💡 10th Grade Geometry: Intersecting Chords Practice Questions

This problem involves the Intersecting Chords Theorem.

• Concept: When two chords intersect inside a circle, the product of the segments of one chord is equal to the product of the segments of the other chord.
• Formula: \(AE \cdot EB = CE \cdot ED\)
• Given: \(AE = 4\), \(EB = 6\), \(CE = 3\)
• Step 1: Substitute the given values into the formula.

\(4 \cdot 6 = 3 \cdot ED\)

• Step 2: Simplify the equation.

\(24 = 3 \cdot ED\)

• Step 3: Solve for \(ED\).

\(ED = \frac{24}{3}\)

\(ED = 8\)

💡 Therefore, the length of ED is 8.

We will use the Intersecting Chords Theorem here.

• Theorem: The product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.
• Equation: \(PT \cdot TQ = RT \cdot TS\)
• Knowns: \(PT = 5\), \(TQ = 10\), \(RT = 4\)
• Calculation:

\(5 \cdot 10 = 4 \cdot TS\)

\(50 = 4 \cdot TS\)

• Solve for TS:

\(TS = \frac{50}{4}\)

\(TS = 12.5\)

✅ The length of TS is 12.5.

Applying the Intersecting Chords Theorem is the key here.

• The Rule: For intersecting chords, the product of the segments of one chord equals the product of the segments of the other.
• Formula: \(MQ \cdot QN = OQ \cdot QP\)
• Given Values: \(MQ = 7\), \(QN = 2\), \(OQ = 4\)
• Substitution:

\(7 \cdot 2 = 4 \cdot QP\)

\(14 = 4 \cdot QP\)

• Solving for QP:

\(QP = \frac{14}{4}\)

\(QP = 3.5\)

👉 Therefore, the length of QP is 3.5.

We use the Intersecting Chords Theorem to solve this.

• Theorem Statement: If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
• Equation: \(XA \cdot AY = ZA \cdot AW\)
• Given Information: \(XA = 9\), \(AY = 6\), \(ZA = 12\)
• Calculation:

\(9 \cdot 6 = 12 \cdot AW\)

\(54 = 12 \cdot AW\)

• Finding AW:

\(AW = \frac{54}{12}\)

\(AW = 4.5\)

💡 The length of AW is 4.5.

This problem requires setting up an equation using the Intersecting Chords Theorem and solving for \(x\).

• Theorem: \(AE \cdot EC = BE \cdot ED\)
• Equation Setup: Substitute the given expressions into the theorem.

\(x(x+2) = (x+1)(x+3)\)

• Expand both sides:

\(x^2 + 2x = x^2 + 3x + x + 3\)

\(x^2 + 2x = x^2 + 4x + 3\)

• Solve for x: Subtract \(x^2\) from both sides.

\(2x = 4x + 3\)

\(2x - 4x = 3\)

\(-2x = 3\)

\(x = -\frac{3}{2}\)

• Analyze the result: A length cannot be negative. Let's re-check the problem statement or our setup. It is possible the problem intended different values or a different configuration. Assuming the setup is correct, this indicates no such circle exists with these segment lengths. However, if we assume a typo and that \(BE = x-1\) and \(ED = x-3\) for instance, we would proceed. Let's assume for the sake of demonstrating the method that the problem intended positive lengths. If we were to ignore the negative result and proceed with calculation for segments: \(AE = -1.5\), \(EC = 0.5\), \(BE = -0.5\), \(ED = 1.5\). This is not geometrically possible.
• Correction/Re-evaluation: Let's assume the problem meant \(AE = x+1\), \(EC = x+3\), \(BE = x\), \(ED = x+2\).

\((x+1)(x+3) = x(x+2)\)

\(x^2 + 4x + 3 = x^2 + 2x\)

\(4x + 3 = 2x\)

\(2x = -3\)

\(x = -1.5\)

• Conclusion for this specific problem statement: With the given expressions, the calculation leads to negative lengths, which are not possible in geometry. This suggests an error in the problem's numerical values or expressions. If the values were different, the steps to solve for \(x\) and then the segment lengths would be as shown. For example, if \(AE=4, EC=6, BE=3\), then \(4 \times 6 = 3 \times ED \Rightarrow 24 = 3ED \Rightarrow ED=8\).

📌 Critical Thinking Point: Always check if your calculated lengths are positive and geometrically feasible. If not, re-examine the problem statement or your calculations.

This scenario can be modeled using the Intersecting Chords Theorem.

• Real-World Application: The theorem applies to any situation where two lines (chords) intersect within a circle, dividing each line into segments.
• The Principle: The product of the segment lengths of one path must equal the product of the segment lengths of the other path.
• Equation: \(15 \cdot 20 = 12 \cdot y\)
• Calculation:

\(300 = 12y\)

• Solving for y:

\(y = \frac{300}{12}\)

\(y = 25\)

💡 So, the length of the second path section is 25 meters.

This problem requires using the Intersecting Chords Theorem and a bit of algebra.

• Theorem: \(FJ \cdot JG = HJ \cdot JI\)
• Given: \(FJ = 8\), \(JG = 5\). The total length of chord HI is 18.
• Let: \(HJ = x\). Then, \(JI = 18 - x\).
• Set up the equation:

\(8 \cdot 5 = x(18 - x)\)

\(40 = 18x - x^2\)

• Rearrange into a quadratic equation:

\(x^2 - 18x + 40 = 0\)

• Solve the quadratic equation by factoring: We need two numbers that multiply to 40 and add to -18. These numbers are -4 and -10.

\((x - 4)(x - 10) = 0\)

• Possible values for x:

\(x - 4 = 0 \Rightarrow x = 4\)

\(x - 10 = 0 \Rightarrow x = 10\)

• Determine the segment lengths:
• If \(HJ = 4\), then \(JI = 18 - 4 = 14\). Check: \(8 \cdot 5 = 40\), \(4 \cdot 14 = 56\). This is incorrect.
• If \(HJ = 10\), then \(JI = 18 - 10 = 8\). Check: \(8 \cdot 5 = 40\), \(10 \cdot 8 = 80\). This is also incorrect.
• Re-check factoring: The numbers that multiply to 40 and add to -18 are indeed -4 and -10. Let's re-verify the equation setup.

The equation \(x^2 - 18x + 40 = 0\) is correct.

Let's use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a=1\), \(b=-18\), \(c=40\).

\(x = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(40)}}{2(1)}\)

\(x = \frac{18 \pm \sqrt{324 - 160}}{2}\)

\(x = \frac{18 \pm \sqrt{164}}{2}\)

\(x = \frac{18 \pm 2\sqrt{41}}{2}\)

\(x = 9 \pm \sqrt{41}\)

• Segment Lengths:
• One segment is \(9 + \sqrt{41}\) and the other is \(9 - \sqrt{41}\).
• Let's check: \((9 + \sqrt{41})(9 - \sqrt{41}) = 9^2 - (\sqrt{41})^2 = 81 - 41 = 40\). This matches \(FJ \cdot JG\).

✅ The lengths of the segments HJ and JI are \(9 + \sqrt{41}\) and \(9 - \sqrt{41}\) (or vice versa).

We will use the Intersecting Chords Theorem.

• Theorem: The product of the segments of one chord equals the product of the segments of the other chord.
• Formula: \(AE \cdot EB = CE \cdot ED\)
• Given: \(AE = 2\), \(EB = 9\), \(CE = 3\)
• Substitute values:

\(2 \cdot 9 = 3 \cdot ED\)

\(18 = 3 \cdot ED\)

• Solve for ED:

\(ED = \frac{18}{3}\)

\(ED = 6\)

👉 The length of ED is 6.

This is a practical application of the Intersecting Chords Theorem.

• Concept: The theorem helps us relate the lengths of segments formed by intersecting chords.
• The Rule: The product of the segments of one stream equals the product of the segments of the other stream.
• Let: The unknown part of the second stream be \(x\) feet.
• Equation: \(10 \cdot 15 = 12 \cdot x\)
• Solve:

\(150 = 12x\)

\(x = \frac{150}{12}\)

\(x = 12.5\)

💡 Therefore, the length of the other part of the second stream is 12.5 feet.