JAMB Mathematics Past Questions & Answers - Page 221

The locus of the mid-point of all chords of length 6cm within a circle of radius 5cm and with centre O is the perpendicular bisectors of the chords.

Base of pyramid of a square of side 8cm vertex directly above the centre edge = \(4\sqrt{3}\)cm

From the diagram, the diagonal of one base is AC2 = 82 + 82

Ac2 = 64 + 64 = 128

AC = \(8\sqrt{2}\)

but OC = \(\frac{1}{2}\)AC = 8\(\sqrt{\frac{2}{2}}\) = \(4\sqrt{2}\)cm

OE = h = height

h2 = (\(4\sqrt{3}\))2

16 x 2 - 16 x 2

48 - 32 = 16

h = \(\sqrt{16}\)

= 4

Circumference of Latitude 0oS where R is the radius of the earth is = 2\(\pi\) R cos \(\theta\)

ABCD is the floor, by pythagoras

AC² = 144 + 81 = \(\sqrt{225}\)

AC = 15cm

Height of room is 8m, diagonal of floor is 15m

∴ The cosine of the angle which a diagonal of the room makes with the floor is EC² = 15²

cosine = \(\frac{\text{adj}}{\text{hyp}}\) = \(\frac{15}{17}\)

From the diagram, WYZ = 60^o, XYW = 180^o - 60^o

= 120^o

LX = 30^o

XWY = 180^o - 120^o + 30^o

XWY = 30^o

WXY = XYW = 30^o

Side XY = YW

YW = 8m, sin 60^o = \(\frac{3}{2}\)

∴ sin 60^o = \(\frac{h}{YW}\), sin 60^o = \(\frac{h}{8}\)

h = 8 x \(\frac{3}{2}\)

= 4\(\sqrt{3}\)