Question

Plz answer this question.....​

Answer 1

Question :)

• In below figure, AE is perpendicular to seg. BC , seg.DF perpendicular to line BC , AE = 6cm , DF = 8 cm , then ratio of area of ΔABC to ΔDBC

Answer :)

$ar(\triangle ABC) =$ $\frac{1}{2} \times BC \times AE$

$ar(\triangle DBC) = \frac{1}{2} \times DF \times BC$

$= \frac{ar(\triangle ABC)}{ar(\triangle DBC)} = \frac{\frac{1}{2} \times BC \times AE}{\frac{1}{2} \times DF \times BC} \\\\= \frac{ar(\triangle ABC)}{ar(\triangle DBC)} = \frac{AE}{DF}\\\\= \frac{ar(\triangle ABC)}{ar(\triangle DBC)} = \frac{6 \: cm}{8 \: cm}\\\\= \frac{ar(\triangle ABC)}{ar(\triangle DBC)} = \frac{2}{3} \\\\= ar(\triangle ABC) : ar(\triangle DBC) = 2 : 3$

Answer 2

Answer:

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Step-by-step explanation:

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