Question

5.) find the missing length indicated. find ki.
6.) find the missing length indicated. find ln.
7.) find the missing side of the triangle. round your answer to the nearest tenth if necessary.
8.) find the missing side of the triangle. leave your answer in simplest radical form.

Explanation:

Problem 5: Find KI (Assuming Similar Triangles or Midline Theorem)

Since the diagram has markings indicating parallel lines (or midsegments), we can assume the midline theorem (or similar triangles) applies. If \( SR \) is a midline or part of similar triangles, and \( SR = 9 \), then \( KI \) would be twice \( SR \) (if \( SR \) is a midline). Wait, maybe the segments are such that \( KI = 2 \times SR \)? Wait, maybe the figure is a trapezoid with midline? Wait, the markings show that \( KS = SJ \) and \( IR = RJ \), so \( SR \) is the midline of the trapezoid \( KIJR \). The midline of a trapezoid is the average of the two bases, but if \( KI \) and \( SR \) are related, maybe \( KI = 2 \times SR \) if \( SR \) is half? Wait, no, midline formula is \( \text{midline} = \frac{\text{base1} + \text{base2}}{2} \). But if \( KI \) and the other base (maybe \( 0 \) or something? No, maybe it's a triangle with a midline. Wait, maybe the figure is a triangle with a line segment parallel to the base, making similar triangles. If \( SR = 9 \) and it's a midline, then \( KI = 2 \times 9 = 18 \)? Wait, maybe the problem is that \( SR \) is a segment, and \( KI \) is equal to \( 2 \times SR \) because of the markings (two marks on \( KS \) and one on \( SJ \)? Wait, maybe the correct approach is: Since the markings show that \( S \) is the midpoint of \( KJ \) and \( R \) is the midpoint of \( IJ \), then \( SR \) is the midline of triangle \( KIJ \), so \( SR = \frac{1}{2} KI \). Therefore, \( KI = 2 \times SR = 2 \times 9 = 18 \).

Problem 6: Find LN

The diagram shows a triangle with \( EF \) parallel to \( LN \), and markings indicating that \( E \) and \( F \) are midpoints (since the segments are marked with equal ticks). By the midline theorem, the midline \( EF \) is half of \( LN \). So:

$$ EF = \frac{1}{2} LN $$

Given \( EF = x + 2 \) and \( LN = 3x - 8 \), substitute:

$$ x + 2 = \frac{1}{2}(3x - 8) $$

Multiply both sides by 2:

$$ 2(x + 2) = 3x - 8 $$

$$ 2x + 4 = 3x - 8 $$

Subtract \( 2x \) from both sides:

$$ 4 = x - 8 $$

Add 8 to both sides:

$$ x = 12 $$

Now, substitute \( x = 12 \) into \( LN = 3x - 8 \):

$$ LN = 3(12) - 8 = 36 - 8 = 28 $$

Problem 7: Find \( x \) (Right Triangle, Pythagorean Theorem)

The triangle is a right triangle with hypotenuse \( 15.4 \) and one leg \( 6.3 \). Let \( x \) be the other leg. By Pythagoras:

$$ x^2 + 6.3^2 = 15.4^2 $$

Calculate \( 6.3^2 = 39.69 \) and \( 15.4^2 = 237.16 \)

$$ x^2 = 237.16 - 39.69 = 197.47 $$

$$ x = \sqrt{197.47} \approx 14.05 \approx 14.1 $$

Problem 8: Find \( x \) (Right Triangle, Pythagorean Theorem)

The triangle is a right triangle with legs \( 11 \) and \( x \), hypotenuse \( 15 \). By Pythagoras:

$$ 11^2 + x^2 = 15^2 $$

$$ 121 + x^2 = 225 $$

$$ x^2 = 225 - 121 = 104 $$

$$ x = \sqrt{104} = \sqrt{4 \times 26} = 2\sqrt{26} $$

Final Answers:

1. \( \boldsymbol{18} \) (assuming midline theorem)
2. \( \boldsymbol{28} \)
3. \( \boldsymbol{\approx 14.1} \)
4. \( \boldsymbol{2\sqrt{26}} \)

Answer:

Problem 5: Find KI (Assuming Similar Triangles or Midline Theorem)

Since the diagram has markings indicating parallel lines (or midsegments), we can assume the midline theorem (or similar triangles) applies. If \( SR \) is a midline or part of similar triangles, and \( SR = 9 \), then \( KI \) would be twice \( SR \) (if \( SR \) is a midline). Wait, maybe the segments are such that \( KI = 2 \times SR \)? Wait, maybe the figure is a trapezoid with midline? Wait, the markings show that \( KS = SJ \) and \( IR = RJ \), so \( SR \) is the midline of the trapezoid \( KIJR \). The midline of a trapezoid is the average of the two bases, but if \( KI \) and \( SR \) are related, maybe \( KI = 2 \times SR \) if \( SR \) is half? Wait, no, midline formula is \( \text{midline} = \frac{\text{base1} + \text{base2}}{2} \). But if \( KI \) and the other base (maybe \( 0 \) or something? No, maybe it's a triangle with a midline. Wait, maybe the figure is a triangle with a line segment parallel to the base, making similar triangles. If \( SR = 9 \) and it's a midline, then \( KI = 2 \times 9 = 18 \)? Wait, maybe the problem is that \( SR \) is a segment, and \( KI \) is equal to \( 2 \times SR \) because of the markings (two marks on \( KS \) and one on \( SJ \)? Wait, maybe the correct approach is: Since the markings show that \( S \) is the midpoint of \( KJ \) and \( R \) is the midpoint of \( IJ \), then \( SR \) is the midline of triangle \( KIJ \), so \( SR = \frac{1}{2} KI \). Therefore, \( KI = 2 \times SR = 2 \times 9 = 18 \).

Problem 6: Find LN

The diagram shows a triangle with \( EF \) parallel to \( LN \), and markings indicating that \( E \) and \( F \) are midpoints (since the segments are marked with equal ticks). By the midline theorem, the midline \( EF \) is half of \( LN \). So:

$$ EF = \frac{1}{2} LN $$

Given \( EF = x + 2 \) and \( LN = 3x - 8 \), substitute:

$$ x + 2 = \frac{1}{2}(3x - 8) $$

Multiply both sides by 2:

$$ 2(x + 2) = 3x - 8 $$

$$ 2x + 4 = 3x - 8 $$

Subtract \( 2x \) from both sides:

$$ 4 = x - 8 $$

Add 8 to both sides:

$$ x = 12 $$

Now, substitute \( x = 12 \) into \( LN = 3x - 8 \):

$$ LN = 3(12) - 8 = 36 - 8 = 28 $$

Problem 7: Find \( x \) (Right Triangle, Pythagorean Theorem)

The triangle is a right triangle with hypotenuse \( 15.4 \) and one leg \( 6.3 \). Let \( x \) be the other leg. By Pythagoras:

$$ x^2 + 6.3^2 = 15.4^2 $$

Calculate \( 6.3^2 = 39.69 \) and \( 15.4^2 = 237.16 \)

$$ x^2 = 237.16 - 39.69 = 197.47 $$

$$ x = \sqrt{197.47} \approx 14.05 \approx 14.1 $$

Problem 8: Find \( x \) (Right Triangle, Pythagorean Theorem)

The triangle is a right triangle with legs \( 11 \) and \( x \), hypotenuse \( 15 \). By Pythagoras:

$$ 11^2 + x^2 = 15^2 $$

$$ 121 + x^2 = 225 $$

$$ x^2 = 225 - 121 = 104 $$

$$ x = \sqrt{104} = \sqrt{4 \times 26} = 2\sqrt{26} $$

Final Answers:

1. \( \boldsymbol{18} \) (assuming midline theorem)
2. \( \boldsymbol{28} \)
3. \( \boldsymbol{\approx 14.1} \)
4. \( \boldsymbol{2\sqrt{26}} \)