IPMAT Question Paper 2019 | IPM Indore Quants

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Question 33: On a circular path of radius 6 m a boy starts from a point A on the circumference and walks along a chord AB of length 3 m. He then walks along another chord BC of length 2 m to reach point C. The point B lies on the minor arc AC. The distance between point C from point A is

1. \\frac{\sqrt{15} + \sqrt{35}}{2}\\) m
2. 8 m
3. √13 m
4. 6 m

Video Explanation

Explanatory Answer

We are given that,
OA = OB = OC = 6m.

AB = 3m.
BC = 2m.

Let’s say angle AOB = ∅ and angle BOC = ∝.
Cos ∅ = \\frac{AO^{2} + BO^{2} - AB^{2}}{2.𝐴𝑂.𝐵𝑂}\\) = \\frac{6^{2} + 6^{2} - 3^{2}}{2.6.6}\\) = \\frac{63}{72}\\) = \\frac{7}{8}\\)

Sin ∅ = \\sqrt{1 − cos^{2} ∅}\\) = \\sqrt{1 − (\frac{7}{8})^{2}}\\) = \\sqrt{\frac{64 - 49}{64}}\\) = \\sqrt{\frac{15}{64}}\\) = \\frac{\sqrt{15}}{8}\\)



Similarly Cos ∝ = \\frac{CO^{2} + BO^{2} - AC^{2}}{2.𝐴𝑂.C𝑂}\\)= \\frac{6^{2} + 6^{2} - 2^{2}}{2.6.6}\\)= \\frac{68}{72}\\) = \\frac{17}{18}\\)

Sin ∝ = \\sqrt{1 − cos^{2} ∝}\\) = \\sqrt{1 − (\frac{17}{18})^{2}}\\) = \\sqrt{\frac{35}{324}}\\) = \\frac{\sqrt{35}}{18}\\)

Cos (∅ + ∝) = Cos ∅ Cos ∝ - Sin ∅ Sin ∝
= \\frac{7}{8}\\) . \\frac{17}{18}\\) - \\frac{\sqrt{15}}{8}\\) . \\frac{\sqrt{35}}{18}\\) = \\frac{119 - 5\sqrt{21}}{8.18}\\)

Also,
Cos (∅ + ∝) = \\frac{AO^{2} + CO^{2} - AC^{2}}{2.𝐴𝑂.C𝑂}\\) =\\frac{6^{2} + 6^{2} - AC^{2}}{2.6.6}\\) = \\frac{72 - AC^{2}}{72}\\)

\\frac{72 - AC^{2}}{72}\\) = \\frac{119 - 5\sqrt{21}}{8*18}\\)

AC²= 72 - \\frac{72(119 - 5\sqrt{21})}{8*18}\\)
= \\frac{72*8*18 - 72*119 + 72*5\sqrt{21}}{8*18}\\)
= \\frac{25 + 5\sqrt{21}}{2}\\)

Going by the options if AC = \\frac{\sqrt{15} + \sqrt{35}}{2}\\)
AC²= \\frac{(\sqrt{15} + \sqrt{35})^{2}}{4}\\) = \\frac{15 + 35 + 2\sqrt{15*35}}{4}\\)

AC²= \\frac{50 + 10\sqrt{21}}{4}\\)
AC²= \\frac{25 + 5\sqrt{21}}{2}\\), which concurs with our calculated value of AC².

Therefore, The distance between C and A is \\frac{\sqrt{15} + \sqrt{35}}{2}\\) m.

The question is"On a circular path of radius 6 m a boy starts from a point A on the circumference and walks along a chord AB of length 3 m. He then walks along another chord BC of length 2 m to reach point C. The point B lies on the minor arc AC. The distance between point C from point A is"

Hence, the answer is \\frac{\sqrt{15} + \sqrt{35}}{2}\\) m

Choice A is the correct answer

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