In the diagram, TR is a diameter of the circle - NSC Mathematics - Question 10 - 2023 - Paper 2
Question 10
In the diagram, TR is a diameter of the circle. PRKT is a cyclic quadrilateral. Chords TP and KR are produced to intersect at S. Chord PK is drawn such that PK = TK.... show full transcript
Worked Solution & Example Answer:In the diagram, TR is a diameter of the circle - NSC Mathematics - Question 10 - 2023 - Paper 2
Step 1
10.1.1 SR is a diameter of a circle passing through points S, P and R
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Answer
To show that SR is a diameter, we utilize the properties of cyclic quadrilaterals. Since S, P, and R lie on the circle and the angle TPR is 90 degrees (as it subtends the diameter TR), by the converse of the angle in a semi-circle theorem, it follows that:
Thus, we conclude that SR is a diameter of the circle.
Step 2
10.1.2 \( \hat{S} = \hat{P}_2 \)
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Answer
In the cyclic quadrilateral PRKT, the angles R1 and R2 are opposite angles. Therefore, due to the property of cyclic quadrilaterals:
R_i = P_{Tk} \\ \hat{S} = P_1 + P_2\
2.
The angle ( \hat{S} ) is equal to ( \hat{P}_2 ) since opposite angles of a cyclic quadrilateral sum up to 180 degrees.
Step 3
10.1.3 \( \Delta SPK \parallel \Delta PRK \)
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Answer
To prove that ( \Delta SPK \parallel \Delta PRK ), we note:
Since ( \hat{S} = \hat{P}_2 ) and the two triangles share the angle K, we can use the corresponding angles' criterion for parallel lines.
2.
Thus, ( SPK \parallel PRK ).
Step 4
10.2 If it is further given that SR = RK, prove that ST = \sqrt{6}RK.
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Answer
Given that SR = RK, let's denote it as RK. By the Pythagorean theorem in the triangles formed:
Consider triangle SRK, we have:
ST2=SK2+RK2ST2=SR2+RK2ST2=2RK2+2RK2=6RK2
2.
Thus, we conclude:
ST=6RK.
