# IPMAT Question Paper 2019 | IPM Indore Quants

###### IPMAT Sample Paper | IPMAT Question Paper | Question 33

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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. $$frac{\sqrt{15} + \sqrt{35}}{2}\\$ m 2. 8 m 3. √13 m 4. 6 m ## Best CAT Online Coaching Try upto 40 hours for free Learn from the best! #### 2IIM : Best Online CAT Coaching. ### Video Explanation ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer We are given that, OA = OB = OC = 6m. AB = 3m. BC = 2m. Let’s say angle AOB = ∅ and angle BOC = ∝. Cos ∅ = $\frac{AO^{2} + BO^{2} - AB^{2}}{2.𝐴𝑂.𝐵𝑂}\\$ = $\frac{6^{2} + 6^{2} - 3^{2}}{2.6.6}\\$ = $\frac{63}{72}\\$ = $\frac{7}{8}\\$ Sin ∅ = $\sqrt{1 − cos^{2} ∅}\\$ = $\sqrt{1 −$$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 ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ###### Already have an Account? ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Mobile:$91) 99626 48484 / 94459 38484
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