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What is the special name for the chord RQ?

Mathematics

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Here's how to solve Question 195:

a) (i) Construct a triangle PQR such that ∣PQ∣=8 cm|PQ|= 8 \text{ cm}, ∠RPQ=90∘\angle RPQ = 90^\circ and ∠PQR=30∘\angle PQR = 30^\circ. Measure ∣RQ∣|RQ|.

  1. Construct the base PQ.

    Draw a line segment PQ‾\overline{PQ} of length 8 cm8 \text{ cm}.

  2. Construct ∠RPQ=90∘\angle RPQ = 90^\circ.

    At point P, use a compass and ruler to construct a perpendicular line to PQ‾\overline{PQ}. This line will contain point R.

  3. Construct ∠PQR=30∘\angle PQR = 30^\circ.

    At point Q, use a compass and ruler to construct an angle of 30∘30^\circ with PQ‾\overline{PQ}. The ray forming this angle should intersect the perpendicular line from P. The intersection point is R.

  4. Measure $|RQ|$.

    Using a ruler, measure the length of the segment RQ‾\overline{RQ}.
    Theoretical calculation for verification:
    In △PQR\triangle PQR, ∠P=90∘\angle P = 90^\circ and ∠Q=30∘\angle Q = 30^\circ.

    cos⁡(∠PQR)=∣PQ∣∣RQ∣\cos(\angle PQR) = \frac{|PQ|}{|RQ|} cos⁡(30∘)=8∣RQ∣\cos(30^\circ) = \frac{8}{|RQ|} ∣RQ∣=8cos⁡(30∘)=832=163=1633 cm|RQ| = \frac{8}{\cos(30^\circ)} = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} \text{ cm} ∣RQ∣≈9.24 cm|RQ| \approx \boxed{9.24 \text{ cm}}

    Your measured value should be close to this.

    a) (ii) Construct the perpendicular bisector (mediator) of RQ. Let it meet RQ at O.

  5. Construct the perpendicular bisector of RQ.

    Place the compass needle at R and open it to more than half the length of RQ‾\overline{RQ}. Draw arcs above and below RQ‾\overline{RQ}. Without changing the compass width, place the needle at Q and draw arcs above and below RQ‾\overline{RQ}, intersecting the previous arcs. Draw a straight line through these two intersection points. This line is the perpendicular bisector of RQ‾\overline{RQ}.

  6. Locate point O.

    Mark the point where the perpendicular bisector intersects RQ‾\overline{RQ} as O.

    b) With O as centre and radius OP, draw a circle. Measure ∣OP∣|OP|.

  7. Draw the circle and measure $|OP|$.

    Place the compass needle at O and open it to the length of OP‾\overline{OP}. Draw a circle. This circle should pass through points P, Q, and R.
    Theoretical calculation for verification:
    Since O is the midpoint of the hypotenuse RQ‾\overline{RQ} in the right-angled triangle PQR, O is the circumcenter, and thus ∣OP∣=∣OQ∣=∣OR∣|OP| = |OQ| = |OR|.

    ∣OP∣=12∣RQ∣=12×1633=833 cm|OP| = \frac{1}{2} |RQ| = \frac{1}{2} \times \frac{16\sqrt{3}}{3} = \frac{8\sqrt{3}}{3} \text{ cm} ∣OP∣≈4.62 cm|OP| \approx \boxed{4.62 \text{ cm}}

    Your measured value should be close to this.

    c) What is the special name for the chord RQ?

  8. Identify the special name for chord RQ.

    Since O is the center of the circle and O is the midpoint of RQ‾\overline{RQ}, the chord RQ‾\overline{RQ} passes through the center of the circle.
    The special name for the chord RQ‾\overline{RQ} is the diameter.

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