The area of the considered 3 triangles is given by: Option B: 13 sq. units.
Suppose the vertices of the considered triangle ABC are on [tex]A(A_x,A_y), B(B_x,B_y), C(C_x,C_y)[/tex], then, the area of the triangle is given by:
[tex]Area = \dfrac{|A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)|}{2}[/tex]
For this case, the missing image is attached below.
The second and third triangle are easy as their base and height are parallel to x or y axis.
We will find area of third triangle by coordinates of its vertices.
Getting area of first triangle:
Thus, we get:
[tex]Area_{ABC} = \dfrac{|4(7-8) + 5(8-10) + 7(10-7)|}{2} = \dfrac{7}{2} = 3.5 \: \rm unit^2[/tex]
Area of second triangle = [tex]\dfrac{|RS| \times h}{2}[/tex]base, and h is the perpendicular line connecting from line RS and the vertex T
From figure we see:
|RS| = 3 units, h = 3 units
Thus, area of second triangle = [tex]3 \times 3 /2 = 4.5 \: \rm unit^2[/tex]
Area of third triangle = [tex]\dfrac{|XY| \times h}{2}[/tex] where XY is base, and h is the perpendicular line connecting from line XY and the vertex Z
From figure we see:
|XY| = 5 units, h = 2 units
Thus, Area of third triangle = [tex]5 \times 2 /2 = 5 \: \rm unit^2[/tex]
Total area = 3.5 + 4.5 + 5 = 13 sq. units
Thus, the area of the considered 3 triangles is given by: Option B: 13 sq. units.
Learn more about area of a triangle from its vertices' coordinates here:
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