Problem 32: A pool holds 450 cubic feet of water and has a height of 4.2 feet. We need to find the area of the bottom of the pool to the nearest square foot, as this is the amount of tile needed to re-tile it. Problem 33: A tent is constructed that is completely enclosed, including the bottom. The ends of the tent are two congruent isosceles triangles. The tent has dimensions $BC = 10$ ft, $CF = 11$ ft, and $AM = 9$ ft. We need to find the surface area of the tent (including the two triangular ends and the rectangular sides and bottom) to the nearest square foot. Problem 34: Find the area of a sector with a central angle of 170 degrees and a diameter of 9.1 cm, rounded to the nearest tenth.

1. Problem Description

Problem 32: A pool holds 450 cubic feet of water and has a height of 4.2 feet. We need to find the area of the bottom of the pool to the nearest square foot, as this is the amount of tile needed to re-tile it.

Problem 33: A tent is constructed that is completely enclosed, including the bottom. The ends of the tent are two congruent isosceles triangles. The tent has dimensions $BC = 10$ ft, $CF = 11$ ft, and $AM = 9$ ft. We need to find the surface area of the tent (including the two triangular ends and the rectangular sides and bottom) to the nearest square foot.

Problem 34: Find the area of a sector with a central angle of 170 degrees and a diameter of 9.1 cm, rounded to the nearest tenth.

2. Solution Steps

Problem 32:

The volume of a cylinder or prism is given by $V = Ah$, where $V$ is the volume, $A$ is the area of the base, and $h$ is the height. In this case, we are given the volume $V = 450$ cubic feet and the height $h = 4.2$ feet. We need to find the area $A$ of the bottom of the pool.

$A = V/h = 450/4.2 \approx 107.142857$ square feet.

Rounding to the nearest square foot, we get $A = 107$ square feet.

Problem 33:

The area of each triangular end is $A_{triangle} = (1/2) * base * height = (1/2) * BC * AM = (1/2) * 10 * 9 = 45$ square feet. Since there are two triangles, the total area of the two ends is $2 * 45 = 90$ square feet.

The bottom of the tent is a rectangle with dimensions $BC = 10$ ft and $CF = 11$ ft, so its area is $A_{bottom} = BC * CF = 10 * 11 = 110$ square feet.

The two sides of the tent are rectangles. Let $AB=AC$. We have to find the length of $AB$ using the Pythagorean theorem. Since $AM$ is the height of the isosceles triangle, $M$ is the midpoint of $BC$. Thus $BM = BC/2 = 10/2 = 5$ feet.

In right triangle $ABM$, we have $AB^2 = AM^2 + BM^2 = 9^2 + 5^2 = 81 + 25 = 106$. Therefore, $AB = \sqrt{106} \approx 10.30$ feet.

The two sides have area $A_{side} = AB * CF = \sqrt{106} * 11$. Since there are two identical sides, their combined area is $2 * 11 * \sqrt{106} \approx 2 * 11 * 10.3 = 226.6$ square feet.

Total surface area of the tent is $A = 2 * A_{triangle} + A_{bottom} + 2 * A_{side} = 90 + 110 + 226.6 \approx 426.6$ square feet.

Rounding to the nearest square foot gives us 427 square feet.

Problem 34:

The area of a sector is given by the formula $A = (\theta/360) * \pi r^2$, where $\theta$ is the central angle in degrees and $r$ is the radius of the circle.

We are given that the central angle $\theta = 170$ degrees and the diameter is $9.1$ cm, so the radius $r = 9.1/2 = 4.55$ cm.

$A = (170/360) * \pi * (4.55)^2 \approx (0.4722) * \pi * 20.7025 \approx 30.6915$ square cm.

Rounding to the nearest tenth, we get $30.7$ square cm.

3. Final Answer

Problem 32: 107 ft$^2$

Problem 33: 427 ft$^2$

Problem 34: 30.7 cm$^2$