Question

How many different (non-congruent) triangles can you draw with two sides 8 and 6 centimetres and area 12 square centimetres? What if the area is to be 24 square centimetres?

Answer 1

For the triangle of area of $12c{m}^2$, the height from the $8cm$ side should be $3cm$.
Then, Area = $\frac{1}{2}\times 8\times 3$ = $12c{m}^2$

Draw a line $8cm$ long. Draw another line parallel to it at a distance of $3cm$. Draw a circle of radius $6cm$ with one end of the first line as the center. The points where this circle cuts the second line are the third vertex of the triangle.
We can draw two triangles ∆OBC and ∆OCA having area $12c{m}^2$.

For the triangle of area $24c{m}^2$, the height from the $8cm$ side should be $6cm$.

We can draw only one triangle of this type.