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Given O is the centre of the circle - NSC Mathematics - Question 11 - 2016 - Paper 2

Question 11

Given O is the centre of the circle. BA ⊥ AC. D is the midpoint of BC. 11.1 Prove that △ABC || △DOC. 11.2 Show that OC = $ \frac{DC \cdot BC}{AC} $ 11.3 Given: ... show full transcript

Worked Solution & Example Answer:Given O is the centre of the circle - NSC Mathematics - Question 11 - 2016 - Paper 2

Step 1

Prove that △ABC || △DOC

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Answer

To prove that two triangles are similar, we need to show that their corresponding angles are equal. Since O is the center of the circle, both OA and OC are radii, making them equal. Given that BA ⊥ AC, we can conclude that angle OAB is 90 degrees. Since D is the midpoint of BC, angle ABD is equal to angle ACD. Thus, by the AA (Angle-Angle) criterion, we can conclude that △ABC || △DOC.

Step 2

Show that OC = DC \cdot \frac{BC}{AC}

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Answer

From the properties of similar triangles (specifically △ABC and △DOC), we have:

$\frac{OC}{DC} = \frac{BC}{AC}$ .

Rearranging gives us:

$OC = DC \cdot \frac{BC}{AC}$ .

Step 3

Calculate the length of OC, rounded off to one decimal unit

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Answer

Given: DC = 15 cm and AB = 18 cm. Since AB = AC (as AB is a radius and thus equal), we can assume BC is also a radius which is connected through point O.

Calculating BC:

BC = 2 * DC = 2 * 15 cm = 30 cm.

Now substituting back into the earlier derived formula:

$OC = 15 \cdot \frac{30}{18} = 15 \cdot \frac{5}{3} = 25$ .

Thus, OC is approximately 25.0 cm.

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