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?

diagram here

  • 1
    $\begingroup$ Try to show your attempt. $\endgroup$
    – user264745
    Commented Oct 10, 2022 at 15:23
  • $\begingroup$ If you haven't already, drawing a picture often helps. $\endgroup$ Commented Oct 10, 2022 at 15:24
  • $\begingroup$ @user264745 I really have no idea how to begin attempting the question. $\endgroup$
    – zxayn
    Commented Oct 10, 2022 at 15:35
  • 1
    $\begingroup$ So you need to find $DE$. Since $DE||BC$, then you have similar triangles. $\endgroup$
    – Andrei
    Commented Oct 10, 2022 at 15:45

1 Answer 1


$\bigtriangleup{AXC} \sim \bigtriangleup{EFC}$ [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 $x[b-\frac{bx}{h}]\Rightarrow xb[1-\frac{x}{h}]$


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