Question

an acute triangle has side lengths 21 cm, x cm, and 2x cm. if 21 is one of the shorter sides of the triangle, what is the greatest possible length of the longest side, rounded to the nearest tenth?
18.8 cm
24.2 cm
42.0 cm
72.7 cm

Explanation:

Step1: Define side order constraints

Since 21 is a shorter side, we have two cases:
Case 1: $21 \leq x < 2x$
Case 2: $x \leq 21 < 2x$
We need the longest side ($2x$) to be as large as possible, so we focus on the condition where the triangle is acute. For an acute triangle, the square of the longest side must be less than the sum of the squares of the other two sides.

Step2: Apply acute triangle rule

When $2x$ is the longest side, the condition is:
$$(2x)^2 < 21^2 + x^2$$
Expand and simplify:
$$4x^2 < 441 + x^2$$
$$3x^2 < 441$$
$$x^2 < 147$$
$$x < \sqrt{147} \approx 12.12$$
But this contradicts $21 \leq x$ in Case 1. Now consider Case 2 where 21 is a shorter side and $2x$ is still the longest side: the condition becomes
$$(2x)^2 < 21^2 + x^2$$
Wait, no—if $x \leq 21 < 2x$, the longest side is still $2x$, so the same inequality applies, but we also have the triangle inequality: $x + 21 > 2x \implies x < 21$.
Now, correct acute condition for longest side $2x$:
$$(2x)^2 < 21^2 + x^2$$
$$4x^2 - x^2 < 441$$
$$3x^2 < 441$$
$$x^2 < 147$$
$$x < \sqrt{147} \approx 12.12$$
This can't be, because if $x < 12.12$, $2x < 24.24$, but $21 > x$, so 21 would be a longer side. Wait, if 21 is one of the shorter sides, the longest side is $2x$, so $21 \leq x < 2x$ is invalid because $x \geq21$ makes $2x \geq42$, then 21 is a shorter side. Now apply acute condition with longest side $2x$, so:
$$(2x)^2 < 21^2 + x^2$$
$$4x^2 < 441 + x^2$$
$$3x^2 < 441$$
$$x^2 < 147$$
$$x < 12.12$$
This contradicts $x \geq21$. So if 21 is a shorter side, the longest side could be 21? No, wait, I mixed up. If $x$ is the shortest side, $x \leq21 <2x$, so $2x$ is longest. Then acute condition is:
$$21^2 + x^2 > (2x)^2$$
$$441 + x^2 > 4x^2$$
$$441 > 3x^2$$
$$x^2 < 147$$
$$x < 12.12$$
But $2x >21 \implies x>10.5$. So $10.5 If $21$ is the middle side, $x \leq21 \leq2x$, so $x \geq10.5$, and longest side is $2x$. Acute condition: $(2x)^2 <21^2 +x^2$, same as above, $2x <24.24$.
If $21$ is the shortest side, $21 \leqx <2x$, longest side is $2x$. Acute condition: $(2x)^2 <21^2 +x^2$, which gives $x <12.12$, contradicting $x \geq21$. So this case is impossible.

Step3: Find maximum $2x$

The maximum $2x$ occurs when $x$ is as large as possible under $x < \sqrt{147} \approx12.12$, so $2x \approx24.24$, rounded to nearest tenth is 24.2.

Answer:

B. 24.2 cm