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Math Help - Similarity..help!

  1. #1
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    Similarity..help!

    In △ABC, <B=90, AB=12cm, and AC=15cm. D and E are points on AB and AC respectively such that <AED=90 and DE=3cm. Find:
    i) Area of △ADE
    ii) Area of quadrilateral BCED : Area of △ABC
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  2. #2
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    well....i managed to work out i) on my own-

    △AED~△ABC->[by A.A. similarity rule]
    So, ar△AED:ar△ABC= (DE/BC)^2
    => ar△AED: 1/2BC.AB=(3/9)^2
    => ar△AED= 54*1/9
    = 6 sq.cm

    So somebody now plz solve the ii) part for me...
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  3. #3
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    Quote Originally Posted by snigdha View Post
    In △ABC, <B=90, AB=12cm, and AC=15cm. D and E are points on AB and AC respectively such that <AED=90 and DE=3cm. Find:
    i) Area of △ADE
    ii) Area of quadrilateral BCED : Area of △ABC
    AB=3AD

    BC=3DE

    Therefore the area of triangle ABC is 3(3)=9 times the area of triangle ADE.

    Therefore the quadrilateral DBCE, being triangle ABC with triangle ADE cut off has area 9(6)-6=8(6)

    The ratio of the areas is 8:9
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