Math  /  Geometry

Question2. The doorsill of a campus building is 5 ft above ground level. To allow sheelchair access, the steps in front of the door are to be replaced by a straight ramp with constant slope 112\frac{1}{12}, as shown in the figure. How long must the ramp be? The answer is not 60 ft .]

Studdy Solution

STEP 1

1. The ramp needs to elevate from ground level to a height of 5 feet.
2. The slope of the ramp is given as 112\frac{1}{12}, meaning for every 1 foot of vertical rise, there are 12 feet of horizontal distance.
3. We need to find the length of the ramp, which is the hypotenuse of the right triangle formed by the ramp, the height of 5 feet, and the horizontal distance.

STEP 2

1. Establish the relationship between the height, the horizontal distance, and the slope.
2. Calculate the horizontal distance using the given slope and height.
3. Use the Pythagorean theorem to find the length of the ramp (hypotenuse).

STEP 3

Establish the relationship between the height, the horizontal distance, and the slope using the slope formula.
The slope 112\frac{1}{12} means: slope=riserun=5 feetx feet \text{slope} = \frac{\text{rise}}{\text{run}} = \frac{5 \text{ feet}}{x \text{ feet}} where xx is the horizontal distance.

STEP 4

Rearrange the equation to solve for xx (horizontal distance).
x=5×12 x = 5 \times 12 x=60 feet x = 60 \text{ feet}

STEP 5

Use the Pythagorean theorem to find the length of the ramp, where the ramp length LL is the hypotenuse of the right triangle formed by the vertical rise of 5 feet and the horizontal run of 60 feet.
L2=(vertical rise)2+(horizontal run)2 L^2 = (\text{vertical rise})^2 + (\text{horizontal run})^2 L2=52+602 L^2 = 5^2 + 60^2 L2=25+3600 L^2 = 25 + 3600

STEP 6

Calculate the sum of the squares.
L2=3625 L^2 = 3625

STEP 7

Take the square root of both sides to find LL.
L=3625 L = \sqrt{3625} L60.21 feet L \approx 60.21 \text{ feet}
The length of the ramp must be approximately 60.21 feet.

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