Question #113556

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.

Expert's answer

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


Recall the Pythagoras theorem c2=a2+b2c^2=a^2+b^2 .


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


3.52=12+y23.5^2=1^2+y^2

12.25=1+y212.25=1+y^2


Subtract 11 from both sides to isolate the variable yy as,


12.251=1+y2112.25-1=1+y^2-1

11.25=y211.25=y^2

y2=454y^2=\frac{45}{4}


Rewrite the fraction 454\frac{45}{4} into perfect square as,


y2=(352)2y^2= (\frac{3\sqrt {5}}{2})^2


Using property of radicals, if x2=a2x^2=a^2 , then x=±a2=±ax=\pm \sqrt {a^2}=\pm a , therefore, the value of yy is,


y=±352y=\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 yy is y=352y=\frac{3\sqrt{5}}{2}


Therefore, the vertical distance that the ladder reaches up the house is y=3523.4y=\frac{3\sqrt{5}}{2} \approx 3.4 m.




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