the longest side of an obtuse triangle measures 20 cm. the two shorter sides meausre x cm and 3x cm. rounded to the nearest tenth , what is the greatest posible of x?
Explanation: Assume that the sides of the triangle are: a = x b = 3x c = 20 We are given that c is the longest side.
For the triangle to be obtuse: c^2 > a^2 + b^2
Substitute with the values of a, b and c and solve for x as follows: c^2 > a^2 + b^2 (20)^2 > (x)^2 + (3x)^2 400 > x^2 + 9x^2 400 > 10 x^2 40 > x^2
Now, we will get the zeros of the function. This means that we will solve for x^2 = 40 as follows: x^2 = 40 x = + or - √40 either x = 6.32 Or x = -6.32
Since we want x^2 < 40, therefore, the greatest possible value for x to satisfy this condition is 6
