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SSC CGL Model Paper 19

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Question : 62 of 100

Marks: +1, -0

Points D, E, F divide the sides of triangle ABC in the ratio 1:3, 1:4 and 1:1 as shown in the figure. What fraction of the area of triangle ABC is the area of triangle DEF.

$\frac{13}{40}$
$\frac{17}{40}$
$\frac{27}{40}$
$\frac{23}{40}$

Solution:

D divides AB into 1:3

E divides BC into 1:4

F divides AC into 1:1

AB = 4c; BD = 3c, DA = c

AC = 2b; AF = b, FC = b

BC = 5a, BE = a, EC = 4a

\frac{\text{ area of }\Delta ADF}{\text{ area of }\Delta ABC}

=\frac{\frac{1}{2}\times c\times b\times \sin A}{\frac{1}{2}\times 4c\times 2b\times \sin A}

=\frac{1}{8}=\frac{5}{40}

\frac{\text{ area of }\Delta FEC}{\text{ area of }\Delta ABC}

=\frac{\frac{1}{2}\times b\times 4a\times \sin C}{\frac{1}{2}\times 2b\times 5a\times \sin C}

=\frac{4}{10}=\frac{16}{40}

\frac{\text{ area of }\Delta BED}{\text{ area of }\Delta ABC}

=\frac{\frac{1}{2}\times 3c\times a\times \sin B}{\frac{1}{2}\times 5a\times 4c\times \sin B}

=\frac{3}{20}=\frac{6}{40}

\frac{\text{ area of }\Delta DEF}{\text{ area of }\Delta ABC}

=1-\left[\frac{5+16+6}{40}\right]=\frac{13}{40}

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