Question

select the correct answer from each drop-down menu. a hotel builds an isosceles trapezoidal pool for children. it orders a tarp to cover the pool when not in use. what is the area of the tarp? diagram of trapezoid with vertices a, b, c, d; bc = 10 ft, cd = 5 ft, ∠c = 120°, ad = 15 ft to find the height, first find that ∠d is dropdown °. this means the height of the trapezoid is approximately dropdown feet. so, the area of the tarp is approximately dropdown square feet. dropdown options: 45, 60, 120 reset next

Explanation:

Step1: Find measure of ∠D

In an isosceles trapezoid, consecutive angles between the bases are supplementary. Since ∠C = 120°, ∠D = 180° - 120° = 60°.

Step2: Calculate the height

The non - parallel side (leg) is 5 ft. We can consider the right triangle formed by the height (h), the leg (5 ft), and the difference in the bases. The difference in the bases is 15 - 10 = 5 ft, so the base of the right triangle (adjacent to ∠D) is $\frac{15 - 10}{2}=2.5$ ft? Wait, no. Wait, in the isosceles trapezoid, when we drop a perpendicular from C to AD, let the foot be E, and from B to AD, let the foot be F. Then the length of DE is $\frac{AD - BC}{2}=\frac{15 - 10}{2}=2.5$ ft. But we know that $\cos D=\frac{DE}{CD}$, so $DE = CD\cos D$. Since CD = 5 ft and ∠D = 60°, $\cos60^{\circ}=\frac{1}{2}$, so $DE = 5\times\frac{1}{2}=2.5$ ft. And $\sin D=\frac{h}{CD}$, so $h = CD\sin D$. Since $\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$, then $h = 5\times\frac{\sqrt{3}}{2}\approx4.33$ ft? Wait, maybe I made a mistake. Wait, the leg is 5 ft, and the angle at D is 60°, so the height $h = 5\times\sin60^{\circ}\approx5\times0.866\approx4.33$? But maybe the options are different. Wait, maybe the first blank is 60, then the height: let's re - check. In an isosceles trapezoid, base angles are equal. Also, consecutive angles between the bases are supplementary. So ∠C and ∠D are supplementary, so ∠D = 60°. Then, to find the height, we can use the leg length (5 ft) and the angle. The height $h = 5\times\sin60^{\circ}\approx4.33$, but maybe the options have 4.33 or something else. Wait, maybe the problem has a typo or I misread. Wait, the leg is 5 ft, angle at D is 60°, so height $h = 5\times\sin60^{\circ}\approx4.33\approx4.3$ or maybe 4.33. Then the area of the trapezoid is $A=\frac{(a + b)}{2}\times h$, where $a = 10$, $b = 15$, $h\approx4.33$. So $A=\frac{(10 + 15)}{2}\times4.33=\frac{25}{2}\times4.33 = 12.5\times4.33\approx54.125$. But let's go step by step.

First, ∠D: since ABCD is an isosceles trapezoid, BC || AD, so ∠C + ∠D = 180°, so ∠D = 180 - 120 = 60 degrees. So first blank: 60.

Then, height: in triangle CDE (right triangle, where E is the foot of perpendicular from C to AD), CD = 5 ft, ∠D = 60°, so $\sin(60^{\circ})=\frac{h}{CD}$, so $h = CD\times\sin(60^{\circ})=5\times\frac{\sqrt{3}}{2}\approx5\times0.866 = 4.33$ ft (approx).

Then area of trapezoid: $A=\frac{(BC + AD)}{2}\times h=\frac{(10 + 15)}{2}\times4.33=\frac{25}{2}\times4.33 = 12.5\times4.33\approx54.125$ square feet.

But let's check the options. Wait, the first drop - down for ∠D: the options are 45, 60, 120. So ∠D is 60. Then height: if we consider the leg is 5 ft, angle 60°, so height is $5\times\sin60^{\circ}\approx4.33$, maybe rounded to 4.3 or 4.33. Then area: $\frac{(10 + 15)}{2}\times4.33\approx54.125$.

Answer:

First drop - down (∠D): 60°

Second drop - down (height): Approximately 4.33 (or 4.3) feet

Third drop - down (area): Approximately 54.13 (or 54.125) square feet

(If we assume the height is calculated as $5\times\sin60^{\circ}\approx4.33$ and area as $\frac{(10 + 15)}{2}\times4.33\approx54.13$)