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As well description:

We have a trapezium $ABCD$ and a circle inside it. all $4$ sides of trapezium are touching the circle. left side AB have length of $26\ cm$, and right side is $22 \ cm$. radius of a circle is $10\ cm$. With that knowledge we have to find the area of trapezium $ABCD$.

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Hint: The area is $20\left(\frac{AD+BC}{2}\right)$. So all we need to know is $AD+BC$.

Claim: This is equal to $AB+CD$. To show that $AD+BC=AB+CD$, use repeatedly the fact that if the line segments $PX$ and $PY$ are tangent to a circle, where $X$ and $Y$ are the points of tangency, then $PX=PY$.

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    $\begingroup$ Is not $area=10(AD+BC)$? $\endgroup$ Nov 23 '15 at 21:46
  • $\begingroup$ @EmilioNovati: Thanks, I thought it was diameter. $\endgroup$ Nov 23 '15 at 21:50

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