Question

find each measure.

1. fg

find the length of $overline{ab}$.

1. a
2. tu

4.

Explanation:

Step1: Set up equation for problem 1

Since the two - triangles are congruent (by Hypotenuse - Leg congruence as the hypotenuses are equal and the right - angles are equal), we set \(5x−17 = 3x + 1\).
\[5x−17=3x + 1\]

Step2: Solve for \(x\) in problem 1

Subtract \(3x\) from both sides: \(5x-3x−17=3x - 3x+1\), which simplifies to \(2x−17 = 1\). Then add 17 to both sides: \(2x-17 + 17=1 + 17\), so \(2x=18\). Divide both sides by 2: \(x = 9\).

Step3: Find \(FG\) in problem 1

Substitute \(x = 9\) into \(5x−17\) (or \(3x + 1\)). \(FG=5\times9−17=45−17 = 28\).

Step4: Set up equation for problem 2

Since the two segments are equal (by the properties of the figure, likely congruent triangles), we set \(2x + 24=5x−30\).
\[2x + 24=5x−30\]

Step5: Solve for \(x\) in problem 2

Subtract \(2x\) from both sides: \(2x-2x + 24=5x-2x−30\), which gives \(24 = 3x−30\). Add 30 to both sides: \(24 + 30=3x-30 + 30\), so \(54 = 3x\). Divide both sides by 3: \(x = 18\).

Step6: Find \(TU\) in problem 2

Substitute \(x = 18\) into \(2x + 24\). \(TU=2\times18+24=36 + 24=60\).

Step7: Set up equation for problem 3

Since the two segments are equal (by the properties of the figure, likely congruent triangles), we set \(3x + 8=7x−16\).
\[3x + 8=7x−16\]

Step8: Solve for \(x\) in problem 3

Subtract \(3x\) from both sides: \(3x-3x + 8=7x-3x−16\), which gives \(8 = 4x−16\). Add 16 to both sides: \(8 + 16=4x-16 + 16\), so \(24 = 4x\). Divide both sides by 4: \(x = 6\).

Step9: Find the length of \(\overline{AB}\) in problem 3

We can use either \(3x + 8\) or \(7x−16\). Substitute \(x = 6\) into \(3x + 8\): \(3\times6+8=18 + 8=26\).

Step10: Set up equation for problem 4

Since the two segments are equal (by the properties of the figure, likely the diagonals of a rhombus bisecting each other), we set \(6x + 11=11x−9\).
\[6x + 11=11x−9\]

Step11: Solve for \(x\) in problem 4

Subtract \(6x\) from both sides: \(6x-6x + 11=11x-6x−9\), which gives \(11 = 5x−9\). Add 9 to both sides: \(11 + 9=5x-9 + 9\), so \(20 = 5x\). Divide both sides by 5: \(x = 4\).

Step12: Find the length of \(\overline{AC}\) (assuming we want to find a relevant length)

We can use \(6x + 11\) (or \(11x−9\)). Substitute \(x = 4\) into \(6x + 11\): \(6\times4+11=24 + 11=35\).

Answer:

1. \(FG = 28\)
2. \(TU = 60\)
3. The length of \(\overline{AB}=26\)
4. Assuming relevant length calculation using \(6x + 11\) with \(x = 4\), the length is \(35\)