Problem: A rectangle inscribed in a triangle has its base coinciding with the base (bbb) of the triangle. If the altitude of the triangle is hhh, and the altitude (xxx) of the rectangle is half the base of the rectangle, then:
Answer Choices:
A. x=12hx=\dfrac{1}{2} hx=21h B. x=bhh+bx=\dfrac{b h}{h+b}x=h+bbh C. x=bh2h+bx=\dfrac{b h}{2 h+b}x=2h+bbh D. x=hb2x=\sqrt{\dfrac{h b}{2}}x=2hb E. x=12bx=\dfrac{1}{2} bx=21b
Solution:
By similar triangles,
h−xh=2xbb(h−x)=h(2x)bh−bx=2hxbh=2hx+bxbh=x(2h+b)bh2h+b=x\begin{aligned} \frac{h-x}{h} &= \frac{2x}{b} \\ b(h-x) &= h(2x) \\ bh - bx &= 2hx \\ bh &= 2hx + bx \\ bh &= x(2h+b) \\ \frac{bh}{2h+b} &= x \end{aligned} hh−xb(h−x)bh−bxbhbh2h+bbh=b2x=h(2x)=2hx=2hx+bx=x(2h+b)=x
∴x=bh2h+b\therefore \boxed{x = \frac{bh}{2h+b}} ∴x=2h+bbh