Photo AI
In die diagram is TR 'n middellyn van die sirkel - NSC Mathematics - Question 10 - 2023 - Paper 2
Question 10
In die diagram is TR 'n middellyn van die sirkel. PRKT is 'n koordevihoek. Koorde TP en KR is verleng en sny by S. Koord PK is getrek sodanig dat PK = TK. 10.1.1 Be... show full transcript
Worked Solution & Example Answer:In die diagram is TR 'n middellyn van die sirkel - NSC Mathematics - Question 10 - 2023 - Paper 2
Step 1
10.1.1 Bewys, met redes, dat:
Answer
To prove that SR is a diameter:
- Since TPR = 90° (angle in semi-circle), it follows that the angle at P is inscribed within the semi-circle.
- Furthermore, S and R are points on a straight line, giving SP + PR = SR.
- Hence, SR is a diameter according to the converses of the angles in a semi-circle.
Step 2
10.1.2 $ar{S} = ar{P}_2$
Answer
In cyclic quadrilateral TRPK:
- The angles at R and P must satisfy the properties of cyclic quadrilaterals, where the opposite angles are equal.
- Thus, ar{S} is equal to ar{P}_2. We conclude that the lengths are equal due to the congruence in the cyclic structure.
Step 3
10.1.3 $ riangle SPK \\parallel riangle PRK$
Answer
To demonstrate that triangles SPK and PRK are parallel:
- By proving that ar{S} is equal to ar{P}_1 and recognizing both triangles share a common angle K, we can apply the angle-angle rule.
- Therefore, we conclude that due to equal corresponding angles.
Step 4
10.2 Indien daar verder gegee word dat SR = RK, bewys dat ST = \sqrt{6RK}.
Answer
Given that SR = RK, and using the Pythagorean theorem:
- From triangles formed, let ST² = SK² + RK², where SK = 2RK and TK = PK.
- Thus, we derive: ST² = 2RK + 2RK = 6RK².
- Taking the square root of both sides gives us ST = \sqrt{6RK}.