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
BDMO 2016: National Junior/2
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Re: BDMO 2016: National Junior/2
[CDE]=225 sq. units
Re: BDMO 2016: National Junior/2
explain the process please
Re: BDMO 2016: National Junior/2
can u explain please?
Re: BDMO 2016: National Junior/2
$F'G'||DE||AB$
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.
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Re: BDMO 2016: National Junior/2
very easy problem