JAMB Mathematics Past Questions & Answers - Page 216

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$\sqrt{3}$