ALGEBRA Find x and y so that the quadrilateral is a parallelogram. a x = 20, y = 12 b x = 20, y = 160 c x = 129, y = 51 d x = 168, y = 12 i need this right now !!

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Ответ:

savannabroyhill24

19.02.2020

Mathematics

3:5

Step-by-step explanation:

$We\ are\ given\ that:\\AB\ is\ a\ Line\ Segment\ that\ has\ Points\ A\ and\ B,\ that\ are\ it's\ two\ End-Points.\\Now,\\Point\ C\ is\ marked\ on\ AB\ that\ divides\ AB\ into\ AC\ and\ CB,\ whose\\ lengths\ are\ in\ a\ Ratio\ 1:2.\\\\Point\ E\ is\ marked\ on\ AB\ that\ divides\ AB\ into\ AE\ and\ EB,\ whose\\ lengths\ are\ in\ a\ Ratio\ 1:3.\\\\Point\ D\ is\ marked\ on\ AB\ that\ divides\ AB\ into\ AD\ and\ DB,\ whose\\ lengths\ are\ in\ a\ Ratio\ 1:4.\\\\$

$Now,\\For\ Point\ C,\\Total\ no.\ of\ parts=1+2=3\\Length\ of\ AC=\frac{1}{3}x\\Length\ of\ CB=\frac{2}{3}x\\\\For\ Point\ E,\\Total\ no.\ of\ parts=1+3=4\\Length\ of\ AE=\frac{1}{4}x\\Length\ of\ EB=\frac{3}{4}x\\\\For\ Point\ D,\\Total\ no.\ of\ parts=1+4=5\\Length\ of\ AD=\frac{1}{5}x\\Length\ of\ DB=\frac{4}{5}x\\$

$Now,\\Lets\ move\ onto\ some\ REAL\ calculations!!\\We\ firstly\ observe\ that,\\AD+DC=AC\\Hence,\\\frac{1}{5}x+DC=\frac{1}{3}x\\DC=\frac{1}{3}x-\frac{1}{5}x\\DC=\frac{5-3}{15}=\frac{2}{15}$

$We\ secondly\ observe\ that,\\AD+DE=AE\\Hence,\\\frac{1}{5}x+DE=\frac{1}{4}x\\DE=\frac{1}{4}x-\frac{1}{5}x\\DE=\frac{5-4}{20}=\frac{1}{20}$

$Now,\\We\ know\ that,\\DE+EC=DC\\Hence,\\\frac{1}{20}x+EC=\frac{2}{15}x \\EC=\frac{2}{15}x-\frac{1}{20}x\\EC=\frac{40-15}{300}x\\EC=\frac{25}{300}x=\frac{1}{12}x\\\\By\ comparing\ DE\ and\ EC,\\Length\ of\ DE=\frac{1}{20}x\\Length\ of\ EC=\frac{1}{12}x\\The\ ratio\ of\ Lengths\ of\ DE\ to\ Length\ of\ EC=\frac{1}{20}x:\frac{1}{12}x\\We\ know\ that,\\Ratio's\ are\ merely\ fractions\ represented\ by\ ':'.\\Hence,\\\frac{1}{20}x:\frac{1}{12}x\\=\frac{\frac{1}{20}x}{\frac{1}{12}x}\\$

$=\frac{1}{20}*12\\=\frac{12}{20}=\frac{3}{5}=3:5$

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