The area of the considered 3 triangles is given by: Option B: 13 sq. units.
<h3>How to find the area of a triangle whose vertices' coordinates are given?</h3>
Suppose the vertices of the considered triangle ABC are on
, then, the area of the triangle is given by:
![Area = \dfrac{|A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)|}{2}](https://tex.z-dn.net/?f=Area%20%3D%20%5Cdfrac%7B%7CA_x%28B_y-C_y%29%20%2B%20B_x%28C_y-A_y%29%20%2B%20C_x%28A_y-B_y%29%7C%7D%7B2%7D)
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:
- Coordinates of A : (4,10)
- Coordinates of B : (5,7)
- Coordinates of C : (7,8)
Thus, we get:
![Area_{ABC} = \dfrac{|4(7-8) + 5(8-10) + 7(10-7)|}{2} = \dfrac{7}{2} = 3.5 \: \rm unit^2](https://tex.z-dn.net/?f=Area_%7BABC%7D%20%3D%20%5Cdfrac%7B%7C4%287-8%29%20%2B%205%288-10%29%20%2B%207%2810-7%29%7C%7D%7B2%7D%20%3D%20%5Cdfrac%7B7%7D%7B2%7D%20%3D%203.5%20%5C%3A%20%5Crm%20unit%5E2)
Area of second triangle =
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 = ![3 \times 3 /2 = 4.5 \: \rm unit^2](https://tex.z-dn.net/?f=3%20%5Ctimes%203%20%2F2%20%3D%204.5%20%5C%3A%20%5Crm%20unit%5E2)
Area of third triangle =
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 = ![5 \times 2 /2 = 5 \: \rm unit^2](https://tex.z-dn.net/?f=5%20%5Ctimes%202%20%2F2%20%3D%205%20%5C%3A%20%5Crm%20unit%5E2)
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:
brainly.com/question/2532574