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?

the greatest possible value for x  is 6

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

Hope this helps :)

What are the options