(a) Using ruler and a pair of compasses only, construct : (i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ ...

Assessment: WAEC SSCE - General Mathematics - 2008 (Essay) Subject: General Mathematics

Question 1 Report

(a) Using ruler and a pair of compasses only, construct :

(i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ = /QR/ ;

(ii) locus, \(l_{1}\) of points equidistant from P and Q ;

(iii) locus, \(l_{2}\) of points equidistant from P and S.

(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect. (ii) With center T and radius /TP/, construct a circle \(l_{3}\).

Answer Details

(a) Construction

  1. Draw the line segment \(PQ=7\text{ cm}\).
  2. At \(P\), construct an angle of \(60^\circ\). On this ray, mark \(S\) such that \(PS=6.5\text{ cm}\).
  3. Join \(Q\) to \(S\). With centre \(Q\) and radius \(QS\), draw an arc. At \(Q\), construct \(\angle PQR=135^\circ\); let this ray meet the arc at \(R\). Join \(R\) to \(S\). Thus \(QR=QS\).
  4. Construct the perpendicular bisector of \(PQ\) and label it \(l_1\). This is the locus of points equidistant from \(P\) and \(Q\).
  5. Construct the perpendicular bisector of \(PS\) and label it \(l_2\). This is the locus of points equidistant from \(P\) and \(S\).
figure
Completed construction of \(PQRS\), the loci \(l_1\) and \(l_2\), their intersection \(T\), and the circle \(l_3\).

(b)

  1. The point of intersection of \(l_1\) and \(l_2\) is \(T\).
  2. With centre \(T\) and radius \(TP\), draw the circle \(l_3\), as shown above. The circle passes through \(P\), \(Q\) and \(S\), since \(TP=TQ=TS\).

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