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A trapezoidal lot ABCD has sides BC parallel to AD. AB = 395m, AD = 530m and BC = CD. Angle CDA = 86 deg. Determine area of the lot.

Sagot :

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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