BDMO National Primary 2018/4

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nahin munkar
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BDMO National Primary 2018/4

Unread post by nahin munkar » Tue Jan 08, 2019 1:27 pm

In square $ABCE$, $AF=3FE$ & $CD=3ED$. What is the ratio of the area of triangle $\triangle BFD$ and square $ABCE$ ?
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samiul_samin
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Re: BDMO National Primary 2018/4

Unread post by samiul_samin » Thu Jan 10, 2019 9:53 am

Answer
$7:32$

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SINAN EXPERT
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Re: BDMO National Primary 2018/4

Unread post by SINAN EXPERT » Sun Jan 20, 2019 8:01 pm

At first let's make the picture easy.
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Here, $AB=BC=CE=EA=a$ and $FE=b,$ $ED=c$
So, $AE=AF+FE=3FE+FE=4FE=4b=a$
Now, $a=4b$ and similarly $a=4c$
Hence, $4b=4c$ $⇒b=c$

The areas of yellow coloured triangles
$=\dfrac {1}{2}\times BC\times CD+\dfrac {1}{2}\times EF\times ED+\dfrac {1}{2}\times AF\times AB$
$=\dfrac {1}{2}\times a\times 3c+\dfrac {1}{2}\times c\times b+\dfrac {1}{2}\times 3b\times a$
$=\dfrac {1}{2}\times 4b\times 3b+\dfrac {1}{2}\times b\times b+\dfrac {1}{2}\times 3b\times 4b$
$=6b^2+\dfrac {b^2}{2}+6b^2$
$=\dfrac {25b^2}{2}$

The area of $□$$ABCE$ $=a*a=4b*4b=16b^2$
$∴$ The area of $△BFD$ $=16b^2-\dfrac {25b^2}{2}=\dfrac {7b^2}{2}$
Therefore, the ratio $△BFD:$$□$ $ABCE=\dfrac {7b^2}{2}:\dfrac {16b^2}{2}=7:16$
$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$

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SINAN EXPERT
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Re: BDMO National Primary 2018/4

Unread post by SINAN EXPERT » Sun Jan 20, 2019 8:04 pm

samiul_samin wrote:
Thu Jan 10, 2019 9:53 am
Answer
$7:32$
Isn't the answer= $7:16$?
$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$

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samiul_samin
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Re: BDMO National Primary 2018/4

Unread post by samiul_samin » Sun Jan 20, 2019 11:32 pm

SINAN EXPERT wrote:
Sun Jan 20, 2019 8:04 pm
samiul_samin wrote:
Thu Jan 10, 2019 9:53 am
Answer
$7:32$
Isn't the answer= $7:16$?

Yes the aunswer is $7:16$

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