9.1 In the diagram, JKLM is a cyclic quadrilateral and the circle has centre O. Prove the theorem which states that $ J + L = 180^{ ext{o}}$. 9.2 In the diagram, a smaller circle ABTS and a bigger circle BDRT are given. BT is a common chord. Straight lines STD and ATR are drawn. Chords AS and DR are produced to meet in C, a point outside the two circles. BS and BD are drawn. $ \hat{A} = x $ and $ \hat{R}_1 = y $. 9.2.1 Name, giving a reason, another angle equal to: (a) x (b) y 9.2.2 Prove that SCDB is a cyclic quadrilateral. 9.2.3 It is further given that $ \hat{D}_1 = 30^{ ext{o}} $ and $ \hat{A}ST = 100^{ ext{o}} $. Prove that SD is not a diameter of circle BDS.

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9.1 In the diagram, JKLM is a cyclic quadrilateral and the circle has centre O - NSC Mathematics - Question 9 - 2018 - Paper 2

Question 9

9.1 In the diagram, JKLM is a cyclic quadrilateral and the circle has centre O. Prove the theorem which states that $ J + L = 180^{ ext{o}}$. 9.2 In the diagram, a ... show full transcript

Worked Solution & Example Answer:9.1 In the diagram, JKLM is a cyclic quadrilateral and the circle has centre O - NSC Mathematics - Question 9 - 2018 - Paper 2

Step 1

Prove the theorem which states that $ J + L = 180^{ ext{o}}$

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Answer

To prove that $J + L = 180^{ ext{o}}$ , we can use the properties of cyclic quadrilaterals. According to the theorem of cyclic quadrilaterals, opposite angles are supplementary. Thus, we can establish:

Since JKLM is cyclic, we know that the angles subtended by points on the circumference at the centre are equal to twice the angle subtended at the circle.
Let ( \hat{O_1} = 2 \hat{J} ) and ( \hat{O_2} = 2 \hat{L} ).
It follows that ( \hat{J} + \hat{L} = \hat{O_1} + \hat{O_2} - 360^{ ext{o}} ).
Therefore, we have ( \hat{J} + \hat{L} = 180^{ ext{o}} ).

Thus, the theorem is proven.

Step 2

Name, giving a reason, another angle equal to: (a) x

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Answer

In triangle ABST, angle ( \hat{A}BT ) is in the same segment as angle ( \hat{A}SR ). Thus, ( \hat{B_1} = x ) because angles in the same segment subtend the same chord.

Step 3

Name, giving a reason, another angle equal to: (b) y

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Answer

In triangle BDRT, angle ( \hat{D}BR ) subtends the same arc as ( \hat{R_1} ) which implies that ( \hat{B_2} = y ) as angles subtended by the same chord are equal.

Step 4

Prove that SCDB is a cyclic quadrilateral.

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Answer

To prove that SCDB is a cyclic quadrilateral, we can use the converse of the cyclic quadrilateral theorem. The following steps outline the proof:

Angle ( \hat{C} = 180^{ ext{o}} - (x+y) ) by cyclic properties.
The angles ( \hat{S}DB + \hat{C} + \hat{B} = 180^{ ext{o}} ) confirms that opposite angles are supplementary.
Therefore, SCDB satisfies the property of cyclic quadrilaterals.

Step 5

Prove that SD is not a diameter of circle BDS.

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Answer

To prove that SD is not a diameter, we will analyze the angles provided:

It is given that ( \hat{D}_1 = 30^{ ext{o}} ) and ( \hat{AST} = 100^{ ext{o}} ).
Thus, the external angle ( \hat{D}ST ) can be represented as ( , \hat{R} + \hat{D} + \hat{AST} ). This gives:

$\hat{S}D + \hat{B} = 110^{ ext{o}}$
For a line to be a diameter, it must subtend a right angle to any point on the circumference. Since( \hat{S}D = 110^{ ext{o}} ) does not fulfill this requirement, SD cannot be the diameter of circle BDS.

9.1 In the diagram, JKLM is a cyclic quadrilateral and the circle has centre O - NSC Mathematics - Question 9 - 2018 - Paper 2

Worked Solution & Example Answer:9.1 In the diagram, JKLM is a cyclic quadrilateral and the circle has centre O - NSC Mathematics - Question 9 - 2018 - Paper 2

Prove the theorem which states that $ J + L = 180^{ ext{o}}$

Name, giving a reason, another angle equal to: (a) x

Name, giving a reason, another angle equal to: (b) y

Prove that SCDB is a cyclic quadrilateral.

Prove that SD is not a diameter of circle BDS.

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