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JAMB Mathematics Past Questions & Answers - Page 159

791.

In $\bigtriangleup$ PQR, PQ = 10cm, QR = 8cm and RP = 6cm, the perpendicular RS is drawn from R to PQ. Find the length of RS

4cm

32cm

$\frac{30}{7}$

$\frac{40}{7}$

4.8cm

View answer & explanation

Correct answer is E

Cos Q = $\frac{r^2 + p^2 - q^2}{2rp}$

= $\frac{10^2 + 8^2 - 6^2}{2(10)(8)}$

= $\frac{164 - 36}{160}$

= $\frac{128}{160}$

= 0.8

Q = Cos^-1 o.8

= 37^o x rep. from rt< RSQ, Let RS = x

$\frac{x}{sin 37^o}$ = $\frac{8}{sin 90^o}$

but sin 90^o = 1

x = 8 sin 37^o

x = 4.8cm

792.

(3.2)² - (1.8)² equals

7.0

2.56

13.48

2.0

0.07

View answer & explanation

Correct answer is A

(3.2)² - (1.8)² = (3.2 + 1.8)(3.2 - 1.8)

= 5 x 1.4

= 7

793.

If the value of $\pi$ is taken to be $\frac{22}{7}$ , the area of a semi-circle of diameter 42m is

5544m²

1386m²

132m²

264m²

693m²

View answer & explanation

Correct answer is E

$\pi$ = $\frac{22}{7}$ , Diameter D = 42m

Area of a circle = $\frac{\pi D^2}{4}$ when a diameter is given

area of a semicircle $\frac{\pi D^2}{4}$ x $\frac{1}{2}$

= $\frac{22}{7}$ x $\frac{4262}{4}$ x $\frac{1}{2}$

= $\frac{11}{7} \times 441$

= 693m $^{2}$