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A painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft from the building. To the nearest foot, how high on the building does the ladder reach?

23 feet

16 feet

8 feet

32 feet



also can ya'll do the questions in the photos

A painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft-example-1
A painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft-example-1
A painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft-example-2
User Rndm
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7.0k points

2 Answers

2 votes

Answer:

Part 1: The building is 16 ft high where the ladder reach.

Part 2: Option A is correct.

Explanation:

Given that a painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft from the building.

we have to find the height of building does the ladder reach.

Let the height of the building be x.

Using Pythagoras theorem,


Hypotenuse^2=Perpendicular^2+Base^2


AC^2=AB^2+BC^2


20^2=x^2+12^2


x^2=400-144=256


x=16 ft

Hence, the building is 16 ft high where the ladder reach.

Part 2: Given a right angles triangle in which

Length of LM=5 units and measure of angle K i.e ∠K=45°

we have to find the length of KL

By trigonometric ratios


\sin \angle K=(LM)/(KL)


\sin 45^(\circ)=(5)/(KL)


(1)/(\sqrt2)=(5)/(KL)


KL=5\sqrt2 units

Option A is correct.

A painter leans a 20-ft ladder against a building. The base of the ladder is 12 ft-example-1
User Mazlor
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7.7k points
3 votes
For the main question, it's basically just the Pythagorean's Theorem, 400-144= 256, square root of 256 is 16, so it reaches 16 feet up the building.
For Picture 1: Another Pythagorean's Theorem question, since the two angles are both 45 degrees, then LM and KM are the same. They are both 5, so they both square to 25, added together 50. So we do square root of 50, and then you can change it to square root of 25 * square root of 2. And then you can change it to 5 * square root of 2.
Picture 2: isn't it the exact same?
User DaGrevis
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7.4k points