Answer:
Area of [tex]\triangle[/tex]ACD = 4 [tex]ft^2[/tex]
Area of [tex]\triangle[/tex]ABC = 16 [tex]ft^2[/tex]
Step-by-step explanation:
Given that:
D is a point on AB.
and ABC is a triangle.
AD:DB = 1 : 3
Area of [tex]\triangle[/tex]CDB = 12 [tex]ft^2[/tex]
Kindly refer to the attached image as per the given dimensions and values.
To find:
Area of [tex]\triangle[/tex]ACD and Area of [tex]\triangle[/tex]ABC = ?
Solution:
Formula for area of a triangle = [tex]\frac{1}{2}\times Base \times Height[/tex]
The altitudes of triangles [tex]\triangle[/tex]CDB and [tex]\triangle[/tex]ACD are equal in dimensions.
Therefore the area of triangles [tex]\triangle[/tex]CDB and [tex]\triangle[/tex]ACD will be equal to the ratio of their bases.
Area of [tex]\triangle[/tex]ACD : Area of [tex]\triangle[/tex]CDB = AD: DB = 1 : 3
[tex]\Rightarrow[/tex] Area of [tex]\triangle[/tex]ACD = [tex]\frac{12}{3} = \bold{4 ft^2}[/tex]
Area of [tex]\triangle[/tex]ABC = Area of [tex]\triangle[/tex]ACD + Area of [tex]\triangle[/tex]CDB = 12 + 4 = 16 [tex]ft^2[/tex]
Therefore, the answer is:
Area of [tex]\triangle[/tex]ACD = 4 [tex]ft^2[/tex]
Area of [tex]\triangle[/tex]ABC = 16 [tex]ft^2[/tex]
