2008 - WAEC Mathematics Past Questions and Answers - page 4

\(\begin{array}{c|c} x & 2 & 3 & 5 & 9 \ \hline 2 & 4 & 6 & 10 & 18 \ \hline 4 & 8 & 12 & 20 & 36 \ \hline 6 & 12 & 18 & 30 & 54 \ \hline 8 & 16 & 24 & 40 & 72 \end{array}\)

Note: A {horizontal}

B {vertical}

Pr (Odd Product) = \(\frac{0}{16}\)

= 0

\(\begin{array}{c|c} - & 2 & 3 & 5 & 9 \ \hline 2 & 0 & 1 & 3 & 7 \ \hline 4 & 2 & 1 & 1 & 5\ \hline 6 & 4 & 3 & 1 & 3 \ \hline 8 & 6 &5 & 3 & 1 \end{array}\)

Note: A {horizontal}

B {vertical}

Pr(Difference of 6 or 7) = \(\frac{2}{16} = \frac{1}{8}\)

\(\begin{array}{c|c} x & d = x - \bar{x} & |d| \ \hline 10 & -4.5 & 4.5 \ 12 & -2.5 & 2.5\ 14 & -0.5 & 0.5\ 15 & 0.5 & 0.5\ 17 & 2.5 & 2.5\ 19 & 4.5 & 4.5\ \hline \sum x = 87 & & \sum|d| = 15\end{array}\)

\(\bar{x} = \frac{\sum x}{n}\)

= \(\frac{87}{6}\)

= 14.5

Given q² = 25 - r².....(1)

from Pythagoras theorem

p² = q² + r²....(2)

put(1) into (2)

p² = 25 - r² + r²

p² = 25

p = \(\sqrt{25}\)

Since |PR| = |RS| = |SP|

\(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70^o

But < PQR + < PSR = 180^o (Opposite interior angles of a cyclic quadrilateral)

< PQR + 60 = 180^o

< PR = 180 - 60 = 120^o

But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)

< QPR + < PRQ + < PQR = 180^o (Angles in a triangle)

2 < QPR + 120 = 18-

2 < QPR = 180 - 120

QPR = \(\frac{60}{2}\) = 30^o

From the diagram, < QRS = < PRQ + < PRS

30 + 60 = 90^o

For a rhombus, all the sides are equal and diagonals bisect each other at 90^o. Hence, the triangles formed are congruent: (under RHS).

Thus: \(\bar{PS}^2\) = 5² + 12²

\(\bar{PS} = \sqrt{25 + 144}\)

= \(\sqrt{169}\)

\(\bar{PS}\) = 13cm

perimeter = 4 x length of a side

= 4 x 13cm

= 52cm

< QPR + < PRQ = < RQS

(Sum of two interior angles of a triangle = Opposite exterior angles)

70^o + < PRQ = 148

< PRQ = 148^o - 70^o

= 78^o

From the diagram, < QPS = x^o (Corresponding angles)

Also, < QPS = < PSR(Alternate angles)

x = 42^o

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A₁, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A₂

Area of \(\bigtriangleup\)PQR = A = A₁ + 2A₂

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A₂ But, A = A₁ + 2A

A₁ = 4A₂ - 2A₂

A₁ = 2A₂

\(\frac{A_1}{A_2}\) = 2

A₁:A₂ = 2:1


Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

x = 65^o (An interior angle of a cyclic quadrilateral = opposite exterior angle).