Solution:
$\dfrac{125}{2}$
Step by step solution:
Applying Pythagoras'Theorem in triangle ADB, we get:
$\quad BD^2 = AD^2 + AB^2$
$\Leftrightarrow BD^2 = 12^2 + 5^2$
$\Leftrightarrow BD^2 = 169$
Square root both side, we get:
$\quad BD = 13\ (cm)$
So that, the area of the quadrilateral $ABCD$ as the following:
$\quad S_{ABCD}= S_{ABD} + S_{DBC}$
$\Leftrightarrow S_{ABCD}=\dfrac12AD.AB + \dfrac12DB.BC$
$\Leftrightarrow S_{ABCD}= \dfrac12\cdot 12\cdot 5 +\dfrac12\cdot 13\cdot 5$
$\Leftrightarrow S_{ABCD}= \dfrac{60}{2} +\dfrac{65}{2}$
$\Leftrightarrow S_{ABCD}=\dfrac{125}{2}\ (cm^2)$
Answer: $\dfrac{125}{2}\ cm^2$
