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olasank [31]
2 years ago
9

A 14-foot ladder is leaning against a building as shown. It touches a point 11 feet up on a building. How far away from the base

of the building does the ladder stand?
Mathematics
1 answer:
Luden [163]2 years ago
6 0
We can assume that the point the ladder creates with the ground and building is a triangle. You can use the Pythagorean theorem to solve this.

A^2 + B^2 = C^2

The ladder is C, and the building can act as A or B, so for the purpose of this explanation, I’ll make it A.

11^2 + B^2 = 14^2
Figure out the squares

121 + B^2 = 196
Subtract 121 from both sides

B^2 = 75
Square root B^2 and 75

B = 5 root3
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Answer: 0.07 (choice B)

===================================================

Explanation:

Let's define two events: B = hitting bullseye, and, T = hitting the third (outermost) ring

To figure out the probabilities of each event happening, we need to find the areas of each region. The bullseye is a circle with radius 6 inches (half of diameter 12), so the area is roughly...

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So P(B) = (36pi)/(256pi) = 9/64 is the probability of hitting bullseye

-----------------

The area of the third ring is equal to the difference of the overall largest circle and the second largest circle.

The second largest circle has a radius of 6+5 = 11 inches with area pi*11^2 = 121pi square inches

So the third ring has an area of 256pi - 121pi = 135pi square inches

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P(T) = (135pi)/(256pi)

P(T) = 135/256

-----------------

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P(B and T) = P(B)*P(T)

P(B and T) = (9/64)*(135/256)

P(B and T) = 0.07415771484376

P(B and T) = 0.07

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