Kushtia '15 \9
Posted: 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 $AB \times CD=?$
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 , $AB \times CD = 16$.
$E$ 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 , $AB \times CD = 16$.
