Kushtia '15 \9
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$ABCD$ 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
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
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Re: Kushtia '15 \9
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