Question

How many different (non-congruent) triangle can yo draw with two sides 8 and 6 cm and area 12squarecentimeters? what if the area is to be 24 square cantimetres

Answer 1

For the question to be interesting, let’s only consider triangles with integer sides.

Using Heron’s formula

A=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√A=s(s−a)(s−b)(s−c)

where a, b, and c are the sides of the triangle and s is the half perimeter.

Making the replacements a=6,a=6, b=8b=8 and A=12,A=12, we get

12=(7+c2)(1+c2)(−1+c2)(7−c2)−−−−−−−−−−−−−−−−−−−−−−−−−−−√=(49−c24)(c24−1)−−−−−−−−−−−−−−−√12=(7+c2)(1+c2)(−1+c2)(7−c2)=(49−c24)(c24−1).

Squaring, 144=(49−c24)(c24−1)144=(49−c24)(c24−1).

Let x=c24x=c24.

Then we solve the quadratic x2−50x+193=0x2−50x+193=0, getting

x=50±2500−772√2x=50±2500−7722

which is not rational. So there are no integer triangle fitting the bill.

If the area is 24, this amounts to solving 576=(49−c24)(c24−1)576=(49−c24)(c24−1)

Again, letting x=c24,x=c24, we consider the quadratic and x2−50x+625=0x2−50x+625=0. This time

x=25x=25 and c=10.c=10.

So the answers in the two cases are 0 and 1.