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WAEC Mathematics Past Questions & Answers - Page 197

981.

In $∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°$ and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place.

3.4

4.3

5.9

6.2

View answer & explanation

Correct answer is B

Using the sine rule
$\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\ \frac{x}{sin43^o}=\frac{5}{sin53^o}\\ x = \frac{5\times sin43^o}{sin53^o}\\ =\frac{5\times 0.6820}{0.7986}\\ =4.3$

982.

Find the value of x in the equation 3x $^2$ - 8x - 3 = 0

$\frac{1}{3},-3$

$-\frac{1}{3},-3$

$-\frac{1}{3},3$

$\frac{1}{3},3$

View answer & explanation

Correct answer is C

3x $^2$ - 8x - 3 = 0

3x $^2$ - 9x + x - 3 = 0

3x(x - 3) + 1(x - 3) = 0

(3x + 1)(x - 3) = 0

3x = -1 $\implies$ x = $-\frac{1}{3}$

or x = 3.

983.

In the diagram, ZM is a straight line. Calculate the value of x.

27^o

30^o

35^o

37^o

View answer & explanation

Correct answer is C

The sum of angles in a triangle is 180^o
2x^o + (2x -21)^o + ∠ZYP = 180^o
Also ∠ZYP = 180^o - (3x+14)^o
∴2x^o + (2x-21)^o + 180^o - (3x+14)^o = 180
x = 35^o

984.

PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?

180^o

150^o

105^o

75^o

View answer & explanation

Correct answer is C

< QPS = 75°

< QRS = 180° - 75° = 105° (opposite angles of a cyclic quadrilateral)

985.

A pyramid of volume 120cm³ has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid

5cm

9cm

12cm

15cm

View answer & explanation

Correct answer is C

$V= \frac{1}{2}\hspace{1mm}base\hspace{1mm}area\hspace{1mm}\times\hspace{1mm}height\\ 120=\frac{1}{3}\times 5 \times 6 \times h; h = \frac{120}{10}=12cm$