+895 John's paint crew knows from experience that its 18-ft ladder is particulary stable when the distance from the ground to the top of the ladder is 5 ft more than the distance from the building to the base of the ladder as shown in the figure. In this position, how far up the building does the ladder reach?
+9466 The wall and the ground form a right angle...so we can use the Pythagorean theorem to find x .
x2 + (x + 5)2 = 182
Multiply out (x + 5)2 .
x2 + (x + 5)(x + 5) = 324
x2 + x2 + 10x + 25 = 324
Combine x2 and x2 , and subtract 324 from both sides.
2x2 + 10x - 299 = 0
Now we can use the quadratic formula to solve for x .
\(x = {-10 \pm \sqrt{10^2-4(2)(-299)} \over 2(2)} \\~\\ x={-10 \pm \sqrt{100+2392} \over 4} \\~\\ x={-10 \pm \sqrt{2492} \over 4} \\~\\ x={-10 \pm 2\sqrt{623} \over 4} \\~\\ x={-5 + \sqrt{623} \over 2}\,\approx\,9.98 \qquad \text{ or }\qquad x={-5 - \sqrt{623} \over 2}\, \approx\, -14.98\)
Since -14.98 causes a negative length for the side on the ground..... x must be ≈ 9.98
So, the length of the side on the wall = \({-5 + \sqrt{623} \over 2}+5\,\approx\,14.98\) feet
+9466 The wall and the ground form a right angle...so we can use the Pythagorean theorem to find x .
x2 + (x + 5)2 = 182
Multiply out (x + 5)2 .
x2 + (x + 5)(x + 5) = 324
x2 + x2 + 10x + 25 = 324
Combine x2 and x2 , and subtract 324 from both sides.
2x2 + 10x - 299 = 0
Now we can use the quadratic formula to solve for x .
\(x = {-10 \pm \sqrt{10^2-4(2)(-299)} \over 2(2)} \\~\\ x={-10 \pm \sqrt{100+2392} \over 4} \\~\\ x={-10 \pm \sqrt{2492} \over 4} \\~\\ x={-10 \pm 2\sqrt{623} \over 4} \\~\\ x={-5 + \sqrt{623} \over 2}\,\approx\,9.98 \qquad \text{ or }\qquad x={-5 - \sqrt{623} \over 2}\, \approx\, -14.98\)
Since -14.98 causes a negative length for the side on the ground..... x must be ≈ 9.98
So, the length of the side on the wall = \({-5 + \sqrt{623} \over 2}+5\,\approx\,14.98\) feet
