Question 16An altitude $h$ of a triangle is increased by a length $m.$ How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle?$\dfrac{bm}{h+m}$$\dfrac{bh}{2h+2m}$$\dfrac{b(2m+h)}{m+h}$$\dfrac{b(m+h)}{2m+h}$$\dfrac{b(2m+h)}{2(h+m)}$