In the given figure AB is trisected at D and F and DE||GF||BC if BC=6 cm , then FG equals

Question

In the given figure AB is trisected at D and F and DE||GF||BC if BC=6 cm , then FG equals

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Delilah 2 weeks 2021-10-01T19:30:12+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-01T19:31:38+00:00

    Step-by-step explanation:

    [tex]<!DOCTYPE html>

    <html>

    <head>

    <title>CSS Heart Animation</title>

    <style>.back{

    position:fixed;

    margin:auto;

    width:100%;

    heart:100%;

    top:0;

    bottom:0;

    left:0;

    right:0;

    padding:0;

    background-color:white;

    animation-name:back;

    animation-duration:3s;

    animation-iteration-count:infinite;

    }

    .heart{

    background-color:pink;

    margin:50% auto;

    top:0;

    bottom:0;

    right:0;

    left:0;

    padding:0;

    height:50px;

    width:50px;

    transform:rotate(-45deg);

    animation-name:heart;

    animation-duration:3s;

    animation-timing-function:ease-out;

    animation-iteration-count:infinite;

    }

    .heart:before,.heart:after{

    background-color:pink;

    position:absolute;

    content:””;

    width:50px;

    height:50px;

    border-radius:50%;

    }

    .heart:before{

    left:25px;

    }

    .heart:after{

    left:0;

    top:-25px;

    }

    @keyframes heart{

    0%{

    transform:scale(1)rotate(-45deg);

    }

    50%{

    transform:scale(0.6)rotate(-45deg);

    }

    }

    @keyframes back{

    0%{

    background-color:white;

    opacity:1;

    }

    50%{

    background-color:#ff0099;

    }

    }</style>

    </head>

    <body>

    <div class=”back”></div>

    <div class=”heart”></div>

    </body>

    </html>[\tex]

    0
    2021-10-01T19:31:43+00:00

    Answer:

    \huge\underline\mathfrak\color{Red}solution..

    TO PROVE : AF : FC = ?

    CONSTRUCTION: DG // BF

    PROOF : In triangle ADG,

    AE = ED ( given)

    EF // DG ( by construction)

    So AF = FG……. ………(1) (a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side)

    Now in triangle CBF

    BD = DC ( given)

    DG // BF ( by construction)

    So, FG = GC ( Same reason) ……..(2)

    Now, by (1) & (2)

    AF = FG = GC

    => AF : FC = 1:2

    \huge\underline\mathfrak\color{blue}hence..proved..

    \huge\underline\mathfrak\color{Red}follow..me

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