Let the length of the common diagonal of the quadrilateral be d = 24 m. The perpendicular heights dropped from the opposite vertices onto this diagonal are h_1 = 13 m and h_2 = 8 m.

Substitute d = 24 m, h_1 = 13 m, and h_2 = 8 m into the formula. First add the heights: 13 + 8 = 21 m. Then multiply: (1)/(2) × 24 × 21 = 12 × 21 = 252 m².

the diagonal of a quadrilateral shaped field is 24 M and the perpendicular dropped on it from the remaining opposite verticals are 8 M 13 M find the area of the field

Answer: $252\text{ m}^2$

Step-by-step solution

Step 1: Identify given measurements

Let the length of the common diagonal of the quadrilateral be $d = 24\text{ m}$ . The perpendicular heights dropped from the opposite vertices onto this diagonal are $h_1 = 13\text{ m}$ and $h_2 = 8\text{ m}$ .

Step 2: State the formula for area of a general quadrilateral

A diagonal divides the quadrilateral into two triangles with a common base $d$ . The area of the quadrilateral is the sum of the areas of these two triangles: $\frac{1}{2} \times d \times h_1 + \frac{1}{2} \times d \times h_2 = \frac{1}{2} \times d \times (h_1 + h_2)$ .

Step 3: Substitute values and calculate the area

Substitute $d = 24\text{ m}$ , $h_1 = 13\text{ m}$ , and $h_2 = 8\text{ m}$ into the formula. First add the heights: $13 + 8 = 21\text{ m}$ . Then multiply: $\frac{1}{2} \times 24 \times 21 = 12 \times 21 = 252\text{ m}^2$ .

Solve your own maths question free →