Question

the longest side of an obtuse triangle measures 20 cm. the two shorter sides measure x cm and 3x cm. rounded to the nearest tenth, what is the greatest possible value of x?
6.3
6.4
7.0
7.1

Explanation:

Step1: Apply obtuse triangle rule

For an obtuse triangle with longest side $c$, $c^2 > a^2 + b^2$. Here $c=20$, $a=x$, $b=3x$:
$$20^2 > x^2 + (3x)^2$$

Step2: Simplify the inequality

Calculate squares and combine terms:
$$400 > x^2 + 9x^2$$
$$400 > 10x^2$$

Step3: Solve for $x^2$

Isolate $x^2$ by dividing:
$$x^2 < \frac{400}{10}$$
$$x^2 < 40$$

Step4: Solve for $x$

Take square root (x>0):
$$x < \sqrt{40} \approx 6.3246$$

Step5: Apply triangle inequality

Sum of shorter sides > longest side:
$$x + 3x > 20$$
$$4x > 20$$
$$x > 5$$

Step6: Round to nearest tenth

The largest $x$ less than ~6.3246, rounded to nearest tenth is 6.3.

Answer:

6.3