Question

in a triangle ABC, D&E are two points AB and AC such that DE// BC and AD: BD = 3:5 . Find the ratio of area of ∆ADE to the area of trapezium BDEC I need step by step explainaton. ​

Answer 1

As, (AD)/(BD)=2/3<br> So,

<br> Since,

are Similar triangles(by using AA)<br>

So,

<br> Hence,

Answer 2

Answer:

$so \: here \: □ABCD \: is \: a \: trapezium \\ moreover \: DE \: || \: BC \\ so \: thus \: considering \: △ADE \: and \: \: △ABC \\ ∠DAE = ∠BAC \: \: \: \: \: \: \: \: \: \: (common \: angle) \\ since \: DE \: || \: BC \\ ∠ADE=∠ABC \: \: \: \: \: \: \: (corresponding \: angles \: theorem) \\ \\ so \: hence \: \: △ADE \: similar \: to \: △ABC \: \: \: \: \: \: \: (AA \: testof \: similarityy$

$so \: hence \: then \\ by \: theorem \: in \: ratios \: of \: areas \: of \: similar \: triangles \\ we \: get \\ \frac{A(△ADE)}{A(△ABC)} = \frac{AD {}^{2} }{AB {}^{2} } \\ \\ so \: here \: \\ \frac{AD}{BD} = \frac{3}{5} \\ so \: hence \: then \\ by \: invertendo \: we \: get \\ \frac{BD}{AD} = \frac{5}{3} \\ \\ so \: so \: applying \: componendo \: \\ we \: get \\ \frac{BD+AD}{AD} = \frac{5 + 3}{3} \\ \\ \frac{AB}{AD} = \frac{8}{3} \\ applying \: invertendo \: we \: get \\ \\ \frac{AD}{AB} = \frac{3}{8}$

$so \: hence \: then \\ \frac{A(△ADE) }{ A(△ABC} = \frac{AD {}^{2} }{AB {}^{2} } \\ \\ = (\frac{3}{8} ) {}^{2} \\ \\ = \frac{(3) {}^{2} }{(8) {}^{2} } \\ \\ = \frac{9}{64} \: \\ \\ so \: let \: the \: common \: multiple \: be \: k \\ ie \: \: A(△ADE) = 9k \\ A(△ABC) = 64k$

$now \: then \: by \: area \: addition \: property \\ we \: get \\ A(△ABC) = A(△ADE) +A (□ABCD) \\ 64k = A (□ABCD) + 9k \\ ie \: A (□ABCD) = 55k \\ \\ so \: hence \: then \\ \frac{A(△ADE) }{A (□ABCD)} = \frac{9k}{55k} \\ \\ = \frac{9}{55}$