Answer:
![6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}](https://tex.z-dn.net/?f=6%2B2%5Csqrt%7B21%7D%5C%3A%5Cmathrm%7Bcm%5E2%7D%5Capprox%2015.17%5C%3A%5Cmathrm%7Bcm%5E2%7D)
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
,
, and
. The two legs, 3 and 4, represent the triangle's height and base, respectively.
The area of a triangle with base
and height
is given by
. Therefore, the area of this right triangle is:
![A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%203%5Ccdot%204%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%2012%3D6%5C%3A%5Cmathrm%7Bcm%5E2%7D)
The other triangle is a bit trickier. Triangle
is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:
, where
,
, and
are three sides of the triangle and
is the semi-perimeter (
).
The semi-perimeter,
, is:
![s=\frac{5+5+4}{2}=\frac{14}{2}=7](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B5%2B5%2B4%7D%7B2%7D%3D%5Cfrac%7B14%7D%7B2%7D%3D7)
Therefore, the area of the isosceles triangle is:
![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}](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B7%287-5%29%287-5%29%287-4%29%7D%2C%5C%5CA%3D%5Csqrt%7B7%5Ccdot%202%5Ccdot%202%5Ccdot%203%7D%2C%5C%5CA%3D%5Csqrt%7B84%7D%2C%20%5C%5CA%3D2%5Csqrt%7B21%7D%5C%3A%5Cmathrm%7Bcm%5E2%7D)
Thus, the area of the quadrilateral is:
![6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}](https://tex.z-dn.net/?f=6%5C%3A%5Cmathrm%7Bcm%5E2%7D%2B2%5Csqrt%7B21%7D%5C%3A%5Cmathrm%7Bcm%5E2%7D%3D%5Cboxed%7B6%2B2%5Csqrt%7B21%7D%5C%3A%5Cmathrm%7Bcm%5E2%7D%7D)