Intersecting Chords Theorem Questions with Solutions

Find \(x\) in the diagram below.

Solution:

• Apply the intersecting chords theorem to \(AB\) and \(ED\) to write: \[ OA \times OB = OE \times OD \]
• Substitute the known quantities: \[ 2 \times 5 = 6 \times x \]
• Solve for \(x\): \[ x = \dfrac{10}{6} = \dfrac{5}{3} \]

Find \(x\) and \(y\) in the diagram below.

Solution:

• Apply the intersecting chords theorem to \(AB\) and \(CD\) to write: \[ OA \times OB = OD \times OC \]
• Substitute the known quantities: \[ 7 \times 10 = 12 \times x \]
• Solve for \(x\): \[ x = \dfrac{70}{12} = \dfrac{35}{6} \]
• Apply the intersecting chords theorem to \(AB\) and \(EF\) to write: \[ OA \times OB = OF \times OE \]
• Substitute the known quantities: \[ 7 \times 10 = 11 \times y \]
• Solve for \(y\): \[ y = \dfrac{70}{11} \]

Find \(x\) in the diagram below.

Solution:

• Apply the intersecting chords theorem to \(AB\) and \(CD\) to write: \[ OA \times OB = OD \times OC \]
• Substitute the known quantities: \[ 10 \times (2x - 1) = (2x + 3) \times 3 \]
• Expand: \[ 20x - 10 = 6x + 9 \]
• Rewrite the above equation with terms in \(x\) on one side of the equation: \[ 20x - 6x = 9 + 10 \]
• Group and solve for \(x\): \[ x = \dfrac{19}{14} \]

Find \(x\) in the diagram below.

Solution:

• Apply the intersecting chords theorem to \(AB\) and \(CD\) to write: \[ OA \times OB = OD \times OC \]
• Substitute by the expressions in \(x\): \[ (x - 1) \times (x + 6) = (2x + 3) \times (x - 2) \]
• Expand: \[ x^2 + 5x - 6 = 2x^2 - x - 6 \]
• Rewrite the above equation in standard form: \[ x^2 - 6x = 0 \]
• Factor the right side: \[ x(x - 6) = 0 \]
• Solve for \(x\) to obtain two solutions: \(x = 0\) and \(x = 6\).
• If you substitute \(x = 0\) into the given algebraic expression \(OC = x - 2\), that will give \(OC = -2\), a negative length which is not allowed.
• Hence, the only valid solution is \(x = 6\).

What is the ratio \(r\) of the area of triangle \(OBD\) to the area of triangle \(OCA\)?

Solution:

• We use the sine rule formula for the area of a triangle.
• Area of \(\triangle OBD = \dfrac{1}{2} \times OD \times 10 \times \sin \angle BOD\)
• Area of \(\triangle OCA = \dfrac{1}{2} \times 2 \times OA \times \sin \angle COA\)
• The ratio \(r\) is given by: \[ r = \dfrac{\dfrac{1}{2} \times OD \times 10 \times \sin \angle BOD}{\dfrac{1}{2} \times 2 \times OA \times \sin \angle COA} \]
• Angles \(\angle BOD\) and \(\angle COA\) have equal sizes because they are vertical angles, and therefore \(\sin \angle BOD = \sin \angle COA\).
• We now simplify the expression for \(r\): \[ r = \dfrac{OD \times 10}{2 \times OA} = 5 \frac{OD}{OA} \]
• Apply the intersecting chords theorem to \(AB\) and \(CD\) to write: \[ OA \times 10 = OD \times 2 \]
• Hence: \[ \dfrac{OD}{OA} = \dfrac{10}{2} = 5 \]
• Substitute to obtain: \[ r = 5 \times 5 = 25 \]