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Standard XII

Mathematics

Section Formula

The points D,...

Question

The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3, then $(\to A D + \to B E + \to C F) : \to C K$ is equal to

1 : 1

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2 : 5

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5 : 2

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None of these

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Solution

The correct option is B 2 : 5
$L e t (\to A B = a, \to A C = b S o, \to A D = \frac{4 a + b}{5}, \to A E = \frac{2 b}{5}, \to A F = \frac{3 a}{10}, a n d \to A K = \frac{a}{4}$

$\frac{\to A D + \to B E + \to C F}{\to C K} = \frac{\frac{b + 4 a}{5} + \frac{2 b - 5 a}{5} + \frac{3 a - 10 b}{10}}{\frac{a - 4 b}{4}} = \frac{6 b - 2 a + 3 a - 10 b}{10 (a - 4 b)} \times 4 = \frac{2}{5} .$

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Similar questions

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Q. If D, E, F are the mid-points of sides BC, CA and AB respectively of ∆ABC, then the ratio of the areas of triangles DEF and ABC is

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The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3, then (→AD+→BE+→CF):→CK is equal to

The correct option is B 2 : 5Let(→AB=a,→AC=bSo,→AD=4a+b5,→AE=2b5,→AF=3a10,and→AK=a4 →AD+→BE+→CF→CK=b+4a5+2b−5a5+3a−10b10a−4b4=6b−2a+3a−10b10(a−4b)×4=25.

The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3, then $(\to A D + \to B E + \to C F) : \to C K$ is equal to