A contractor leans a 21-foot ladder against a building. The distance from the ground to the top of the ladder is 9 feet more than the distance from the building to the base of the ladder. How far up the building is the ladder to the nearest tenth of a foot

Respuesta :

Answer:

The distance from the ground to the top of the ladder = 18.7 feet

Step-by-step explanation:

The distance from the ground to the top of the ladder is 9 feet more than the distance from the building to the base of the ladder.

distance from the building to the base of the ladder.= X

The distance from the ground to the top of the ladder = 9+x

Solving the triangle using Pythagorean theorem

X² +(x+9)²= 21²

X² + x² + 18x +81= 441

2x² +18x -360 = 0

X² +9x -180 = 0

X= (-9-28.3)/2 or (-9+28.3)/2

Definitely x is going to be a positive number

So

X= (-9+28.3)/2

X=9.65 feet

The distance from the ground to the top of the ladder = 9+x

The distance from the ground to the top of the ladder = 9+9.65

The distance from the ground to the top of the ladder = 18.65

The distance from the ground to the top of the ladder = 18.7 feet

Ver imagen martin58853

Building is 9.65 foot far from ladder.

Length of ladder = 21 foot. This diagonal ladder is Hypotenuse to the right angled triangle formed between building & ladder.

Let the distance from building to base of ladder = x. This is the base of right angled triangle.

So, distance from ground to top of ladder = x + 9. This is the height of right angled triangle.

As per Pythagorus Theorem,

[tex]Hypotenuse^2 = Base^2 + Height^2[/tex]

[tex]21^2 = x^2 + (x+9)^2[/tex]

441 = [tex]x^2 + x^2 + 81 + 18x[/tex]  

360 = [tex]2x^2 + 18x[/tex]

[tex]2x^2 + 18x[/tex] - 360 = 0

x = -b + [tex]\sqrt{b^2 - 4ac[/tex] / 2a  

x = 9.65 { Building distance to ladder base }

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