Right triangle
AKA very cool use of quadratic equation
Task.
Construct from an index card a right triangle that has area 9 square centimeters, and perimeter 15 centimeters.
General approach.
Let A be the area of a right triangle and let B be its perimeter. We want to find the legs x and y of the triangle, and its hypotenuse z.
Equations:
A = x*y/2
B = x + y + z
z2 = x2 + y2
Because z = B - x - y, we have (B - x - y)2 = x2 + y2.
And
(B - x - y)2 = B2 -
2*B*x - 2*B*y + x2 + 2*x*y +
y2.
So,
B2 - 2*B*(x + y) + 2*x*y = 0.
Substituting x*y = 2*A, we have
B2 - 2*B*(x + y) + 4*A = 0,
and solving for x + y, we have
x +
y = B/2 + 2*A/B.
So
-x - y = -B/2 - 2*A/B.
And since
z = B - x - y,
we have
z = B/2 - 2*A/B.
To find x and y we need to solve the following equation for t:
t2 - (B/2 + 2*A/B)*t + 2*A = 0,
because x + y = B/2 + 2*A/B, and x*y = 2*A.
(So by solving the equation above, we are actually solving the equation
t2 - (x+y)*t + 2*x*y = 0 = (t-x)*(t-y), and its solutions are t = x and t = y. Do you see it?)
You can solve this equation however you want: by hand, with the TI-34, or with the TI-83/84. For the values A = 9 cm2 and B = 15 cm, we have
x = 5.3104686 and y = 3.3895314. Here is the triangle:
Remark.
The required triangle can be drawn without solving the quadratic equation, and thus without getting numerical values, if we draw the line x + y = B/2 + 2*A/B, in the coordinate system x, y, and intersect it with a circle of radius z = B/2 - 2*A/B with center at (0, 0). (z is the length of the hypotenuse of the desired triangle.) The point of intersection (x, y), together with (0, 0) and (x, 0), form the three vertices of the required triangle. See the picture below:
