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SSC Selection Post 2022 PYQ — Question 57

SSC Selection Post - Higher Secondary Level 2022 - 15 March 2022 - Shift 2 • Quantitative Aptitude • Q57

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In ∆ABC, D and E are points on sides AB and BC, respectively, such that DE ǁ AC. If BD=46BD= 4{\sqrt{6}}  and  AB=123AB= 12{\sqrt{3}}, then the ratio of the area of ∆BDE to that of quadrilateral ACED is:

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Correct answer: C

Explanation

The correct answer is Option 3 i.e. 2 : 7.

Given:  DE∥AC,  BD=46,  AB=123\ DE \parallel AC, \; BD = 4\sqrt{6}, \; AB = 12\sqrt{3}

Since △BDE∼△BAC, ratio of sides: BDAB=46123=13⋅2=23 \\ \frac{BD}{AB} = \frac{4\sqrt{6}}{12\sqrt{3}} = \frac{1}{3} \cdot \sqrt{2} = \frac{\sqrt{2}}{3}

Area ratio of similar triangles:Area(△BDE)Area(△ABC)=(23)2=29\text{Area ratio of similar triangles:} \\ \frac{\text{Area}(\triangle BDE)}{\text{Area}(\triangle ABC)} = \left( \frac{\sqrt{2}}{3} \right)^2 = \frac{2}{9}

So,Area(△BDE)Area(ACED)=29−2=27\text{So,} \quad \frac{\text{Area}(\triangle BDE)}{\text{Area}(ACED)} = \frac{2}{9 - 2} = \frac{2}{7}

Required ratio is 2 : 7.