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Thread: Ratio of areas of similar figures

  1. #1
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    Ratio of areas of similar figures

    Hi all,
    Here goes the question:
    Ratio of areas of similar figures-figure1.jpg
    In the diagram, AC,RQ and BP are parallel lines. AR=9cm,RQ=6cm,BP=8cm, and RB=3cm. Calculate the ratio of:
    (i)area of triangle ARQ:area of trapezium RBPQ'
    (ii)area of triangle BQR:area of triangle ARQ
    (iii)area of triangle ABC:area of triangle BQR

    Another question:
    Ratio of areas of similar figures-figure2.jpg
    In the diagram, AB=5cm,BE=4cm,EC=2cm,CD=3cm and DA=1cm. If DE=xcm, using a geometrical argument, deduce the value of x.

    Thanks in advance
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  2. #2
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    In Q2 are the two triangles similar?
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  3. #3
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    nope it isn't stated in the question but anyway it doesn't look similar
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  4. #4
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    please can someone help me because i need it for my assisgnment. thanks again.
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  5. #5
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    We're not supposed to do work that contributes to a student's final grade, so you'll need to show us what you've tried...
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  6. #6
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    you got me wrong, by saying assignment i am referring to normal homework. I mean i have no idea how to start it off so can anyone provide me with hints?
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  7. #7
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    Hello, AeroScizor!

    Click image for larger version. 

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    $\displaystyle \text{In the diagram: }\:AC \parallel RQ \parallel BP.$

    $\displaystyle AR=9,\; RQ=6,\;BP=8,\;RB=3.$

    $\displaystyle \text{Calculate the ratios of the areas of:}$

    $\displaystyle (i)\;\Delta ARQ : \text{quad }RBPQ$

    We have: .$\displaystyle \Delta ARQ \sim \Delta ABP$

    Hence: .$\displaystyle AR:AB \:=\:9:12 \;=\;3:4$

    Then: .$\displaystyle \text{area }\Delta ARQ:\text{area }\Delta ABP \;=\;9:16$

    Therefore: .$\displaystyle \Delta ARQ:\text{quad }RBPQ \;=\;9:7$




    $\displaystyle (ii)\;\Delta BQR : \Delta ARQ$

    $\displaystyle \Delta BQR$ has a base of 3.
    $\displaystyle \Delta ARQ$ has a base of 9.
    . . Their bases are in the ratio $\displaystyle 3:9 \:=\:1:3$

    They have the same height.

    Therefore, their areas are in the ratio $\displaystyle 1:3.$




    $\displaystyle (iii)\; \Delta ABC : \Delta BQR$

    We have: .$\displaystyle \Delta ABC \sim \Delta BQR$

    Their bases are in the ratio $\displaystyle 12:3 \:=\:4:1$
    Their heights are also in the ratio $\displaystyle 4:1$

    Therefore, their areas are in the rato $\displaystyle 16:1$

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  8. #8
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    thanks a million. i really wanted to know the thought process
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