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SSC Selection Post 2019 PYQ — Question 69

SSC Selection Post - Graduation Level 2019 - 16 Oct 2019 • Quantitative Aptitude • Q69

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In trapezium ABCD, AB || DC. The sides AD and BC are produced to meet at E. If AB = 15cm CD = 10cm AD = 4cm and BC = 5cm then DE = ?

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Correct answer: D

Explanation

The correct answer is Option 4 i.e. 8 cm.

Using similar triangles:  

AEDE=ABCD=1510⇒AE−DEDE=15−1010=510⇒ADDE=12⇒4DE=12⇒DE=4×2=8\frac{AE}{DE} = \frac{AB}{CD} = \frac{15}{10} \Rightarrow \frac{AE - DE}{DE} = \frac{15 - 10}{10} = \frac{5}{10} \Rightarrow \frac{AD}{DE} = \frac{1}{2} \Rightarrow \frac{4}{DE} = \frac{1}{2} \Rightarrow DE = 4 \times 2 = 8