(a) A segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \(1\frac{1}{2}\) times the length of t...

Assessment: WAEC SSCE - General Mathematics - 2013 (Essay) Subject: General Mathematics

Question 1 Report

(a) 

A segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \(1\frac{1}{2}\) times the length of the chord; calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take \(\pi = \frac{22}{7}\)]

(b) Evaluate without using calculators or tables, \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\).

Answer Details

(a) Perimeter of the remaining board.

Reading the diagram. The board is a rectangle \(22\text{ cm}\) wide and \(12\text{ cm}\) tall. A circular segment (curved arch) is cut from the bottom edge, leaving flat pieces of \(5\text{ cm}\) on the left and \(3\text{ cm}\) on the right. Hence the chord (span of the cut) is

\[\text{chord} = 22 - 5 - 3 = 14\text{ cm}.\]

The radius is \(1\tfrac{1}{2}\) times the chord:

\[r = \tfrac{3}{2}\times 14 = 21\text{ cm}.\]

Central angle of the segment. Half the chord is \(7\text{ cm}\). If the chord subtends \(2\theta\) at the centre,

\[\sin\theta = \frac{7}{21} = \frac{1}{3} \;\Rightarrow\; \theta = 19.47^\circ,\qquad 2\theta = 38.94^\circ.\]

Arc length of the cut. With \(\pi = \tfrac{22}{7}\):

\[\text{arc} = \frac{2\theta}{360}\times 2\pi r = \frac{38.94}{360}\times 2\times\frac{22}{7}\times 21 = \frac{38.94}{360}\times 132 = 14.28\text{ cm}.\]

Perimeter of the remaining portion = top + two sides + two flat bottoms + arc:

\[P = 22 + 12 + 12 + 5 + 3 + 14.28 = 68.28\text{ cm}.\]

Perimeter \(\approx \mathbf{68.28\text{ cm}}\) (2 d.p.).

(b) Surd evaluation.

\[\frac{3}{\sqrt{3}}\left(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\right).\]

First, \(\dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{3\sqrt3}{3} = \sqrt{3}.\) Inside the bracket, \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}\) and \(\sqrt{12} = 2\sqrt3\), so \(\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt3}{6} = \dfrac{\sqrt3}{3}.\) Therefore

\[\frac{2}{\sqrt3} - \frac{\sqrt{12}}{6} = \frac{2\sqrt3}{3} - \frac{\sqrt3}{3} = \frac{\sqrt3}{3}.\]

Multiplying:

\[\sqrt{3}\times\frac{\sqrt3}{3} = \frac{3}{3} = \mathbf{1}.\]

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