Question

for the triangles to be congruent by hl, what must be the value of x? 2 3 4 7

Explanation:

Step1: Identify HL Congruence Requirements

For right triangles to be congruent by HL (Hypotenuse - Leg), the hypotenuses and one pair of corresponding legs must be equal. In $\triangle ABC$ and $\triangle FGH$ (right triangles at $B$ and $G$ respectively):
- Hypotenuse of $\triangle ABC$ is $AC = 15$, so hypotenuse of $\triangle FGH$ (which is $FH=3x + 3$) must equal $15$.
- Leg of $\triangle ABC$ is $BC = 12$, so leg of $\triangle FGH$ (which is $FG = 2x+1$) must equal $12$. Let's use the leg equality first (either can be used, but let's check the leg $FG = 12$).

Step2: Solve for \(x\) using Leg Equality

Set $2x + 1=12$? Wait, no, wait $\triangle ABC$ has leg $BC = 12$, and $\triangle FGH$ has leg $FG=2x + 1$. Wait, also $\triangle ABC$ has leg $AB = 9$, and $\triangle FGH$ has leg $HG$ (not given, but $FH$ is hypotenuse). Wait, maybe better to use hypotenuse and the other leg. Wait, $\triangle ABC$: legs $AB = 9$, $BC = 12$; hypotenuse $AC = 15$. $\triangle FGH$: let's assume $HG = AB = 9$ (corresponding legs), so $FH$ (hypotenuse) should be $15$, and $FG$ (leg) should be $12$. So set $FG=12$: $2x + 1=12$? No, wait $2x + 1$ should equal $BC = 12$? Wait, $BC = 12$, $FG = 2x + 1$. So $2x+1 = 12$? No, wait $2x + 1=12$ gives $2x=11$, $x = 5.5$ which is not an option. Wait, maybe $FH$ (hypotenuse) is $15$, so $3x + 3=15$. Let's solve that:
$3x+3 = 15$
Subtract 3 from both sides: $3x=15 - 3=12$
Divide by 3: $x = \frac{12}{3}=4$? Wait, no, wait $3x + 3 = 15$ gives $x = 4$? But wait, let's check the leg. If $x = 3$, then $3x + 3=12$, and $2x + 1=7$? No, wait maybe I mixed up. Wait $\triangle ABC$: right angle at $B$, so legs $AB = 9$, $BC = 12$; hypotenuse $AC = 15$. $\triangle FGH$: right angle at $G$, so legs $FG$ and $HG$, hypotenuse $FH$. For HL, we need one leg and hypotenuse equal. So either:
- Case 1: $HG = AB = 9$ (leg), and $FH = AC = 15$ (hypotenuse), and $FG = BC = 12$ (leg). So set $FG = 12$: $2x + 1=12$ → $2x=11$ → $x = 5.5$ (not option).
- Case 2: $HG = BC = 12$ (leg), and $FH = AC = 15$ (hypotenuse), and $FG = AB = 9$ (leg). So set $FG = 9$: $2x + 1=9$ → $2x=8$ → $x = 4$? No, wait $3x + 3=15$ (hypotenuse) → $3x=12$ → $x = 4$. Wait, but let's check the options. Wait the options are 2,3,4,7. Wait if $x = 3$: $3x + 3=12$, $2x + 1=7$. No. If $x = 4$: $3x + 3=15$, $2x + 1=9$. Ah! There we go. So $\triangle ABC$ has leg $AB = 9$, leg $BC = 12$, hypotenuse $15$. $\triangle FGH$: leg $HG = AB = 9$ (so $HG = 9$), leg $FG = 9$? No, wait $FG = 2x + 1=9$ when $x = 4$, and $FH = 3x + 3=15$. So then $\triangle FGH$ has leg $FG = 9$, leg $HG = 9$? No, wait no, $\triangle ABC$ has legs 9 and 12, hypotenuse 15. $\triangle FGH$: if $FH = 15$ (hypotenuse), and $FG = 9$ (leg), then $HG$ should be 12 (by Pythagoras: $15^2=9^2 + 12^2$ → $225 = 81 + 144=225$). So yes, so $FG = 9$ (so $2x + 1=9$ → $2x=8$ → $x = 4$? Wait no, $2x + 1=9$ → $x = 4$? Wait $2x + 1=9$ → $x = 4$? Wait $2*4 + 1=9$, yes. And $3x + 3=15$ → $3*4 + 3=15$, yes. So that works. Wait but earlier I thought $FG$ was 12, but no, $FG$ is the other leg. So the correct value is $x = 4$? Wait no, wait the options have 3 as an option? Wait maybe I made a mistake. Wait let's check the hypotenuse: $3x + 3$ should equal $AC = 15$. So $3x + 3=15$ → $3x=12$ → $x = 4$. Yes, that's correct. So the answer is 4. Wait but let's check again. If $x = 3$: $3x + 3=12$, $2x + 1=7$. Then $\triangle FGH$ would have hypotenuse 12, leg 7, so the other leg would be $\sqrt{12^2 - 7^2}=\sqrt{144 - 49}=\sqrt{95}\approx9.7$, not 9. If $x = 4$: hypotenuse 15, leg 9, other leg $\sqrt{15^2 - 9^2}=\sqrt{225 - 81}=\sqrt{144}=12$, which matches $BC = 12$. So yes, $x = 4$. Wait but the options include 4, so that's the answer.

Answer:

3