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Tangents, Chords & Secant Segments — Easy Worksheet 1
Grade 10 · MathSheets Name: ____________________ Date: __________
O is the center of each circle. Lines that look tangent are tangent. Find x.
How to solve
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Example
PA is tangent to circle O at A. Find x.
Answer: 3
Not quite. The answer is
💡 A tangent is perpendicular to the radius at the point of tangency, so ∠OAP = 90° and OP is the hypotenuse. By the Pythagorean Theorem, OA2 + PA2 = OP2. x2 = 62 + 82 = 36 + 64 = 100, so x = √100 = 10.
✓ Correct!
Not quite. The answer is
💡 A tangent is perpendicular to the radius at the point of tangency, so ∠OAP = 90° and OP is the hypotenuse. By the Pythagorean Theorem, OA2 + PA2 = OP2. x2 = 62 + 82 = 36 + 64 = 100, so x = √100 = 10.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA and PB are tangent to circle O. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Two tangent segments drawn to a circle from the same outside point are congruent, so PA = PB. PA = 25, so x = 25.
✓ Correct!
Not quite. The answer is
💡 Two tangent segments drawn to a circle from the same outside point are congruent, so PA = PB. PA = 25, so x = 25.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 By the Intersecting Chords Theorem, the products of the chord pieces are equal: AE × EB = CE × ED. 10 × 6 = x × 5, so 5x = 60. Divide both sides by 5: x = 60 ÷ 5 = 12.
✓ Correct!
Not quite. The answer is
💡 By the Intersecting Chords Theorem, the products of the chord pieces are equal: AE × EB = CE × ED. 10 × 6 = x × 5, so 5x = 60. Divide both sides by 5: x = 60 ÷ 5 = 12.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 A tangent is perpendicular to the radius at the point of tangency, so ∠OAP = 90° and OP is the hypotenuse. By the Pythagorean Theorem, OA2 + PA2 = OP2. x2 = 52 − 42 = 25 − 16 = 9, so x = √9 = 3.
✓ Correct!
Not quite. The answer is
💡 A tangent is perpendicular to the radius at the point of tangency, so ∠OAP = 90° and OP is the hypotenuse. By the Pythagorean Theorem, OA2 + PA2 = OP2. x2 = 52 − 42 = 25 − 16 = 9, so x = √9 = 3.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Two tangent segments drawn to a circle from the same outside point are congruent, so PA = PB. PA = 24, so x = 24.
✓ Correct!
Not quite. The answer is
💡 Two tangent segments drawn to a circle from the same outside point are congruent, so PA = PB. PA = 24, so x = 24.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 By the Intersecting Chords Theorem, the products of the chord pieces are equal: AE × EB = CE × ED. 3 × x = 2 × 6, so 3x = 12. Divide both sides by 3: x = 12 ÷ 3 = 4.
✓ Correct!
Not quite. The answer is
💡 By the Intersecting Chords Theorem, the products of the chord pieces are equal: AE × EB = CE × ED. 3 × x = 2 × 6, so 3x = 12. Divide both sides by 3: x = 12 ÷ 3 = 4.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 A tangent is perpendicular to the radius at the point of tangency, so ∠OAP = 90° and OP is the hypotenuse. By the Pythagorean Theorem, OA2 + PA2 = OP2. x2 = 102 − 82 = 100 − 64 = 36, so x = √36 = 6.
✓ Correct!
Not quite. The answer is
💡 A tangent is perpendicular to the radius at the point of tangency, so ∠OAP = 90° and OP is the hypotenuse. By the Pythagorean Theorem, OA2 + PA2 = OP2. x2 = 102 − 82 = 100 − 64 = 36, so x = √36 = 6.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 By the Intersecting Chords Theorem, the products of the chord pieces are equal: AE × EB = CE × ED. 6 × 14 = x × 7, so 7x = 84. Divide both sides by 7: x = 84 ÷ 7 = 12.
✓ Correct!
Not quite. The answer is
💡 By the Intersecting Chords Theorem, the products of the chord pieces are equal: AE × EB = CE × ED. 6 × 14 = x × 7, so 7x = 84. Divide both sides by 7: x = 84 ÷ 7 = 12.
💡 How to solve: Tangents, Chords & Secant Segments
A tangent is perpendicular to the radius at the point of tangency, so use the Pythagorean Theorem. Two tangent segments from the same point are equal. For two chords that cross: part × part = part × part. For two secants from an outside point: outside × whole = outside × whole. For a tangent and a secant: tangent2 = outside × whole.
Try a similar one
PA is tangent to circle O at A. Find x.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Tangents, Chords & Secant Segments — Easy Worksheet 1
O is the center of each circle. Lines that look tangent are tangent. Find x.