Answer to Question #113556 in Trigonometry for Angelina Squires

Question

Louise is painting around a window frame on the outside of her house. Her ladder is 3.5m long and she placed it 1 m from the house. What is the vertical distance that the ladder reaches up the house? Round to one decimal place. Use the Pythagorean theorem.

Answer 1

Consider the ladder leans against the wall of the house as shown in the figure below:

The figure above shown is a right angled triangle.

Here, $a=1,b=y,c=3.5$

Recall the Pythagoras theorem $c^2=a^2+b^2$ .

Plug $a=1,b=y,c=3.5$ into above formula as,

$3.5^2=1^2+y^2$

$12.25=1+y^2$

Subtract $1$ from both sides to isolate the variable $y$ as,

$12.25-1=1+y^2-1$

$11.25=y^2$

$y^2=\frac{45}{4}$

Rewrite the fraction $\frac{45}{4}$ into perfect square as,

$y^2= (\frac{3\sqrt {5}}{2})^2$

Using property of radicals, if $x^2=a^2$ , then $x=\pm \sqrt {a^2}=\pm a$ , therefore, the value of $y$ is,

$y=\pm \frac{3\sqrt{5}}{2}$

Since, negative value has no significant as it is a distance, so negative value neglected.

So, the value of $y$ is $y=\frac{3\sqrt{5}}{2}$

Therefore, the vertical distance that the ladder reaches up the house is $y=\frac{3\sqrt{5}}{2} \approx 3.4$ m.