Processing math: 100%

WAEC Mathematics Past Questions & Answers - Page 147

731.

PQR is a sector of a circle centre O, radius 4cm. If PQR = 30^o, find, correct to 3 significant figures, the area of sector PQR. [Take $\pi = \frac{22}{7}$ ]

4.19cm²

8.38cm²

10.5cm²

20.9cm²

View answer & explanation

Correct answer is A

Given q² = 25 - r².....(1)

from pythagora's theorem

P² = q² + r².......(2)

put (1) into (2)

p² = 25 - p² + x²

p² = 25

p = $\sqrt{25}$

= 5

732.

Simplify: $\frac{2x^2 - 5x - 12}{4x^2 - 9}$

$\frac{x + 4}{2x + 3}$

$\frac{x + 4}{2x - 3}$

$\frac{x - 4}{2x + 3}$

$\frac{x - 4}{2x - 3}$

View answer & explanation

Correct answer is D

$\frac{2x^2 - 5x - 12}{4x^2 - 9}$ = $\frac{3x^2 - 8x + 3x - 12}{(2x)^2 - 3^2}$

= $\frac{32(x - 4) + 3(x - 4)}{(2x - 3)(2x + 3)} - \frac{(x - 4) + (2x + 3)}{(2x - 3) (2x + 3)}$

= $\frac{x - 4}{2x - 3}$

733.

If p = $\frac{3}{5} \sqrt{\frac{q}{r}}$ , express q in terms of p and r

$\frac{9}{25} pr^2$

$\frac{9}{25} p^2r$

$\frac{25}{9} p^2r$

$\frac{25}{9} pr^2$

View answer & explanation

Correct answer is C

p = $\frac{3}{5} \sqrt{\frac{q}{r}}; \frac{5p}{3} = \sqrt{\frac{q}{r}}$

= ( $\frac{5}{3}p$ )²

= $\frac{q}{r}$

= $\frac{25p^2}{9} = \frac{q}{r}$

q = $\frac{25}{9} p^2 r$