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In the diagram, TSR is a diameter of the larger circle having centre S - NSC Mathematics - Question 10 - 2021 - Paper 2
Question 10
In the diagram, TSR is a diameter of the larger circle having centre S. Chord TQ of the larger circle cuts the smaller circle at M. PQ is a common tangent to the two... show full transcript
Worked Solution & Example Answer:In the diagram, TSR is a diameter of the larger circle having centre S - NSC Mathematics - Question 10 - 2021 - Paper 2
Step 1
10.1.1 SQ is the diameter of the smaller circle.
Answer
To prove that SQ is the diameter of the smaller circle, we can use the properties of angles in a semicircle. Since TS is a diameter of the larger circle, then angle Q₁ + angle Q₂ = 90° (Inscribed angle theorem).
Given that angle M₂ = 90° (as co-interior angles when MS || QR), we can infer that TM is equal to MQ because these segments are equal by the property of intersecting chords.
Thus, since angle M₁ = angle MQ and angle M₂ = 90°, we conclude:
- SQ is a diameter because line segments TM and SQ bisect each other (line in semi-circle).
- Hence, SQ is the diameter of the smaller circle.
Step 2
10.1.2 RT = RQ² / RP.
Answer
In triangles ARTQ and AROP, applying the tangent-chord theorem gives us that tan of angle θ (where θ is the angle between the tangent QR and the radius R) equals the ratio of the opposite side RQ to the adjacent side RP.
Using Pythagorean identities, we find that:
- RT/RQ = RP/RQ,
- Thus RT = RQ² / RP by rearranging this equation.
Step 3
10.2 Calculate the length of the radius of the larger circle.
Answer
Given that MS = 14 units and PQ = √640 units, we can use the midpoint theorem which states that if QR is drawn, then QR² = 28² - (√640)².
- Setting RP = 12 units, use Pythagoras's theorem to find RP² = 28² - (√640)².
- Calculate that RP = 12 units.
- Finally, the radius of the larger circle is determined to be 98/3 units.