Từ $A$ kẻ đường cao $AH$ của hình thang

Ta có:

$S_{ABC}=\dfrac12AB.AH$

$S_{ACD}=\dfrac12CD.AH$

$\to \dfrac{S_{ABC}}{S_{ACD}}=\dfrac{AB}{CD}=\dfrac47$

Từ $B$ kẻ $BK\perp AC\quad (K\in AC)$

Từ $D$ kẻ $DI\perp AC\quad (I\in AC)$

Ta được:

$S_{ABC}=\dfrac12AC.BK$

$S_{ACD}=\dfrac12AC.DI$

$\to \dfrac{S_{ABC}}{S_{ACD}}=\dfrac{BK}{DI}$

mà $\dfrac{S_{ABC}}{S_{ACD}}=\dfrac47$

nên $\dfrac{BK}{DI}=\dfrac47$

Mặt khác:

$S_{BMC}=\dfrac12BK.MC$

$S_{DMC}=\dfrac12DI.MC$

$\to \dfrac{S_{BMC}}{S_{DMC}}=\dfrac{BK}{DI} =\dfrac47$

$\to S_{DMC}=\dfrac74S_{BMC}$

$\to S_{DMC}=\dfrac74\cdot 15 =\dfrac{105}{4}\, cm^2$

$\to S_{BCD}=S_{BMC} + S_{DMC}=15 +\dfrac{105}{4} =\dfrac{165}{4}\, cm^2$

Ta có:

$S_{BCD}=\dfrac12AH.CD$

$S_{ACD}=\dfrac12AH.CD$

$\to S_{ACD}= S_{BCD}=\dfrac{165}{4}\, cm^2$

Ta lại có:

$\dfrac{S_{ABC}}{S_{ACD}}=\dfrac47$

$\to S_{ABC} = \dfrac47S_{ACD}$

$\to S_{ABC}=\dfrac47\cdot \dfrac{165}{4}=\dfrac{165}{7}\, cm^2$

Diện tích hình thang $ABCD$

$S_{ABCD}=S_{ABC} + S_{ACD}$

$\to S_{ABCD}=\dfrac{165}{7} + \dfrac{165}{4}$

$\to S_{ABCD}=\dfrac{1815}{28}\approx 64,82\,cm^2$