Question

directions: solve each problem and color the object that corresponds with your answer.

1. solve for x... (a) if x = 100° color the hair red. (b) if x = 90° color the hair black.
2. solve for x... (a) if x = 70° color the center of the eyes red. (b) if x = 110° color the center of the eyes black.
3. solve for x... (a) if x = 40° color this part of the... (b) if x = 70° color this part of the...
4. solve for x... (a) if x = 20° leave the teeth white. (b) if x = 40° color the teeth yellow.
5. solve for x... (a) if x = 10° color the mouth red. (b) if x = (19)° color the mouth black.
6. solve for x... (a) if x = 90° color the ears, face, nose, neck, and arms brown. (b) if x = 45° color the ears, face, nose, neck, and arms gray.
7. solve for x... (a) if x = 25° outline the nose, eyes, eyebrows, scar, and mouth in red. (b) if x = 65° outline the nose, eyes, eyebrows, scar, and mouth in black.
8. solve for x... (a) if x = 65° color the shirt collar red. (b) if x = 90° color the shirt collar green.
9. solve for x... (a) if x = 6 color the buttons on the shirt black. (b) if x = 12 color the buttons on the shirt purple.
10. solve for x... (a) if x = 27√7 color the shirt area with the buttons green. (b) if x = 3√21 color the shirt area with the buttons yellow.
11. solve for x... (a) if x = 4 color the rest of the shirt orange. (b) if x = 8 color the rest of the shirt black.
12. solve for x... (a) if x = 4 color the mountains/rocks brown. (b) if x = 5 color the mountains/rocks gray.
13. solve for x... (a) if x = 4 color the bats/birds gray. (b) if x = 4√2 color the bats/birds blue.
14. solve for x... (a) if x = 10√3 color the moon yellow. (b) if x = 20 color the moon red.
15. solve for x... (a) if x = 8 color the rest of the background blue. (b) if x = 4 color the rest of the background purple.

artistic tip: when you are done coloring, it looks nice to outline the major features using a black crayon or marker.

Answer:

To solve these problems, we'll use the Power of a Point theorem, specifically the Tangent-Secant Theorem or Pythagorean Theorem (for right triangles formed with radii and tangents) as needed. Let's tackle a few examples (e.g., Problem 21, 22, 25, 26, 27):

Example 1: Problem 21

Diagram: A tangent segment of length \( 8 \), a secant segment with external part \( x \) and internal part \( 5 \) (radius), and the circle has radius \( 5 \).
By the Power of a Point (Tangent-Secant):
\( \text{Tangent}^2 = \text{External part} \times (\text{External part} + \text{Internal part}) \)
Wait, actually, the tangent is perpendicular to the radius, so we can use the Pythagorean Theorem:
Let the tangent length be \( 8 \), radius \( 5 \), and the distance from the external point to the center be \( x + 5 \)? Wait, no—re-examining the diagram (from the image: tangent \( 8 \), radius \( 5 \), and the other segment is \( x \)). Wait, the correct formula for a tangent (\( t \)) and a secant (\( s \)) from a common external point: \( t^2 = s \times (s + 2r) \)? No, better to use the right triangle: tangent is perpendicular to radius, so \( (\text{tangent})^2 + (\text{radius})^2 = (\text{distance from external point to center})^2 \). Wait, the diagram shows a tangent of length \( 8 \), radius \( 5 \), and the other segment (from external point to center) is \( x \). Wait, no—maybe the tangent is \( 8 \), the external segment is \( x \), and the radius is \( 5 \). Wait, the correct Power of a Point: if a tangent of length \( t \) and a secant with external segment \( a \) and internal segment \( b \) (where \( b = 2r \) if it’s a diameter), then \( t^2 = a(a + b) \). But in Problem 21, the diagram has a tangent of length \( 8 \), radius \( 5 \), and the external segment is \( x \), and the secant goes through the center (so \( b = 2 \times 5 = 10 \)). Wait, no—let’s use the Pythagorean Theorem: the tangent is perpendicular to the radius, so the triangle is right-angled with legs \( 8 \) (tangent) and \( 5 \) (radius), and hypotenuse \( x + 5 \)? No, that doesn’t fit. Wait, maybe the tangent is \( 8 \), the external segment is \( x \), and the secant has length \( x + 5 + 5 = x + 10 \) (since radius is \( 5 \), so the secant goes from external point to center to the other side, total length \( x + 10 \)). Then by Power of a Point:
\( 8^2 = x(x + 10) \)? No, that would be \( 64 = x^2 + 10x \), which is not matching the options. Wait, the options are \( x = 6 \) or \( x = 12 \). Let’s check \( x = 12 \): \( 8^2 + 5^2 = 64 + 25 = 89 \), not a square. Wait, maybe the diagram is a tangent and a secant where the secant is \( x \) and the tangent is \( 8 \), and the radius is \( 5 \). Wait, the correct formula for a tangent (\( t \)) and a secant (\( s \)) from a common external point: \( t^2 = s \times (s + 2r) \) only if the secant is a diameter. Wait, maybe the problem is simpler: the triangle is right-angled with legs \( x \) and \( 5 \), hypotenuse \( 8 + 5 \)? No. Wait, the options are \( x = 6 \) or \( x = 12 \). Let’s test \( x = 12 \): if \( x = 12 \), then \( 8^2 + 5^2 = 64 + 25 = 89 \), no. Wait, maybe the tangent is \( 8 \), the external segment is \( x \), and the secant has length \( x + 5 \) (no, radius is \( 5 \), so the secant is from external point to the circle, length \( x \), and then through the circle, length \( 5 + 5 = 10 \)). Then Power of a Point: \( 8^2 = x(x + 10) \)? \( 64 = x^2 + 10x \). Solving \( x^2 + 10x - 64 = 0 \), discriminant \( 100 + 256 = 356 \), not a perfect square. So maybe the diagram is a tangent and a radius, with the other segment being \( x \), and the tangent is \( 8 \), radius \( 5 \), so \( x = \sqrt{8^2 - 5^2} \)? No, that would be \( \sqrt{39} \), not matching. Wait, maybe the problem is a secant-secant or tangent-secant with different numbers. Let’s move to Problem 22.

Example 2: Problem 22

Diagram: A secant segment with external part \( 10 \), internal part \( x \) (wait, no—radius is \( 17 \)? Wait, the diagram shows a tangent? No, a secant? Wait, the diagram has a segment of length \( 10 \), a radius \( 17 \), and the other segment is \( x \). Wait, using the Pythagorean Theorem: if the tangent is \( 10 \), radius \( 17 \), then the distance from external point to center is \( \sqrt{10^2 + 17^2} \), no. Wait, the options are \( x = 27\sqrt{7} \) or \( x = 3\sqrt{21} \). Let’s compute \( 17^2 - 10^2 = 289 - 100 = 189 = 9 \times 21 \), so \( \sqrt{189} = 3\sqrt{21} \). Ah! So here, the radius is \( 17 \), the external segment is \( 10 \), and \( x \) is the other leg (since it’s a right triangle: radius is hypotenuse? No, wait—if the segment \( 10 \) is the tangent, and \( x \) is the radius, and \( 17 \) is the distance from external point to center, then by Pythagoras: \( 10^2 + x^2 = 17^2 \). So \( x^2 = 289 - 100 = 189 \), so \( x = \sqrt{189} = 3\sqrt{21} \). So option (b) is correct: \( x = 3\sqrt{21} \), so we color the shirt area with buttons yellow.

Example 3: Problem 25

Diagram: A tangent of length \( 4\sqrt{2} \), radius \( x \), and the distance from external point to center is \( x + \) (tangent)? No, using Pythagoras: tangent is \( 4\sqrt{2} \), radius \( x \), and the distance from external point to center is \( x + \) (external segment)? Wait, the tangent is perpendicular to the radius, so \( (4\sqrt{2})^2 + x^2 = (x + \text{external segment})^2 \)? No, the diagram shows the tangent as \( 4\sqrt{2} \), radius \( x \), and the external segment is \( x \) (wait, no—let’s use the right triangle: tangent \( t = 4\sqrt{2} \), radius \( r = x \), and the hypotenuse (distance from external point to center) is \( x + x = 2x \)? No, that would mean \( (4\sqrt{2})^2 + x^2 = (2x)^2 \). So \( 32 + x^2 = 4x^2 \), so \( 3x^2 = 32 \), \( x^2 = 32/3 \), not matching. Wait, the options are \( x = 4 \) or \( x = 4\sqrt{2} \). Let’s check \( x = 4 \): then \( (4\sqrt{2})^2 + 4^2 = 32 + 16 = 48 \), not a square. \( x = 4\sqrt{2} \): \( (4\sqrt{2})^2 + (4\sqrt{2})^2 = 32 + 32 = 64 = 8^2 \). So if the distance from external point to center is \( 8 \), then \( x = 4\sqrt{2} \) (radius) would make the hypotenuse \( 8 \). So option (b) is correct: \( x = 4\sqrt{2} \), color bats/birds blue.

Example 4: Problem 26

Diagram: A secant with external part \( x \), internal part \( 10 \) (diameter, since radius is \( 10 \), so diameter \( 20 \)? Wait, the diagram shows a secant from external point to the circle, with the external segment \( x \), and the secant passes through the center (so internal segment is \( 2 \times 10 = 20 \)). By Power of a Point: \( (\text{tangent})^2 = x(x + 20) \), but there’s no tangent—wait, the diagram has a segment of length \( 10 \) (maybe the tangent? No, the radius is \( 10 \), and the other segment is \( x \). Wait, using Pythagoras: if the tangent is \( 10 \), radius \( 10 \), then distance from external point to center is \( \sqrt{10^2 + 10^2} = 10\sqrt{2} \), not matching. Wait, the options are \( x = 10\sqrt{3} \) or \( x = 20 \). Let’s use the Power of a Point for a secant-secant: if two secants, but here it’s a secant and a tangent? No, the diagram shows a secant from external point \( x \) to the circle, passing through the center (so length \( x + 20 \)) and another segment? Wait, maybe the tangent is \( 10 \), and the secant is \( x \), so \( 10^2 = x(x - 20) \)? No, that would be \( x^2 - 20x - 100 = 0 \), solution \( x = 10 \pm 10\sqrt{2} \), not matching. Wait, the correct approach for Problem 26: the diagram has a secant with external segment \( x \), and the secant goes through the center (so internal segment is \( 2 \times 10 = 20 \)), and the tangent is \( 10 \)? No, the options are \( x = 10\sqrt{3} \) or \( x = 20 \). Let’s test \( x = 20 \): if the external segment is \( 20 \), and the internal segment is \( 20 \) (diameter), then by Power of a Point: \( t^2 = 20(20 + 20) = 800 \), so \( t = 20\sqrt{2} \), not matching. Wait, maybe it’s a right triangle with legs \( 10 \) and \( 10\sqrt{3} \), hypotenuse \( 20 \) (since \( 10^2 + (10\sqrt{3})^2 = 100 + 300 = 400 = 20^2 \)). So if the radius is \( 10 \), and the external segment is \( 10\sqrt{3} \), then the distance from external point to center is \( 20 \), so \( x = 20 \)? No, the options are \( x = 10\sqrt{3} \) or \( x = 20 \). Wait, the correct answer is \( x = 20 \)? Let’s check: if \( x = 20 \), then the secant length is \( 20 + 20 = 40 \)? No, this is getting confusing. Let’s summarize:

For each problem, we use the Power of a Point theorem (or Pythagorean Theorem for tangent-radius right triangles) to solve for \( x \), then choose the correct option (a or b) to determine the coloring.

Key Steps for Any Problem:

1. Identify the type of segment (tangent, secant, radius) and the right triangle or Power of a Point relationship.
2. Apply the Pythagorean Theorem (\( a^2 + b^2 = c^2 \)) or Power of a Point (\( t^2 = s \times (s + 2r) \) for tangent and secant from a common external point).
3. Solve for \( x \) and match with the given options (a or b) to determine the coloring.

For example, in Problem 22, we found \( x = 3\sqrt{21} \) (option b), so we color the shirt area with buttons yellow. In Problem 25, \( x = 4\sqrt{2} \) (option b), so we color bats/birds blue.

To solve a specific problem, follow these steps:

• Identify the segments (tangent, radius, external segment).
• Apply the appropriate theorem (Pythagorean or Power of a Point).
• Solve for \( x \) and select the correct option (a or b) to color as instructed.

If you need help with a specific problem (e.g., 19, 20, 23, etc.), share the diagram details or problem number, and we can walk through it!