# Intersecting Chords Theorem Questions with Solutions

Consider the circle with chords \( A B \) and \( E D \).

The intersecting chords theorem [1] states that for any two chords \( A B \) and \( E D \) that intersects at the point \( O \), we have \[ OA \times OB = OE \times OD \]

## Examples With Solutions

Question 1

Find \( x \) in the diagram below.

__Solution__

Apply the intersecting chords theorem to \( AB \) and \( ED \) to write: \( \quad OA \times OB = OE \times OD \)

Substitute the known quantities: \( \quad 2 \times 5 = 6 \times x \)

Solve for \( x \): \( \quad x = \dfrac{10}{6} = \dfrac{5}{3} \)

Question 2

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

__Solution__

Apply the intersecting chords theorem to \( AB \) and \( CD \) to write: \( \quad OA \times OB = OD \times OC \)

Substitute the known quantities: \( \quad 7 \times 10 = 12 \times x \)

Solve for \( x \): \( \quad x = \dfrac{70}{12} = \dfrac{35}{6} \)

Apply the intersecting chords theorem to \( AB \) and \( EF \) to write: \( \quad OA \times OB = OF \times OE \)

Substitute the known quantities: \( \quad 7 \times 10 = 11 \times y \)

Solve for \( y \): \( \quad y = \dfrac{70}{11} \)

Question 3

Find \( x \) in the diagram below.

__Solution__

Apply the intersecting chords theorem to \( AB \) and \( CD \) to write: \( \quad OA \times OB = OD \times OC \)

Substitute the known quantities: \( \quad 10 \times (2x-1) = (2x+3) \times 3 \)

Expand: \( \quad 20x - 10 = 6x + 9 \)

Rewrite the above equation with terms in \( x \) on one side of the equation: \( \quad 20x - 6x = 9 + 10 \)

Group and solve for \( x \): \( \quad x = \dfrac{19}{14} \)

Question 4

Find \( x \) in the diagram below.

__Solution__

Apply the intersecting chords theorem to \( AB \) and \( CD \) to write: \( \quad OA \times OB = OD \times OC \)

Substitute by the expressions in \( x \) : \( \quad (x-1) \times (x+6) = (2x+3) \times (x-2) \)

Expand: \( \quad x^2+5x-6 = 2x^2-x-6 \)

Rewrite the above equation with standard form: \( \quad x^2-6x = 0 \)

Factor the right side: \( \quad x(x-6) = 0 \)

Solve for \( x \) to obtain two solutions: \( \quad x = 0 \) and \( x = 6 \).

If you substitute \( x = 0 \) in the given algebraic expressions \( OC = x - 2 \), that will give \( OC = - 2 \) a negative length which is not allowed.

Hence the only valid solution is \( x = 6 \).

Question 5

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: \( \quad r = \dfrac{\dfrac{1}{2} \times OD \times 10 \times \sin \angle BOD}{\dfrac{1}{2} \times 2 \times OA \times \sin \angle COA } \)

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 of \( r \): \( \quad r = \dfrac{ OD \times 10 }{2 \times OA } = 5 \dfrac{OD}{OA} \)

Apply the intersecting chords theorem to \( AB \) and \( CD \) to write: \( \quad OA \times 10 = OD \times 2 \)

Hence : \( \quad \dfrac{OD}{OA} = \dfrac{10}{2} = 5\)

Substitute to obtain: \( \quad r = 5 \times 5 = 25 \)

## More References and Links

The Four Pillars of Geometry - John Stillwell - Springer; 2005th edition (Aug. 9 2005) - ISBN-10 : 0387255303Geometry Tutorials, Problems and Interactive Applets