Find the area and perimeter of the figure.

Perimeter = 70.75 units

Area = 233.78 units²

Step-by-step explanation:

(Area) The first step in finding the area of this trapezoid is identifying its height. You can find the height of the trapezoid by using sin with the given angle and hypotenuse of the right triangle in the figure (sin x° = opposite leg/hypotenuse):

sin 60° = h/12 (where h = height)

12(sin 60°) = h

h = 12(sin 60°) or ≈10.39

Now, you need to find the third leg of the right triangle:

12^2 - (12 × sin 60°)^2 = l^2

144 - (12 × sin 60°)^2 = l^2

36 = l^2

6 = l

Now that you know the third leg is 6, you also know that the length of the base of the right trapezoid to the right of the right triangle is 24 since the length of the base of the whole trapezoid is 30. You can then deduct that if you split that right trapezoid into a rectangle and right triangle, the base of the triangle would be 9 because 24 (base 2) - 15 (base 1) = 9. At this point, you have all the necessary measures to find the area of the trapezoid:

Right triangle on the left: (6 × 10.39)/2 ≈ 31.17

Rectangle formed in the middle: 15 × 10.39 ≈ 155.85

Right triangle on the right: (9 × 10.39)/2 ≈ 46.755

If you add these all together, you'll find that the area is ≈233.775 units^2. Since you have to round to the nearest hundredth, it's ≈233.78 units².

(Perimeter) In order to find the perimeter or this figure, you need to find the hypotenuse of the aforementioned right triangle created from dividing the right trapezoid into a rectangle and right triangle. We already found that the triangle's legs are 9 and (12×sin 60°), so you just need to use the Pythagorean theorem to solve for c, or the hypotenuse:

c^2 = 9^2 + (12 × sin 60°)^2

c^2 = 81 + 108

c^2 = 189

c = 3√21

Now that you know that the missing side length is 3√21, you can add them all together to find the perimeter:

30 + 12 + 15 + 3√21

= 57 + 3√21

≈ 70.75 units (rounding to the hundredths)

a framing bias

step-by-step explanation:

rodrigo is subject to a framing bias i believe.