I want to find those tri-angle whose area is 100 square cm and having following features :
- Any height drawn from any top point of the tri-angle is not more than 1 cm ?
My trying :
I know that the area of tri-angle is $ \frac{1}{2}*base*height$ So if the height is 1 cm then all the base should be 200 cm .
THen how should I go ahead ?
Take any triangle $\triangle ABC$ of area $100 \,\text{cm}^2$ with $B,C$ on a fixed line $\mathbf{L}\,$. Flatten the triangle by pushing vertex $A$ down towards $\mathbf{L}$ while sliding $B,C$ outwards along the line as to keep the area constant. In the limit, the distances from any point inside the triangle to either of the three sides tend to $0\,$, so all heights will eventually become $\le 1 \,\text{cm}$ for $\,A\,$ sufficiently close to $\,\mathbf{L}\,$.
[ EDIT ] For a concrete example of such a triangle, consider a triangle $\triangle ABC$ isosceles at $A$ with base $400 \,\text{cm}$ and height $\frac{1}{2} \,\text{cm}\,$, so that thea area is $\,\frac{1}{2} \cdot \frac{1}{2} \cdot 400 = 100 \,\text{cm}^2\,$. Then the equal sides will be $\sqrt{200^2+\left(\frac{1}{2}\right)^2} \,\text{cm}\,$ each, and the corresponding heights are $\,\frac{2 \,\cdot\, 100}{\sqrt{200^2+\left(\frac{1}{2}\right)^2}} \lt 1 \, \text{cm}\,$.