Question

triangle congruence: sss and hl
for the triangles to be congruent by hl, what must be the value of x?
image of two right triangles, first with legs 9, 12 and hypotenuse 15; second with leg 2x+1, hypotenuse 3x+3, and multiple choice options 2, 3, 4, 7

Explanation:

Step1: Identify congruent sides

Triangles \( \triangle ABC \) and \( \triangle FGH \) are right - angled. For HL (Hypotenuse - Leg) congruence, the hypotenuse and one leg of one right triangle must be equal to the hypotenuse and one leg of the other right triangle. In \( \triangle ABC \), leg \( AB = 9 \), leg \( BC=12 \), hypotenuse \( AC = 15 \). In \( \triangle FGH \), leg \( FG=2x + 1 \), leg \( GH\) (we can assume the corresponding leg to \( AB \) or \( BC \)), and hypotenuse \( FH=3x + 3 \). Let's assume the legs \( AB \) and \( FG \) are corresponding, and hypotenuses \( AC \) and \( FH \) are corresponding. Wait, actually, let's check the ratios. The sides of \( \triangle ABC \) are 9, 12, 15. The ratio of 9:12:15 simplifies to 3:4:5. Let's see the expressions for the sides of \( \triangle FGH \). Let's assume that \( 2x + 1 \) corresponds to 9 and \( 3x+3 \) corresponds to 15? Wait, no, maybe the legs: \( AB = 9 \), \( BC = 12 \); in \( \triangle FGH \), \( FG=2x + 1 \), \( GH \) (let's say \( GH \) corresponds to \( AB \) and \( FG \) corresponds to \( BC \))? Wait, maybe a better approach: since the triangles are congruent by HL, the hypotenuse and a leg of \( \triangle ABC \) must equal the hypotenuse and a leg of \( \triangle FGH \). Let's take the leg \( AB = 9 \) and leg \( FG=2x + 1 \), hypotenuse \( AC = 15 \) and hypotenuse \( FH=3x + 3 \). Wait, but maybe the ratio of the sides. Let's see the sides of \( \triangle ABC \): \( AB = 9 \), \( BC = 12 \), \( AC=15 \). Let's check the expressions: if we set \( 2x + 1 \) equal to 9? No, wait, maybe the leg \( AB = 9 \) and the leg \( FG \) should be equal to \( AB \) if they are corresponding, or \( BC \). Wait, let's try to find \( x \) such that the triangles are congruent. Let's assume that \( FG \) (leg) corresponds to \( AB \) (leg) and \( FH \) (hypotenuse) corresponds to \( AC \) (hypotenuse). So \( 2x+1 = 9 \) and \( 3x + 3=15 \). Let's solve \( 3x+3 = 15 \). Subtract 3 from both sides: \( 3x=15 - 3=12 \), then \( x = 4 \). Let's check \( 2x + 1 \) when \( x = 4 \): \( 2(4)+1=9 \), which is equal to \( AB = 9 \). So that works.

Step2: Solve for \( x \) from hypotenuse equation

We have the hypotenuse of \( \triangle ABC \) is \( AC = 15 \) and the hypotenuse of \( \triangle FGH \) is \( FH=3x + 3 \). For HL congruence, \( 3x+3=15 \).
Subtract 3 from both sides: \( 3x=15 - 3=12 \).
Divide both sides by 3: \( x=\frac{12}{3}=4 \).
We can also check the leg: when \( x = 4 \), \( 2x + 1=2(4)+1 = 9 \), which is equal to the leg \( AB = 9 \) of \( \triangle ABC \). So this satisfies HL congruence.

Answer:

\( x = 4 \)