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Geometry and Measurement - Circle Geometry: Chords, Tangents, Arcs, Sectors, and Segments

Grade 8IB

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

🔑Concepts

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A chord is a straight line joining any two points on the circumference. The perpendicular bisector of a chord always passes through the center of the circle.

Circle with a chord and a perpendicular line from the center bisecting it.
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A tangent is a line that touches the circle at exactly one point. The radius drawn to the point of contact is always perpendicular (90∘90^\circ) to the tangent line.

A circle with a tangent line and a radius meeting at a right angle.
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An arc is a part of the circumference. A sector is the 'pie-slice' region bounded by two radii and an arc. The central angle θ\theta determines their size relative to the whole circle.

A circle highlighting a sector with central angle theta.
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A segment is the region bounded by a chord and its corresponding arc. Its area is found by subtracting the area of the triangle formed by the radii and the chord from the sector's area.

📐Formulae

Circumference(C)=2πrCircumference (C) = 2\pi r

Area of Circle(A)=πr2Area\ of\ Circle (A) = \pi r^2

Arc Length(L)=θ360∘×2πrArc\ Length (L) = \frac{\theta}{360^\circ} \times 2\pi r

Area of Sector=θ360∘×πr2Area\ of\ Sector = \frac{\theta}{360^\circ} \times \pi r^2

Area of Segment=Area of Sector−Area of Triangle (12r2sin⁡θ)Area\ of\ Segment = Area\ of\ Sector - Area\ of\ Triangle\ (\frac{1}{2}r^2 \sin\theta)

💡Examples

Problem 1:

Calculate the length of an arc that subtends an angle of 120∘120^\circ at the center of a circle with a radius of 2121 cm. (Take π=227\pi = \frac{22}{7})

Solution:

L=θ360∘×2πrL = \frac{\theta}{360^\circ} \times 2\pi r L=120360×2×227×21L = \frac{120}{360} \times 2 \times \frac{22}{7} \times 21 L=13×2×22×3L = \frac{1}{3} \times 2 \times 22 \times 3 L=44 cmL = 44\text{ cm}

Explanation:

To find the arc length, substitute the given angle θ=120∘\theta = 120^\circ and r=21r = 21 cm into the formula. Simplify the fraction and solve.

Problem 2:

Find the area of a sector with a radius of 66 cm and a central angle of 60∘60^\circ.

Solution:

Area=θ360∘×πr2Area = \frac{\theta}{360^\circ} \times \pi r^2 Area=60360×π×62Area = \frac{60}{360} \times \pi \times 6^2 Area=16×36πArea = \frac{1}{6} \times 36\pi Area=6π≈18.85 cm2Area = 6\pi \approx 18.85\text{ cm}^2

Explanation:

The sector area is a fraction of the total area of the circle. We multiply the ratio of the angle to 360∘360^\circ by the full area formula πr2\pi r^2.

Problem 3:

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

Solution:

Let rr be the radius. The perpendicular from the center bisects the chord into two 88 cm segments. Using Pythagoras' Theorem in the right-angled triangle formed: r2=62+82r^2 = 6^2 + 8^2 r2=36+64r^2 = 36 + 64 r2=100r^2 = 100 r=100=10 cmr = \sqrt{100} = 10\text{ cm}

Explanation:

The radius, the distance from the center, and half the chord length form a right-angled triangle. We solve for the hypotenuse (radius) using the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2.

Problem 4:

A tangent PTPT is drawn from an external point PP to a circle with center OO. If the radius OT=7OT = 7 cm and the distance OP=25OP = 25 cm, calculate the length of the tangent segment PTPT.

Right-angled triangle OTP with radius 7 and hypotenuse 25.

Solution:

  1. In △OTP\triangle OTP, the angle ∠OTP=90∘\angle OTP = 90^\circ because the radius is perpendicular to the tangent.
  2. By Pythagoras' Theorem: PT2+OT2=OP2PT^2 + OT^2 = OP^2
  3. Substitute values: PT2+72=252PT^2 + 7^2 = 25^2
  4. PT2+49=625PT^2 + 49 = 625
  5. PT2=576PT^2 = 576
  6. PT=576=24PT = \sqrt{576} = 24 cm.

Explanation:

Since the tangent is perpendicular to the radius at the point of tangency, we apply the Pythagorean theorem to the right-angled triangle formed by the center, the external point, and the point of contact.

Problem 5:

Find the area of the segment cut off by a chord that subtends a 90∘90^\circ angle at the center of a circle with radius 1010 cm. (Use π=3.14\pi = 3.14)

A sector with a 90 degree angle and a chord connecting the endpoints of the arc.

Solution:

  1. Area of sector =θ360×πr2= \frac{\theta}{360} \times \pi r^2
  2. Area of sector =90360×3.14×102=0.25×314=78.5= \frac{90}{360} \times 3.14 \times 10^2 = 0.25 \times 314 = 78.5 cm2^2
  3. Area of triangle =12r2sin⁡(90∘)=12×10×10×1=50= \frac{1}{2} r^2 \sin(90^\circ) = \frac{1}{2} \times 10 \times 10 \times 1 = 50 cm2^2
  4. Area of segment == Area of sector −- Area of triangle
  5. Area of segment =78.5−50=28.5= 78.5 - 50 = 28.5 cm2^2

Explanation:

To find the area of a segment, we calculate the area of the circular sector and subtract the area of the isosceles triangle formed by the chord and the two radii.