1989 - WAEC Mathematics Past Questions and Answers - page 1
a2 = 64 a = 8
ar = 8 .....(1)
ar3 = 32 .....(2)
divide (2) by (1)
r2 = 4 r = 2 ∴ a = 4
In the diagram above, TPS is a straight line, PQRS is a parallelogram with base QR and height 8cm. |QR| = 6cm and the area of triangle QST is 52cm2. Find the area of ΔQPT
Area of \(\Delta\)QST = 52cm\(^2\); height = 8cm
b=base length; but A = 1/2 b x h
=52=1/2 x b x 8cm = 104 =8b
b=104/8 = 13cm; ST= 13cm; PS = 6cm
TP = 13 - 6 = 7cm;
area of \(\Delta\)QST = 1/2 b x h
= 1/2 x 7 x 8/1 = 28cm\(^2\)
In the diagram O is the center of the circle, if ∠QRS = 62o, find the value of ∠SQR.
< QSR = 90° (angle in a semi-circle)
\(\therefore\) < SQR = 180° - (90° + 62°)
= 28°
Evaluate, using logarithm tables \(\frac{5.34 \times 67.4}{2.7}\)
| No | Log |
| 5.34 | 0.7275 + |
| 67.4 | 1.8287 |
| 2.5562 - | |
| 2.7 | 0.4314 |
| Antilog = 133.2 | 2.1248 |
\(\therefore \frac{5.34 \times 67.4}{2.7} = 133.2\)
35 - 7b + 5b - b 2
7(5 - b) + 5(5 - b)
(7 + b)(5 - b)
Evaluate \(\frac{0.009}{0.012}\), leaving your answer in standard form.
\(\frac{0.009}{0.012}\)
=\(\frac{0.009}{0.012}\) = \(\frac{9}{12}\) = 0.75
=\(7.5 \times 10^{-1}\)
The roots of the equation are
(2x 2 + 6x) - ( 5x + 15)
2x(x + 3) - 5(x + 3)
(2x - 5)(x + 3)