How to calculate the perpendicular on the diagonal of a rectangle with sides $2$ and $\sqrt2$ ?
I've already calculated the diagonal by using Pythagorean theorem which is $\sqrt 6$. Then I didn't know which technique I need to use to calculate the perpendicular that goes from the right angle to the diagonal.
Can someone give me a hint?
On
Not the most elegant approach, but here is a way: You can place the rectangle in the first quadrant of the xy plane, with one of the vertices in the Origin. Choose the horizontal segment to be 2. The diagonal has a slope of $\frac{1}{\sqrt{2}}$ The perpendicular slope to the diagonal is $-\sqrt{2}$ The perpendicular has to pass through $(0,\sqrt{2})$ and so the equation of the perpendicular is $y=-\sqrt{2}x+\sqrt{2} $ The equation of the diagonal is $y=\frac{x}{\sqrt{2}}$ Finding the point of intersection can be found by equating the lines, to find $(\frac{2}{3},\frac{2}{3\sqrt{2}})$ You got two points marking the segment of that diagonal. Now can you use the distance formula here? Note, it can be done quicker, like David's approach...
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By analytical geometry, with natural coordinate axes, taking $A(2,0)$ and $B(0,\sqrt{2})$ you are looking for the distance of point $B$ to straight line with equation $y=\frac{x}{\sqrt{2}}$, i.e., $\frac{x}{\sqrt{2}}-y=0$, which is known to be $\frac{|\frac{0}{\sqrt{2}}-\sqrt{2}|}{\frac{1}{2}+1}=\frac{2\sqrt{2}}{3}$ using the classical formula that can be found there.
Draw a picture. Label the perp, $x$. The short distance for along the diagonal from corner to the point where the perp interseccts the diagonal as $y$, the rest of the diagonal as $z$ so that $y + z = \sqrt{6}$.
By similar triangles you have $x/\sqrt{2} = y/2 = \sqrt{2}/\sqrt{6}$. And $y/\sqrt{2} = z/2 = 2/\sqrt{6}$ Solve for $x$.
By pythagorean theorem you have $x^2 + y^2 = \sqrt{2}^2$; $y^2 + z^2 = 2^2$; (and $2^2 + \sqrt{2}^2 = \sqrt{6}^2$-- but you already did that...). Solve for $x$.
You have way more than enough to finish.