Answer:
[tex]6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}[/tex]
Step-by-step explanation:
The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.
Vertices A, B, and C form a right triangle with legs [tex]AB=3[/tex], [tex]BC=4[/tex], and [tex]AC=5[/tex]. The two legs, 3 and 4, represent the triangle's height and base, respectively.
The area of a triangle with base [tex]b[/tex] and height [tex]h[/tex] is given by [tex]A=\frac{1}{2}bh[/tex]. Therefore, the area of this right triangle is:
[tex]A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}[/tex]
The other triangle is a bit trickier. Triangle [tex]\triangle ADC[/tex] is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:
[tex]A=\sqrt{s(s-a)(s-b)(s-c)}[/tex], where [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] are three sides of the triangle and [tex]s[/tex] is the semi-perimeter ([tex]s=\frac{a+b+c}{2}[/tex]).
The semi-perimeter, [tex]s[/tex], is:
[tex]s=\frac{5+5+4}{2}=\frac{14}{2}=7[/tex]
Therefore, the area of the isosceles triangle is:
[tex]A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}[/tex]
Thus, the area of the quadrilateral is:
[tex]6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}[/tex]
