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SSC Selection Post 2019 PYQ — Question 59

SSC Selection Post - Higher Secondary Level 2019 - 15 Oct 2019 • Quantitative Aptitude • Q59

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It is given that ΔABC ~ ΔEDF and Area ΔABC : Area ΔDEF = 64 : 25. If AD = 16 cm, BC = 18 cm, CA = 20 cm. What is the value of EF (in cm)?

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Correct answer: B

Explanation

The correct answer is option 2 i.e. 12.5.

Given, ΔABC ~ ΔEDF

So, ar.ABCar.EDF=(ABED)2=(BCDF)2=(ACEF)2\frac{ar. ABC}{ar. EDF} = (\frac{AB}{ED})^2 = (\frac{BC}{DF})^2 = (\frac{AC}{EF})^2

⇒ 6425=(20EF)2\frac{64}{25} = (\frac{20}{EF})^2

⇒ 20/EF = 8/5

⇒ EF = 100/5

⇒ EF = 12.5