Exercise 9.3 — Segments of a Secant

Exercise 9.3 — Areas of Sectors and Segments of a Circle

Exercise 9.3, from the chapter Tangents and Secants to a Circle in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus), moves from tangent lengths to a completely practical skill: calculating the area of a sector and the area of a segment of a circle. These ideas show up everywhere — from a car's wiper blade sweeping the windscreen to decorative "leaf" patterns cut into floor tiles.

This exercise has eight questions, ranging from a direct application of the segment-area formula to multi-step shaded-region problems involving squares, semicircles and quadrants. Every single question is built from just two formulas: the area of a sector, and the relationship between a sector, a segment, and a triangle.

What is a Segment of a Circle?

When a secant (a line that cuts a circle at two points) is drawn, the chord it creates divides the circle into two regions called segments. Recognising the difference between a minor and a major segment is the first step in every question of this exercise.

• Minor segment: The smaller region enclosed between the chord and the minor (shorter) arc.
• Major segment: The larger region enclosed between the chord and the major (longer) arc.
• Special case: When the chord passes through the centre (a diameter), the "minor" and "major" segments become identical — each is a semicircle.

Formula — Finding the Area of a Segment

The key trick is to relate the segment to two shapes you already know how to measure: a sector (a pizza-slice shape) and a triangle. Look at the shaded minor segment APB below — it is simply the sector OAPB with the triangle OAB removed.

Question 1 — Minor and Major Segment for a 90° Chord

In a circle of radius 10 cm, a chord subtends a right angle (90°) at the centre. We are asked to find the areas of both the minor segment and the major segment, using π = 3.14.

Question 2 — Minor Segment for a 120° Chord

Here the angle is no longer 90°, so the triangle OAB is not right-angled at O. We instead drop a perpendicular OM from the centre to the chord — this perpendicular always bisects both the chord and the angle at the centre, splitting the problem into two easy right triangles.

Question 3 — Area Cleaned by a Car's Wiper Blades

This is a direct, real-world use of the sector-area formula — no segment or triangle needed. Each wiper blade traces out a sector as it sweeps across the windscreen, and since the two wipers "do not overlap," the total area cleaned is simply twice the area of one sector.

Question 4 — The Four-Petal "Flower" in a Square

ABCD is a square of side 10 cm, and a semicircle is drawn on each of its four sides as diameter. The four semicircles overlap to create a flower-like shaded pattern in the middle of the square. We need the total area of this shaded design.

Question 5 — The "Hourglass" Pattern in a Square

This time ABCD is a square of side 7 cm, but only two opposite semicircles are drawn — APD on the left side and BPC on the right side, each meeting at the centre point P. The shaded region is whatever is left of the square once these two semicircles are removed.

Question 6 — Shaded Region in a Quadrant

OACB is a quadrant (a quarter circle) of radius 3.5 cm with centre O. A point D lies on OA such that OD = 2 cm. We need the area of the region that remains after removing triangle ODB from the quadrant.

Question 7 — Region Between Two Concentric Sector Arcs

AB and CD are arcs of two concentric circles (radii 21 cm and 7 cm) with the same centre O, and ∠AOB = 30°. The shaded region is the "ring-slice" trapped between the two arcs — like a slice cut from a cone-shaped lampshade.

Question 8 — The "Leaf" Common to Two Quadrants

ABCD is a square, and two quadrants are drawn inside it — one centred at A (passing through D and B) and one centred at C (also passing through D and B). The shaded "designed region" is the lens-shaped area that is common to both quadrants, running diagonally across the square.

Common Mistakes to Avoid

• Confusing minor and major segments: The minor segment is always the smaller region cut off by the chord — when in doubt, compare it to half the circle. If the central angle is less than 180°, the segment on the same side as that angle is the minor one.
• Using the wrong radius for the triangle's area: When the sector angle is not 90° (as in Question 2), do not assume the triangle is right-angled at O — instead drop a perpendicular from O and use sine/cosine ratios.
• Forgetting to double the semicircle area: In Questions 4 and 5, each "side" contributes one semicircle of radius (side/2) — always check whether the question removes 2 or 4 semicircles before subtracting.
• Mixing up R and r in concentric circle problems: In Question 7, always subtract the smaller radius squared from the larger (R² − r²), never the reverse, since the shaded region lies outside the inner arc.
• Skipping the construction step: In Question 8, joining diagonal BD first is what allows the "leaf" to be split into two manageable segments — without this step, the problem looks far harder than it is.

Quick Reference — All Answers at a Glance

What This Lesson Prepares You For

Exercise 9.3 brings together the circle theorems from earlier in this chapter with practical area calculations. If the idea of a tangent length or the perpendicular-from-centre property felt unfamiliar while drawing these figures, it helps to revisit Exercise 9.1 and Exercise 9.2 on tangents and equal tangent lengths, since several constructions here (like the perpendicular bisector of a chord) reuse the same logic.

The sector and segment area formulas practised here are also the foundation for combined mensuration problems involving cones, cylinders and spheres later in Class 10, where curved surfaces are often unrolled into sectors. Strengthening your speed with π-based calculations now — including switching comfortably between π = 3.14 and π = 22/7 — will save valuable time in board exams.