KVPY 2019 SA Exam Previous Papers

Question : 5 of 80

Marks: +1, -0

In a triangle ABC, ∠BAC = 90° ; AD is the altitude from A on to BC. Draw DE perpendicular to AC and DF perpendicular to AB. Suppose AB = 15 and BC = 25. Then the length of EF is

12
10
$5 √3$
$5√ 5$

Validate

Solution:

AD = {2(\Area\of \ABC)} /{BC}

={20×15}/25=12

Note that AFDE is a rectangle
Hence AD = EF.

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