Question

triangle congruence: sss and hl
using the hl theorem
in the diagram, \\( \angle a \\) and \\( \angle d \\) are right angles.
for the triangles to be congruent by hl, what must be the value of \\( x \\)?
options: 17, 31, 8, 9

Explanation:

Step1: Identify Congruent Parts

Since the triangles are congruent by HL (Hypotenuse - Leg) Theorem, the corresponding legs should be equal. Here, \( AB = 34 \) (hypotenuse of first triangle) and \( ED = 4x + 2 \) (leg of second triangle)? Wait, no, wait. Wait, the right - angled triangles: in \( \triangle ABC \) (right - angled at \( A \)) and \( \triangle EDF \) (right - angled at \( D \)). The hypotenuses are equal (marked with \( x \) on the hypotenuse, so \( BC = EF \)), and one leg should be equal. Wait, the leg \( AC \) and \( DF \)? Wait, no, the given leg in the first triangle: wait, the length of \( AB \) is 34? Wait, no, the first triangle has a leg (the vertical or horizontal) and hypotenuse. Wait, the second triangle has a leg \( ED = 4x + 2 \). Wait, maybe the legs that are equal are \( AB \) (wait, no, \( A \) is right angle, so \( AC \) and \( AB \) are legs? Wait, no, in right - triangle \( \triangle BAC \), right - angled at \( A \), so legs are \( AB \) and \( AC \), hypotenuse \( BC \). In \( \triangle EDF \), right - angled at \( D \), legs are \( ED \) and \( DF \), hypotenuse \( EF \). Since they are congruent by HL, hypotenuse \( BC = EF \) and one leg \( AB = ED \) (assuming \( AB \) and \( ED \) are the corresponding legs). Wait, \( AB = 34 \)? No, wait the first triangle: the side \( AB \) is 34? Wait, no, the first triangle has a side labeled 34, and the second triangle has a leg \( ED = 4x + 2 \). Wait, maybe \( AB = ED \), so \( 4x+2 = 34 \)? Wait, no, that would be if \( AB \) and \( ED \) are corresponding legs. Wait, let's re - examine.

Wait, the HL Theorem states that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent. So, let's assume that the hypotenuses are equal (marked with the same symbol) and the legs \( AB \) (length 34? No, wait the first triangle: the side adjacent to the right angle \( A \) and the hypotenuse. Wait, maybe the leg of the first triangle is 34? No, the problem is to find \( x \) such that the triangles are congruent by HL. Wait, the options are 17, 31, 8, 9. Wait, maybe the leg of the first triangle is 34, and the leg of the second triangle is \( 4x + 2 \), and they are equal. Wait, no, let's solve \( 4x+2=34 \):

Step2: Solve for \( x \)

If \( 4x + 2=34 \), subtract 2 from both sides: \( 4x=34 - 2=32 \). Then divide by 4: \( x = \frac{32}{4}=8 \). Wait, but let's check. If \( x = 8 \), then \( 4x + 2=4\times8+2 = 32 + 2=34 \), which matches the length of the corresponding leg in the first triangle. So that makes sense.

Answer:

\( x = 8 \)