BDMO 2016: National Junior/2

For students of class 6-8 (age 12 to 14)
Tasnova Novera
Posts: 1
Joined: Fri Feb 03, 2017 10:32 pm

BDMO 2016: National Junior/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.

Zahin Hasin Rudro
Posts: 11
Joined: Sun Jul 12, 2015 3:52 am

Re: BDMO 2016: National Junior/2

[CDE]=225 sq. units

abrarfiaz
Posts: 14
Joined: Mon Mar 20, 2017 4:53 pm

Re: BDMO 2016: National Junior/2

abrarfiaz
Posts: 14
Joined: Mon Mar 20, 2017 4:53 pm

Re: BDMO 2016: National Junior/2

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

Re: BDMO 2016: National Junior/2

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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.

prottoydas
Posts: 8
Joined: Thu Feb 01, 2018 11:56 am

Re: BDMO 2016: National Junior/2

very easy problem