Kushtia '15 \9

For students of class 6-8 (age 12 to 14)
Absur Khan Siam
Posts:65
Joined:Tue Dec 08, 2015 4:25 pm
Location:Bashaboo , Dhaka
Kushtia '15 \9

Unread post by Absur Khan Siam » Wed Jan 18, 2017 10:50 pm

$ABCD$ is a trapezium. Both $AB$ and $CD$ is perpendicular to $AD$. If $AB<CD$, $AD=8, BC = AB + CD$ ; then $AB \times CD=?$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
Posts:65
Joined:Tue Dec 08, 2015 4:25 pm
Location:Bashaboo , Dhaka

Re: Kushtia '15 \9

Unread post by Absur Khan Siam » Wed Jan 18, 2017 11:01 pm

My solution:
Just the use of pythagorus' theorem and $(a+b)^2 - (a-b)^2 = 4ab$.
$E$ is a point on $CD$ such that $BE \perp CD$.Thus $CE = CD - AB,BE = AD$.
Using the pythagorus' theorem,$BC^2 = BE^2 + CE^2 \rightarrow (CD+AB)^2 = (CD-AB)^2 + AD^2 $
$\rightarrow (CD+AB)^2 - (CD-AB)^2 = 8^2 \rightarrow 4 \times AB \times CD = 64$.
Thus we get the solution , $AB \times CD = 16$. :)
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

aritra barua
Posts:57
Joined:Sun Dec 11, 2016 2:01 pm

Re: Kushtia '15 \9

Unread post by aritra barua » Thu Jan 19, 2017 2:33 pm

We may also find AB*CD by finding their respective values.In that case,we draw a perpendicular from B to CD which divides CD into DQ+QC.Then we may notice that by joining A with C,we get a right angled triangle DAC.AD=8 units.Again by connecting D with B,we get a right angled triangle DAB.But if we try to apply Pythagorean formula,we find DC has 2 options (6,8)units & AB has 1option of 6 units.But we also get a right angled quadrilateral ABQD.Then we may ponder over that since BQD is not bound by any condition,QC is of 6 units,by applying Pythagorean theorem.Which necessarily cuts off the option of 6units for CD and 6 units for AB.Then the lines of ABQD are parallel.For which DQ =2,which follows AB=2 units.So,AB*CD =2*8 units =16 units :D

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