Algebra homework

Question

A 20 m ladder and a 15 m ladder we’re leaned against a building. The bottom of the longer ladder was 7m farther from the building than the bottom of the shorter ladder, but both ladders reached the same distance up the building. Find the distance.

Answer 1

Let x = horizontal distance of the shorter ladder

x+7 = horizontal distance of the longer ladder

Let y = vertical distance of both ladders

The equations are:

x²+y²=15²

(x+7)²+y²=20²

This is a system of equations that can be solved using substitution.

Solve the first equation for y²

y²=15²-x²

Plug it into the second equation:

(x+7)²+(15²-x²)=20² Now this can be simplified and solved. First step is FOIL and simplifying the exponents.

x²+14x+49+225-x²=400 Combine like terms (the x² cancel out!)

14x+274=400 Subtract 274 from both sides

14x+274-274=400-274 Simplify

14x=126 Divide both sides by 14

14x/14=126/14

x=9 ft This is the horizontal distance, they actually are asking to find the vertical distance, which is y.

So we can plug this into the first equation and solve for y

x²+y²=15²

9²+y²=15² Simplify the exponents

81+y²=225 Subtract 81 from both sides

81+y²-81=225-81

y²=144 Take the square root of both sides

y=12 ft

The height both ladders go up the side of the building is 12 ft.

Answer 2

Let x=the horizontal distance of the 20' ladder that touches the ground. Then x^2 + y^2 = 20^2

Then x-7 is the horizontal distance of the 15' ladder that touches the ground, and (x-7)^2 + y^2 = 13^2

Solving these by any manner yields x-=16.61 m.