# Finding area of rectangle with height 'x' enclosed within triangle of height 'h' and base 'b'

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?

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

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