A trapezoidal lot ABCD has sides BC parallel to AD. AB = 395 m, AD = 530 m and BC = CD. Angle CDA = 86 deg.
The area of the lot ABCD is 180,860.625 m²
To solve this problem, you can use the formula for the area of a plane figure.
The Area Formula
1. A square
The formula for the area of a square is to multiply the lengths of its sides.
Area = s x s
2. The formula for the area of a triangle
The area = [tex]\frac{1}{2}[/tex] x (base x height)
3. A area of the trapezoid, the formula is:
The area = ½ × sum of parallel sides × height.
Step-by-step explanation:
Given:
A trapezoidal lot ABCD has sides BC parallel to AD. AB = 395 m, AD = 530 m and BC = CD. Angle CDA = 86 deg.
Question:
Determine area of the lot.
Solution:
Step 1
Find the height of the trapezoid.
Sin 86⁰ = height ÷ hypotenuse
height = Sin 86⁰ x hypotenuse
height = 0.99 x 395 m
height = 391.05 m
Step 2
Find the area of the trapezoid.
The area = ½ × sum of parallel sides × height
The area = ½ × (395 m + 530 m) × 391.05 m
The area = ½ × 925 m × 391.05 m
The area = ½ × 361,721.25 m²
The area = 180,860.625 m²
So, the area of the lot ABCD is 180,860.625 m²
Learn more about:
- Area=_sq. What is the area of this trapezoid?: https://brainly.ph/question/27763705
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