular to a chord bisects the chord if drawn from the center of the circle.)

From ∆OPB

Using Pythagoras theorem

⇒ \(13^2 = 12^2 + x^2\)

⇒ \(169 = 144 + x^2\)

⇒ \(169 - 144 = x^2\)

⇒ \(x^2 = 25\)

⇒ \(x = \sqrt25 = 5 cm\)

∴ Length of the chord |AB| = \(x  + x  = 5 + 5 = 10 cm\)

On the \(y\)-axis, each box is \(\frac{1 - 0}{5} = \frac{1}{5}\) = 0.2unit

On the \(x\)-axis, each box is \(\frac{90 - 0}{6} = \frac{90}{6} = 15^o\)

⇒ \(θ_1 = 180^o + (2.5\times15^o) = 180^o + 37.5^o = 217.5^o ≃ 218^o \)(2 and half boxes were counted to the right of 180\(^o\))

⇒ \(θ_2 = 270^o + (3.5\times15^o) = 270^o + 52.5^o = 322.5^o ≃ 323^o \)(3 and half boxes were counted to the right of 270\(^o\))

∴ \(θ = 218^o, 323^o\)

AB = \(\frac{θ}{360}\times 2\pi Rcos\propto\) (distance on small circle)

= 64 - 56 = 8\(^o\)

\(\propto = 86^o\)

⇒ AB = \(\frac{8}{360}\) x 2 x 3.142 x 6370 x cos 86

⇒ AB = \(\frac{22,338.29974}{360}\)

∴ AB = 62km (to the nearest km)

Perimeter of a triangle = sum of all sides

⇒ \(P = y + x + x = 2\)

⇒ \(y + 2x = 2\)

⇒ \(y= 2 - 2x\)-----(i)

Using Pythagoras theorem

\(y^2 = x^2 + x^2\)

⇒ \(y^2 = 2x^2\)

⇒ \(y = \sqrt2x^2\)

⇒ \(y = x\sqrt2\)-----(ii)

Equate \(y\)

⇒ \(2 - 2x = x\sqrt2\)

Square both sides

⇒ \((2 -2x) ^2 = (x\sqrt2)^2\)

⇒ \(4 - 8x + 4x^2 = 2x^2\)

⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\)

⇒ \(2x^2 - 8x + 4 = 0\)

⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\)

⇒ \(x = \frac{8\pm\sqrt32}{4}\)

⇒ \(x = \frac{8\pm4\sqrt2}{4}\)

⇒ \(x = 2\pm\sqrt2\)

⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\)

∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter)

∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\)

∴ The length of the longer side = -2 + 2\(\sqrt2\)m

Speed = \(\frac{Distance}{Time}\)

⇒ Time = \(\frac{Distance}{Time}\)

Let D = distance between the two airports

∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\)

Since total time of flight= 3hours,

⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3

⇒ \(\frac{15D + 12D}{1800}\) = 3

⇒ \(\frac{27D}{1800}\) = 3

⇒ \(\frac{3D}{200} = \frac{3}{1}\)

⇒ 3D = 200 x 3

∴ D =\(\frac{ 200\times3}{3}\)= 200km