9514 1404 393
Explanation:
The length CD can be found from the Pythagorean theorem.
CF² + CD² = DF²
2² + CD² = (2√10)²
CD² = 40 -4 = 36
CD = √36 = 6
From here, any of several methods could be used to find the area of ΔDEF.
One of them is to subtract the areas of the triangles that are not ΔDEF from the area of the rectangle ABCD.
area ABCD = bh = (6 cm)(5 cm) = 30 cm²
area ΔADE = 1/2bh = 1/2(4 cm)(5 cm) = 10 cm²
area ΔBEF = 1/2bh = 1/2(2 cm)(3 cm) = 3 cm²
area ΔCDF = 1/2bh = 1/2(6 cm)(2 cm) = 6 cm²
Then the area of interest is ...
area ΔDEF = area ABCD -area ΔADE -area ΔBEF -area ΔCDF
area ΔDEF = (30 -10 -3 -6) cm²
area ΔDEF = 11 cm²
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Alternate solution
If we define A as the origin of a coordinate system, then the coordinates of the vertices of ΔDEF are ...
D(0, 5), E(4, 0), F(6, 3)
The area can be computed as half the absolute value of the sum of the determinants of 2×2 matrices consisting of adjacent points.
[tex]A=\dfrac{1}{2}\left|\left|\begin{array}{cc}0&5\\4&0\end{array}\right|+\left|\begin{array}{cc}4&0\\6&3\end{array}\right|+\left|\begin{array}{cc}6&3\\0&5\end{array}\right| \right|\\\\=\dfrac{1}{2}|(0-20)+(12-0)+(30-0)|=\dfrac{1}{2}(22) = \boxed{11}[/tex]
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Additional comment
The solution using coordinates is effectively equivalent to the computation of the area of trapezoid ABFD, less the areas of triangles EBF and AED.
