JAMB Mathematics Past Questions & Answers

661.

In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS

60^o

90^o

120^o

30^o

80^o

View answer & explanation

Correct answer is A

From the diagram, PQRSTW is a regular hexagon.

Hexagon is a six sided polygon.

Sum of interior angles of polygon = (2n - 4)90^o = [2 x 6 - 4] x 90 = 8 x 90 = 720^o

each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)

662.

Find the area of the shaded portion of the semicircular figure.

\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)

\(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)

\(\frac{1}{2}r^2 \pi\)

\(\frac{1}{8}r^2 \sqrt{3}\)

\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)

View answer & explanation

Correct answer is B

A_sector = \(\frac{60}{360} \times \pi r^2\)

= \(\frac{1}{6} \pi r^2\)

A_{\(\bigtriangleup\)} = \(\frac{1}{2}r^2 \sin 60^o\)

\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)

A_{\(\text{shaded portion}\)} = A_sector -
A_{\(\bigtriangleup\)}

= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)

= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)

= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)

663.

Using \(\bigtriangleup\)XYZ in the figure, find XYZ

29^o

31^o

31^o 20'

31^o 18'

View answer & explanation

Correct answer is C

\(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\)

sin 120^o = sin 60^o

5 sin y = 3 sin 60^o

sin y = \(\frac{3 \sin 60^o}{5}\)

\(\frac{3 \times 0.866}{5}\)

= \(\frac{2.598}{5}\)

y = sin^-1 0.5196 = 30^o 18'