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a wall is 15 ft high and 10 ft from a house. find the length x of the s…

Question

a wall is 15 ft high and 10 ft from a house. find the length x of the shortest ladder which can go from the top of the wall to a window 20.5 ft above the ground. x = ? ft. round to the nearest hundredth.

Explanation:

Step1: Use similar triangles

Let the distance from the base of the wall to the base of the ladder be \(a\) and the distance from the base of the house to the base of the ladder be \(a + 10\).
From similar triangles, \(\frac{15}{a}=\frac{20.5}{a + 10}\).
Cross - multiply: \(15(a + 10)=20.5a\).
Expand: \(15a+150 = 20.5a\).
Subtract \(15a\) from both sides: \(150=20.5a - 15a\).
Simplify: \(150 = 5.5a\).
Solve for \(a\): \(a=\frac{150}{5.5}=\frac{300}{11}\approx27.27\).

Step2: Use the Pythagorean theorem

The length of the ladder \(x\) is the hypotenuse of a right - triangle with legs \(a + 10\) and \(20.5\).
\(a + 10=\frac{300}{11}+ 10=\frac{300 + 110}{11}=\frac{410}{11}\).
By the Pythagorean theorem \(x=\sqrt{(20.5)^{2}+(\frac{410}{11})^{2}}\).
\(20.5=\frac{41}{2}\), so \(x=\sqrt{(\frac{41}{2})^{2}+(\frac{410}{11})^{2}}=\sqrt{\frac{41^{2}}{4}+\frac{410^{2}}{121}}\).
Factor out \(41^{2}\): \(x = 41\sqrt{\frac{1}{4}+\frac{100}{121}}\).
\(\frac{1}{4}+\frac{100}{121}=\frac{121 + 400}{484}=\frac{521}{484}\).
\(x = 41\sqrt{\frac{521}{484}}=\frac{41\sqrt{521}}{22}\approx\frac{41\times22.83}{22}\approx42.62\).

Answer:

\(42.62\)