Hydrostatic pressure on a triangle

Question

I am attempting to follow Paul's calculus notes, but am having trouble, in particular at this page: http://tutorial.math.lamar.edu/Classes/CalcII/HydrostaticPressure.aspx

I get to the part with the "The height of this strip is $\Delta x$ and the width is $2a$. We can use similar triangles to determine a as follows,"

and I am completely lost, I do not follow at all what is happening. Is there a mistake? The calculation seems impossible. Is he writing things backwards? $\frac{3}{4}$ is a constant how is it equal to that weird statement? I don't get it, is that solving for $a$?

Answer 1

$\hskip 1in$

In the above, the base and height of the larger right triangle are $A$ and $B$ respectively, and the base and height of the inside, smaller triangle is $a$ and $b$ respectively. The two triangles are similar, which is a term from geometry meaning the ratios between the sides of one triangle are equal to the ratios between the sides of the other triangle. That is, in the above, we have that

$$\frac{a}{b}=\frac{A}{B}.$$

(Equivalently, the inner triangle is the bigger one scaled down in size.) Hence we have:

$\hskip 1in$ $\huge\displaystyle \frac{3}{4}=\frac{a}{4-x_i^*}$

Keep in mind we've approximated the triangle by covering it with thin rectangular strips, and intend to calculate the hydrostatic force associated to each strip and then add them up. As we make the strips smaller and smaller our calculation will be a Riemann sum for an integral, and hence this integral will be our desired answer. In order to calculate the force for a strip, we first denote its dimensions as $\Delta x\times 2a$. Note that $a$ is not really a constant: it varies with choice $x_i^*$. In order to do the Riemann sum though we need to write the hydrostatic force (of each strip) purely in terms of $x_i^*$, which means we need to solve for $a$ in terms of $x_i^*$, and similarity is what allows us to do that.

Answer 2

The big half-triangle has width $3$ and height $4$ while the similar small half-triangle has width $a$ and height $4-x_i^{^.}$

Since you have similar triangles, you can then say $\frac{3}{4}=\frac{a}{4-x_i^{^.}}$ and so $a=3 -\frac{3}{4}x_i^{^.}$

Answer 3

Note that the height of the triangle is $4$. Because the small triangle is similar (i.e. same angles) to the big one, we know the ratios of side lengths are the same.

On the big triangle, one side is $3$ and the adjacent is $4$. On the smaller one, the similar side has the length of $a$ and its adjacent is $4-x_i^*$. Thus $3$ is to $4$ as $a$ is to $4-x_i^*$. Thus

$$\frac{3}{4}=\frac{a}{4-x_i^*}$$