Question

hw 7: relationships in triangles
if the sides of the triangle have the following lengths, find the range of the possible values for x.
pq = 7x + 13, qr = 10x - 2, pr = x + 27
< x <

Explanation:

To find the range of \( x \) for the triangle with sides \( PQ = 7x + 13 \), \( QR = 10x - 2 \), and \( PR = x + 27 \), we use the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. We need to consider all three inequalities:

Step 1: \( PQ + QR > PR \)

Substitute the expressions for the sides:

$$ (7x + 13) + (10x - 2) > x + 27 $$

Simplify the left side:

$$ 17x + 11 > x + 27 $$

Subtract \( x \) from both sides:

$$ 16x + 11 > 27 $$

Subtract 11 from both sides:

$$ 16x > 16 $$

Divide both sides by 16:

$$ x > 1 $$

Step 2: \( PQ + PR > QR \)

Substitute the expressions for the sides:

$$ (7x + 13) + (x + 27) > 10x - 2 $$

Simplify the left side:

$$ 8x + 40 > 10x - 2 $$

Subtract \( 8x \) from both sides:

$$ 40 > 2x - 2 $$

Add 2 to both sides:

$$ 42 > 2x $$

Divide both sides by 2:

$$ 21 > x \quad \text{or} \quad x < 21 $$

Step 3: \( QR + PR > PQ \)

Substitute the expressions for the sides:

$$ (10x - 2) + (x + 27) > 7x + 13 $$

Simplify the left side:

$$ 11x + 25 > 7x + 13 $$

Subtract \( 7x \) from both sides:

$$ 4x + 25 > 13 $$

Subtract 25 from both sides:

$$ 4x > -12 $$

Divide both sides by 4:

$$ x > -3 $$

Step 4: Combine the Inequalities

From Step 1, \( x > 1 \); from Step 2, \( x < 21 \); from Step 3, \( x > -3 \). The most restrictive lower bound is \( x > 1 \), and the upper bound is \( x < 21 \). Thus, the range of \( x \) is \( 1 < x < 21 \).

Answer:

\( 1 < x < 21 \)