BDMO 2016: National Junior/2

For students of class 6-8 (age 12 to 14)
Tasnova Novera
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Joined: Fri Feb 03, 2017 10:32 pm

BDMO 2016: National Junior/2

Unread post by Tasnova Novera » Mon Feb 06, 2017 6:56 pm

Please help me solving BDMO 2016: National Junior/ Question No. 2

The Question is:
ABC is a triangular piece of paper with an area of 500 square units and AB = 20 units. DE is parallel to AB. The triangle is folded along DE. The part of the triangle below AB has an area of 80 square units. What is the area of CDE?

Ref.
http://www.matholympiad.org.bd/question ... 6-national

Zahin Hasin Rudro
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Joined: Sun Jul 12, 2015 3:52 am

Re: BDMO 2016: National Junior/2

Unread post by Zahin Hasin Rudro » Thu Aug 03, 2017 11:13 pm

[CDE]=225 sq. units

abrarfiaz
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Re: BDMO 2016: National Junior/2

Unread post by abrarfiaz » Thu Oct 12, 2017 11:41 am

explain the process please

abrarfiaz
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Joined: Mon Mar 20, 2017 4:53 pm

Re: BDMO 2016: National Junior/2

Unread post by abrarfiaz » Sat Nov 18, 2017 8:02 pm

can u explain please?

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Tasnood
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Re: BDMO 2016: National Junior/2

Unread post by Tasnood » Fri Dec 01, 2017 2:07 pm

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Let $CF'G'$ be the reflection of $C'FG$. And as $CDE$ is the reflection of $C'DE$, we can say,
$F'G'||DE||AB$
So, $(CF'G')=(C'FG)=80$.
(ABC)=$\frac{1}{2}AB.CP=500$ or, $10CP=500$ or,$CP=50$
We know, $\triangle ABC~\triangle CF'G'$
$\frac {AB}{F'G'}=\frac {CP}{CQ}=k$
So, $F'G'=\frac{AB}{k}=\frac{20}{k}$, $CQ=\frac{CP}{k}=\frac{50}{k}$
$(CF'G')=\frac{CQ.F'G'}{2}=\frac{20*50}{2.k.k}=\frac{500}{k.k}=80$ or,$k=\frac{5}{2}$
$FG=F'G'=\frac{20}{k}=8$, $CQ=\frac{50}{k}=20$
Then, $CP+C'P'=50+20=70$ or,$CQ+C'P'+PR+RQ=70$ or,$2CR=70$ or,$CR=35$
We proved, $\triangle CF'G'~\triangle CDE$
So, $\frac{DE}{F'G'}=\frac{CR}{CQ}=\frac{35}{20}=\frac{7}{4}$
$DE=\frac{7F'G'}{4}=\frac{7*8}{4}=14$
So, $(CDE)=\frac{1}{2}DE.CR=\frac{14*35}{2}=245$; that we need. :P

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