krit.club logo

Geometry and Trigonometry - Chords and their perpendicular bisectors-extended

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The perpendicular bisector of a chord always passes through the center of the circle. This implies that any line from the center that bisects a chord must be perpendicular to it.

Circle with a chord AB and a perpendicular line from center O to midpoint M.
•

The distance of a chord from the center is the length of the perpendicular segment from the center to the chord. In a circle, equal chords are equidistant from the center.

Two parallel equal chords at equal distance d from the center.
•

A right-angled triangle is formed by the radius (rr), the distance from the center (dd), and half the chord length (L2\frac{L}{2}). Applying the Pythagorean theorem: r2=d2+(L2)2r^2 = d^2 + (\frac{L}{2})^2.

•

If two chords are not equal in length, the longer chord is closer to the center of the circle than the shorter chord.

📐Formulae

r2=d2+(L2)2r^2 = d^2 + \left(\frac{L}{2}\right)^2

d=r2−(L2)2d = \sqrt{r^2 - \left(\frac{L}{2}\right)^2}

L=2r2−d2L = 2\sqrt{r^2 - d^2}

💡Examples

Problem 1:

A chord of length 1616 cm is at a distance of 66 cm from the center of a circle. Find the radius of the circle.

Solution:

Let L=16L = 16 cm and d=6d = 6 cm. The radius rr can be found using the Pythagorean theorem on the triangle formed by the radius, the distance from the center, and half the chord.

r2=d2+(L2)2r2=62+(162)2r2=36+82r2=36+64r2=100r=100r=10 cm\begin{aligned} r^2 &= d^2 + \left(\frac{L}{2}\right)^2 \\ r^2 &= 6^2 + \left(\frac{16}{2}\right)^2 \\ r^2 &= 36 + 8^2 \\ r^2 &= 36 + 64 \\ r^2 &= 100 \\ r &= \sqrt{100} \\ r &= 10 \text{ cm} \end{aligned}

Explanation:

We use the property that the perpendicular from the center bisects the chord into two 88 cm segments. This creates a right-angled triangle with sides 66 cm and 88 cm, with the radius as the hypotenuse.

Problem 2:

A circle has a radius of 1313 cm. Calculate the length of a chord that is 55 cm away from the center.

Solution:

Given r=13r = 13 cm and d=5d = 5 cm. Let xx be half the length of the chord.

x2=r2−d2x2=132−52x2=169−25x2=144x=144=12 cm\begin{aligned} x^2 &= r^2 - d^2 \\ x^2 &= 13^2 - 5^2 \\ x^2 &= 169 - 25 \\ x^2 &= 144 \\ x &= \sqrt{144} = 12 \text{ cm} \end{aligned}

The total length of the chord L=2xL = 2x: L=2×12=24 cmL = 2 \times 12 = 24 \text{ cm}

Explanation:

Applying the Pythagorean theorem allows us to find half the chord length. We must multiply by 22 at the end because the perpendicular from the center bisects the chord.

Problem 3:

Two parallel chords of lengths 1010 cm and 2424 cm lie on opposite sides of the center of a circle of radius 1313 cm. Find the distance between the two chords.

Solution:

First, find the distance of each chord from the center.

For the 1010 cm chord (L1=10L_1 = 10): d1=132−52=169−25=144=12 cmd_1 = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}

For the 2424 cm chord (L2=24L_2 = 24): d2=132−122=169−144=25=5 cmd_2 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}

Since the chords are on opposite sides of the center, the distance between them is: Distance=d1+d2=12+5=17 cm\text{Distance} = d_1 + d_2 = 12 + 5 = 17 \text{ cm}

Explanation:

We calculate the perpendicular distance from the center for both chords independently. Because they are on opposite sides of the center, we add the distances to find the total gap between them.

Problem 4:

A chord of length 3030 cm is drawn in a circle of radius 1717 cm. Find the distance of the chord from the center of the circle.

Triangle showing radius 17, half-chord 15, and distance d.

Solution:

  1. Let L=30L = 30 cm and r=17r = 17 cm.
  2. Half the length of the chord is L2=302=15\frac{L}{2} = \frac{30}{2} = 15 cm.
  3. Using Pythagoras' theorem: d2+152=172d^2 + 15^2 = 17^2
  4. d2+225=289d^2 + 225 = 289
  5. d2=289−225=64d^2 = 289 - 225 = 64
  6. d=64=8d = \sqrt{64} = 8 cm.

Explanation:

Since the perpendicular from the center bisects the chord, we form a right triangle with legs dd and 1515, and hypotenuse 1717. Solving for dd gives the distance.

Problem 5:

Two parallel chords of length 1616 cm and 1212 cm lie on the same side of the center of a circle with radius 1010 cm. Calculate the distance between the two chords.

Two parallel chords on the same side of the center showing their respective distances.

Solution:

  1. For the 1616 cm chord: d1=102−82=100−64=36=6d_1 = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 cm.
  2. For the 1212 cm chord: d2=102−62=100−36=64=8d_2 = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 cm.
  3. Since they are on the same side, the distance between them is d2−d1=8−6=2d_2 - d_1 = 8 - 6 = 2 cm.

Explanation:

We find the distance of each chord from the center separately using the radius and half-chord lengths. Subtracting these distances gives the gap between the parallel lines.