1. Problem (a)(i): Construct the perpendicular bisector of chord [AB] using only a compass and straight edge. - To construct the perpendicular bisector of [AB], place the compass at point A and draw arcs above and below the chord. - Without changing the compass width, repeat from point B to intersect the previous arcs. - Draw a straight line through the two intersection points of the arcs. This line is the perpendicular bisector of [AB]. 2. Problem (a)(ii): Find O, the center of circle k. - Repeat the perpendicular bisector construction for chord [AC]. - The intersection of the two perpendicular bisectors is point O, the center of the circle. - Measure |∠AOB| using a protractor or geometric tools. - Suppose the measured angle is $x$ degrees; write the answer as $|\angle AOB| = x^\circ$. 3. Problem (a)(iii): Find |∠ACB| and justify. - By the Inscribed Angle Theorem, the angle subtended by chord AB at the center (|∠AOB|) is twice the angle subtended at the circumference (|∠ACB|). - Therefore, $$|\angle ACB| = \frac{1}{2} |\angle AOB| = \frac{x}{2}^\circ.$$ - Justification: The inscribed angle is half the central angle subtending the same chord. 4. Problem (b)(i): Given |PQ| = 15 cm and |OQ| = 17 cm, find |OP|. - Since PQ is tangent at P, triangle OPQ is right angled at P. - By Pythagoras theorem: $$|OQ|^2 = |OP|^2 + |PQ|^2.$$ - Substitute values: $$17^2 = |OP|^2 + 15^2$$ - Simplify: $$289 = |OP|^2 + 225$$ - Subtract 225: $$|OP|^2 = 289 - 225 = 64$$ - Take square root: $$|OP| = \sqrt{64} = 8 \text{ cm}.$$ 5. Problem (b)(ii): Find |QT|. - Since OT is radius and OP is radius, |OP| = |OT| = 8 cm. - In right triangle OQT, by Pythagoras theorem: $$|OQ|^2 = |OT|^2 + |QT|^2.$$ - Substitute values: $$17^2 = 8^2 + |QT|^2$$ - Simplify: $$289 = 64 + |QT|^2$$ - Subtract 64: $$|QT|^2 = 225$$ - Take square root: $$|QT| = \sqrt{225} = 15 \text{ cm}.$$