Answer:
(B)
Step-by-step explanation:
If we re draw the given figure which is not so scale, we get that ABC is right angled triangle as:
[tex](AC)^{2}=(AB)^{2}+(CB)^{2}[/tex]
[tex]5^{2}=3^{2}+4^{2}[/tex]
[tex]25=9+16[/tex]
[tex]25=25[/tex]
which holds the Pythagoras theorem, thus ABC is right angle triangle which is right angled at B.
Now, it can also be seen from the drawn figure, that ADC is an isosceles triangle, therefore
Area of triangle ABC=[tex]\frac{1}{2}{\times}AB{\times}4[/tex]
=[tex]\frac{1}{2}{\times}3[\times}4[/tex]
=[tex]6sq units[/tex]
Now, Area of triangle ADC is given by heron's formula that is:
A=[tex]\sqrt{s(s-a)(s-b)(s-c)}[/tex]
and s=[tex]\frac{a+b+c}{2}[/tex]
Thus, s=[tex]\frac{6+6+5}{2}=\frac{17}{2}[/tex]
Area=[tex]\sqrt{\frac{17}{2}(\frac{17}{2}-6)(\frac{17}{2}-6)(\frac{17}{2}-5) }[/tex]
=[tex]\sqrt{\frac{17}{2}(\frac{5}{2})(\frac{5}{2})(\frac{7}{2})}[/tex]
=[tex]\sqrt{\frac{2975}{16}}[/tex]
=[tex]13.63sq units[/tex]
Now, area of the Quadrilateral ABCD is =Area of triangle ABC+ area of triangle ADC
=[tex]6+13.63[/tex]
=[tex]19.64[/tex]
therefore, area of the Quadrilateral ABCD is 19.64 sq units.
