Consider the length of diagonal is 8.5 cm instead of 8.5 m because length of perpendiculars are in cm.
Given:
Length of the diagonal of a quadrilateral = 8.5 cm
Lengths of the perpendiculars dropped on it from the remaining opposite vertices are 3.5 cm and 4.5 cm.
To find:
The area of the quadrilateral.
Solution:
Diagonal divides the quadrilateral in 2 triangles. If diagonal is the base of both triangles then the lengths of the perpendiculars dropped on it from the remaining opposite vertices are heights of those triangles.
According to the question,
Triangle 1 : Base = 8.5 cm and Height = 3.5 cm
Triangle 2 : Base = 8.5 cm and Height = 4.5 cm
Area of a triangle is
[tex]Area=\dfrac{1}{2}\times base \times height[/tex]
Using this formula, we get
[tex]Area(\Delta 1)=\dfrac{1}{2}\times 8.5\times 3.5[/tex]
[tex]Area(\Delta 1)=14.875[/tex]
and
[tex]Area(\Delta 2)=\dfrac{1}{2}\times 8.5\times 4.5[/tex]
[tex]Area(\Delta 2)=19.125[/tex]
Now, area of the quadrilateral is
[tex]Area=Area(\Delta 1)+Area(\Delta 2)[/tex]
[tex]Area=14.875+19.125[/tex]
[tex]Area=34[/tex]
Therefore, the area of the quadrilateral is 34 cm².
