Question

The diagonal of a quadrilateral shaped field is $25$ $cm$ and the perpendiculars dropped on it from the remaining opposite vertices are $8$ $cm$ and $12$ $cm$. Find the area of the field(in $c{m}^2$)

• A. 100
• B. 200
• C. 250
• D. 300

Answer 1

The correct option is C 250

We can calculate the area of this quadrilateral as the sum of the areas of two triangles taking diagonal length $25$ $cm$ as the base for both triangles and perpendicular length $8$ $cm$ as the height of one triangle and the perpendicular length $12$ $cm$ as the height of another triangle.

Area of triangle = $\frac{1}{2}\times base\times height$

Area of the first triangle = $\frac{1}{2}\times 25\times 8$

Area of the second triangle = $\frac{1}{2}\times 25\times 12$

Area of quadrilateral = Area of the first triangle + Area of the second triangle

Area of quadrilateral = $\frac{1}{2}\left(25\right)(8+12)=25\times 10=250$ ${cm}^2$