It’s quite challenging to only use AM to find angle AOM because we need two known lengths to obtain a trigonometric ratio.

i cant seem to find the second length. i only know that the base is 4

Only way is OT plus OM equals 8 but then we must express OM in pyth.

Let OT = r for simplification

Can share the answer ? It is 5.66?

Hmmm...I have no paper and calculator with me at the moment though

Draw a line from B to T

∠TMB = 90° since TM ∟ AB. So ∆TMB is a right angled triangle.

tan ∠MTB = MB/MT = 4/8 = ½
so ∠MTB = tan ‾¹ (½) = 26.57° (2d.p)


Now, OB = OT (both are radius of circle)
So ∆ TOB is isosceles.

∠MTB = ∠OTB = 26.57° and thus
∠OTB = ∠ OBT = 26.57° (base ∠s of isosceles ∆)

∠MOB = ∠OTB + ∠OBT = 26.57° + 26.57°
= 53.14°
(exterior angle = sum of two interior opposite angles)

∆MOB is also right angled.

So sin ∠MOB = MB/OB = 4/OB

OB = 4/sin∠MOB = 4/sin53.14°
≈ 4.999m = 5m

OB is a radius of the circle so radius of the cross sectional area is 5m.

(Note that we got 4.999m because the value of ∠MOB was truncated. If we had used OB = 4/sin(2tan‾¹(½) ) and keyed.in directly into the calculator, you will get exactly 5.)

Alternatively,


OT + OM = MT
OT + OM = 8
Thus OM = 8 - OT

As TM ∟ AB, TMB = 90° and thus OMB = 90°. So ∆OMB is right angled.

Then OM² + MB² = OB² (Pythagoras' Theorem)

Since OB = OT as both are radius of circle,
OM² + MB² = OT²


Since MB = 4, OM² + 4² = OT²

Sub OM = 8 - OT,

(8 - OT)² + 16 = OT²

64 - 16OT + OT² + 16 = OT²

80 = 16OT
OT = 5

This method utilises Pythagoras Theorem, substitution and the algebraic multiplication formula (a - b)² = a² - 2ab + b² . No need to find angles.

I suspect that the author intends the candidate to use the Pythagoras theorem approach to solve this question.

Yes, and also the property that the perpendicular bisector of the chord passes through the centre of the circle.

Eric and J .. great answer and explanation!

thank you so much!