The sum of the two shorter sides of a right angled triangle is 18cm. Calculate. A.) The least possible length of the hypotenuse. B.) The greatest possible area of the triangle

Answer 1

sum of sides of a right triangle = 18 cm

the hypotenuse is anywhere from maximum 18 cm to minimum 9sqr2 = about 9(1.414 = about 12.7

minimum hypotenuse is when the other 2 sides are equal or 18/2 = 9 cm. then hypotenuse = 9 x sqr2 cm

9sqr2 < hypotenuse < 18 cm

12.7 cm < hypotenuse < 18 cm

maximum area = 9x9/2 = 81/2 = 40.5 cm^2

Answer 2

Let x: length of 1st leg ; (18 - x) is length of 2nd leg ; h(x): length of hypotenuse ; A(x): area of triangle

h(x) = √[x² + (18 - x)²] = √(2x² - 36x + 324)

h^'(x) = (4x - 36) / 2√(2x² - 36x + 324) = 0

x = 9 will minimize the length of the hypotenuse (a right, isosceles triangle with legs = 9, hyp = 9√2)

A(x) = 1/2·x·(18 - x) = - 1/2x² + 9x

A^'(x) = -x + 9 = 0

x = 9

The same dimensions that minimize the length of the hypotenuse maximize the enclosed area.