WAEC Mathematics Past Questions & Answers - Page 143

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. 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

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

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

Note: A {horizontal}

B {vertical}

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

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

Note: A {horizontal}

B {vertical}

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

= 0

A = [2, 4, 6, 8}

B = {2, 3, 5, 9}

Pr = (Prime in B) = $\frac{3}{4}$