Question

In the given figure AB||CD||EF||GH and GH=100cm.Find the value of DF

Answer 1

Answer:

BH =

$\frac{1}{2} bh$

= DF .

= If BH = 100 cm then , DF =

$\frac{1}{2} bh$

=

$\frac{1}{2} 100$

= 50 cm .

= Therefore DF = 50 cm .

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\sf l\ =\ \Large {\displaystyle \sf \int \limits^{1/2}_ {0}\ \footnotesize \dfrac{\dfrac{-dt}{t^2}}{\bigg(1\ +\ \dfrac{1}{t^2} \bigg)\ \sqrt{1\ -\ \dfrac{1}{t^2}}}}$