The of a 15-foot ladder is 3 feet farther up

percibaa8 2021-12-23 Answered
The of a 15-foot ladder is 3 feet farther up a wall than the foot is from the bottom of the wall. How far is the ladder from the bottom of the wall?

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Expert Answer

Navreaiw
Answered 2021-12-24 Author has 3306 answers
Step 1
Let:
\(\displaystyle{x}=\text{distance of the ladder from the bottom of the wall}\)
\(\displaystyle{y}=\text{height of the ladder from the bottom of the wall}\)
\(\displaystyle{L}=\text{length of the ladder}\)
Using Pythagorean theorem
\(\displaystyle{L}^{{{2}}}={x}^{{{2}}}+{y}^{{{2}}}\)
Since the height is 3 feet farther up than the distance:
\(\displaystyle{y}={x}+{3}\)
\(\displaystyle{L}^{{{2}}}={x}^{{{2}}}+{\left({x}+{3}\right)}^{{{2}}}\)
\(\displaystyle{\left({15}\right)}^{{{2}}}={x}^{{{2}}}+{x}^{{{2}}}+{6}{x}+{9}\)
\(\displaystyle{225}={2}{x}^{{{2}}}+{6}{x}+{9}\)
\(\displaystyle{2}{x}^{{{2}}}+{6}{x}-{216}={0}\)
\(\displaystyle{\left({x}-{9}\right)}{\left({x}+{12}\right)}={0}\)
\(\displaystyle{x}{1}={9}\) and \(\displaystyle{x}{2}=-{12}\) (neglect the negative value)
Answer: \(\displaystyle{x}={9}\text{feet}\)
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veiga34
Answered 2021-12-25 Author has 803 answers
Step 1
\(\displaystyle{15}^{{{2}}}={\left({x}+{3}\right)}^{{{2}}}+{x}^{{{2}}}\)
\(\displaystyle{225}={x}^{{{2}}}+{6}{x}+{9}+{x}^{{{2}}}\)
\(\displaystyle{2}{x}^{{{2}}}+{6}+{9}-{225}={0}\)
\(\displaystyle{2}{x}^{{{2}}}+{6}{x}-{216}={0}\)
\(\displaystyle{2}{\left({x}^{{{2}}}={3}{x}-{108}\right)}={0}\)
\(\displaystyle{2}{\left({X}={12}\right)}{\left({x}-{9}\right)}={0}\)
\(\displaystyle{x}-{9}={0}\)
\(\displaystyle{x}={9}\) feet is the distance from the wall to the bottom of the ladder.
Proof:
\(\displaystyle{9}^{{{2}}}+{\left({9}+{3}\right)}^{{{2}}}={15}^{{{2}}}\)
\(\displaystyle{81}+{12}^{{{2}}}={225}\)
\(\displaystyle{81}+{144}={225}\)
\(\displaystyle{225}={225}\)
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user_27qwe
Answered 2021-12-30 Author has 9558 answers

Step 1
This is a right triangle with a hypotenuse =15, the base x & the other side =x+3
\(x^{2}+(x+3)^{2}=15^{2}\)
\(x^{2}+x^{2}+6x+9=225\)
\(2x^{2}+6x+9-225=0\)
\(2x^{2}+6x-216=0\)
\(2(X^{2}+3X-108)=0\)
\(2(X-9)(X+12)=0\)
\(X-9=0\)
X=9ft. Distance from the wall and the base of the ladder.
9+3=12ft. Distance of the ladder up the wall.

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