areageometryrectanglestriangles
A rectangle with a height x is drawn with its base lying on the base of the triangle. The triangle has an altitude with height h and the length of its base is b. How can I calculate the area of the enclosed rectangle in terms of these three variables?
Let $d(Q,DC)$ the distance between point $Q$ and the line defined by $DC$, and $d(R,DC)$ the distance between point $R$ and the line defined by $DC$.
Note that: $\triangle DQS \sim \triangle BQP$ and $\triangle BRP \sim \triangle DRT$, hence: $$\frac{d(Q,DC)}{d(Q,AB)}= \frac{1x}{\frac{3}{2}x}= \frac{2}{3} \quad(1)$$ and $$\frac{d(R,DC)}{d(R,AB)}= \frac{2x}{\frac{3}{2}x}= \frac{4}{3}. \quad(2)$$ From $(1)$ and $(2)$ we get: $$d(Q,AB)=\frac{3}{5}y \quad(3)$$ and $$d(R,AB)=\frac{3}{7}y, \quad(4)$$ which are the heights of triangles $\triangle BQP$ and $\triangle BRP$.
We can get the area of $\triangle PQR$ from: $$A_{\triangle PQR}= A_{\triangle PQB}- A_{\triangle PRB}$$ Therefore $$A_{\triangle PQR}= \frac{9}{70}xy=3.$$
Let BC = 1 unit. Then, $4 = y = [\triangle CBQ] = \frac {1.QC}{2}$ yields QC = 8 units
Let DQ = s units. Then AB = s + 8 units
$5 = z = [\triangle ABP] = \frac {(s+8).AP}{2}$ yields $AP = \frac {10}{s + 8}$ units
$DP = 1 – AP = … = \frac {s – 2}{s + 8}$
Similarly, $3 = (\frac {1}{2}) {s}{\frac{ s – 2}{s + 8}}$.
The above result is a solvable quadratic with roots $s = 12$ or $–4$ (rejected)
Therefore, the rectangle occupies 20 square units and x, y, z occupy …. Giving f = ...
Best Answer
$\bigtriangleup{AXC} \sim \bigtriangleup{EFC}$ [by $AA$ corollary]
Thus $\frac{AX}{XC}=\frac{EF}{FC} \Rightarrow \frac{h}{XC} =\frac{x}{FC} \Rightarrow \frac{XC}{h} =\frac{FC}{x} $
Similarly for $\bigtriangleup{AXB} \sim \bigtriangleup{DGB}$:-
$\frac{AX}{XB} = \frac{DG}{GB} \Rightarrow \frac{h}{XB}=\frac{x}{GB} \Rightarrow \frac{XB}{h}=\frac{GB}{x}$
Now add the above two equations:- $$\frac{XB+XC}{h}=\frac{FC+GB}{x}$$ $$\Rightarrow \frac{b}{h} = \frac{b-GF}{x}$$ $$\Rightarrow b-GF = \frac{bx}{h}$$ $$\Rightarrow GF = b-\frac{bx}{h}$$ So the area of the rectangle is $x[b-\frac{bx}{h}]\Rightarrow xb[1-\frac{x}{h}]$