Question

1. Draw acute Δ GHI and acute Δ JKL below to where they look congruent.
2. Write a congruent statement for the two triangles. 3. Label ∠D as 6x + 6, ∠E as x 4x + 8, ∠F as 2x + 12 and ∠B as 3y + 6 4. Solve for x and solve for y.

Answer 1

Let's break down your request step by step.

### Step 1: Draw the Triangles

Since I can't physically draw, I will describe how you can draw the two congruent acute triangles \( \triangle GHI \) and \( \triangle JKL \).

- **Step-by-step guide to drawing the triangles**:
1. Draw an acute triangle \( \triangle GHI \). Make sure the angles are less than 90°.
2. The lengths of the sides should be such that they are equal to those of \( \triangle JKL \). You might label the vertices as follows:
- \( G \) at the top (∠G),
- \( H \) at the left bottom (∠H),
- \( I \) at the right bottom (∠I).
3. Draw the second triangle \( \triangle JKL \), making sure it has the same side lengths and angles. Label it:
- \( J \) at the top (∠J),
- \( K \) at the left bottom (∠K),
- \( L \) at the right bottom (∠L).
4. Ensure that both triangles appear congruent.

### Step 2: Write a Congruent Statement

Once you have drawn both triangles, you can write the congruent statement for them:

\[
\triangle GHI \cong \triangle JKL
\]

### Step 3: Label the Angles

You mentioned to label angles as follows:
- \( \angle D = 6x + 6 \)
- \( \angle E = 4x + 8 \)
- \( \angle F = 2x + 12 \)
- \( \angle B = 3y + 6 \)

Since angles D, E, and F should correspond to angles J, K, and L in \( \triangle JKL \), assign them correspondingly:
- \( \angle G = 6x + 6 \) (corresponds to \( \angle J \))
- \( \angle H = 4x + 8 \) (corresponds to \( \angle K \))
- \( \angle I = 2x + 12 \) (corresponds to \( \angle L \))

### Step 4: Solve for \( x \) and \( y \)

The sum of angles in a triangle is \( 180^\circ \), thus:

For \( \triangle GHI \):
\[
(6x + 6) + (4x + 8) + (2x + 12) = 180
\]

Combine like terms:
\[
6x + 4x + 2x + 6 + 8 + 12 = 180
\]
\[
12x + 26 = 180
\]
Subtract 26 from both sides:
\[
12x = 154
\]
Divide by 12:
\[
x = \frac{154}{12} = 12.8333 \text{ (or } 12 \frac{10}{12} \text{ or } 12 \frac{5}{6} \text{ when simplified)}
\]

Next, to solve for \( y \), we need \( \angle B \) to be one of the other angles from triangle \( JKL \) (assuming there are angles related to \( y \), you can adjust your angles accordingly). However, since we only have one \( y \) angle and it wasn't explicitly related:

For now, let \( 3y + 6 \) also equal some angle in our triangle:
Using \( (3y + 6) = (4x + 8) \)

Replace \( x \):
\[
3y + 6 = 4 \times 12.8333 + 8
\]
Calculate \( 4 \times 12.8333 + 8 = 51.3333 + 8 = 59.3333 \):
\[
3y + 6 = 59.3333
\]
Subtract 6:
\[
3y = 53.3333
\]
Divide by 3:
\[
y = \frac{53.3333}{3} = 17.7777 \text{ (or } 17 \frac{7}{9} \text{ when simplified)}
\]

### Conclusion

Values of \( x \) and \( y \):
- \( x \approx 12.83 \)
- \( y \approx 17.78 \)

These solutions depend on the relationships you've set up between the angles. Please clarify if there are specific conditions or relationships for angle \( y \).