Question

In the figure shown above DE is parallel BC. If the area of triangle ADE is half that of trapezoid DECB what is the ratio of AE to AC?

Answer 1

Answer:

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refer the reference diagram uploaded above for better understanding

Step-by-step explanation:

$so \: here \: quadrilateral \: decb \: is \: a \: trapezoid \: ie \: is \: a \: trapezium \\ so \: then \: de \: parallel \: to \: bc$

$moreover \: here \\ area \: of \: triangle \: ade = 1/2×area \: of \: trapezoid \: decb \\ ie \: area \: of \: trapezoid \: decb = 2.area \: of \: traiangle \: ade$

$thus \: then \: by \: area \: addition \: property \: \\ we \: get \\ area \: of \: triangle \: abc = area \: of \: triangle \: ade + area \: of \: trapzoid \: decb$

$now \: so \: we \: get \ \\ area \: of \: triangle \: abc = area \: of \: triangle \: ade + 2.area \: of \:triangle \: ade \\ ie \: area \: of \: triangle \: abc = 3.area \: of \: triangle \: ade \\ \\ so \: henceforth \: w \: get \\ area \: of \: triangle \: ade \div area \: of \: triangle \: abc = 1 \div 3 \: \: \: \: \: \: (1)$

$now \: considering \: triangle \: ade \: and \: triangle \: abc \\ angle \: bac = angle \: bac \: \: : (common \: angle) \\ since \: de \: parallel \: to \: bc \\ angle \: ade = angle \: abc \: \: \: \: \: \: \: (corresponding \: angles) \\ \\ hence \: triangle \: ade \: similar \: to \: triangle \: abc \: \: \: \: \: (a.a \: test \: of \: similarity)$

$so \: as \: triangle \: ade \: similar \: to \: triangle \: abc \\ by \: theorem \: on \: ratios \: of \: areas \: of \: similar \: triangles \\ we \: get \\ area \: of \: triangle \: ade \div area \: of \: triangle \: abc = ae \: square \div ac \: square \\ \\ so \: from \: (1) \: we \: get \\ ae \: square \div ac \: square = 1 \div 3 \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ ae \div ac = 1 \div \sqrt{3}$