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SSC CGL Maths · Mensuration 2D

Combined and Shaded Regions

A shaded region is a whole shape with known pieces taken away or added on, so its area is a sum or difference of areas from the earlier pages.

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Why this matters

Sixteen questions in the bank, fifteen of them from the 2025 shifts, and half are MODERATE: the hardest page of the chapter. Nothing here is a new formula. The work is seeing which pieces make up the region: a segment is a sector minus a triangle, and a figure inside another touches it at points that fix its size.

Concept 1 of 3: Area of a segment: sector minus triangle

A chord cuts a slice of the circle away from the centre. The sector reaches all the way to the centre; take away the triangle between the chord and the centre, and what remains is the thin piece between the chord and the arc.

Definition

  • A segment is the region between a chord and its arc. The minor segment is the smaller piece.
  • Segment == sector −- triangle formed by the two radii and the chord.
  • The triangle has two sides rr with angle θ\theta between them, so its area is 12r2sin⁡θ\frac{1}{2}r^2\sin\theta:
  • 60∘60^\circ: the triangle is equilateral, area 34r2\frac{\sqrt{3}}{4}r^2. A chord equal to the radius means this case.
  • 90∘90^\circ: a right triangle, area r22\frac{r^2}{2}.
  • 120∘120^\circ: area 34r2\frac{\sqrt{3}}{4}r^2 again, since sin⁡120∘=sin⁡60∘\sin 120^\circ = \sin 60^\circ. 150∘150^\circ: area r24\frac{r^2}{4}.
  • Major segment == whole circle −- minor segment.
  • Fast route: a segment's area is positive and smaller than its sector. Work out each option roughly (π≈3.14\pi \approx 3.14, 3≈1.73\sqrt{3} \approx 1.73) and strike out any that is negative or larger than the sector. Safe for elimination; it does not replace the working when two options survive.

Area of a segment

Asegment=θ360∘πr2−12r2sin⁡θA_{\text{segment}} = \frac{\theta}{360^\circ} \pi r^2 - \frac{1}{2}r^2\sin\theta

Worked example

A chord of a circle of radius 14 cm makes an angle of 90° at the centre. Find the area of the minor segment. (Take π=227\pi = \frac{22}{7}.)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q73Moderate

Example 1 · Mensuration 2D · Combined and Shaded Regions

A circle of radius of 12 cm. Two radii are drawn such that the length of the chord connecting their endpoints is 12 cm. What is the area of the minor segment formed?

Giving the sector as the segment

The sector runs to the centre; the segment stops at the chord. Subtract the triangle between the chord and the centre.

Using the equilateral triangle at every angle

34r2\frac{\sqrt{3}}{4}r^2 is the triangle only at 60∘60^\circ (and, by the same sine, 120∘120^\circ). At 90∘90^\circ it is r22\frac{r^2}{2}; in general use 12r2sin⁡θ\frac{1}{2}r^2\sin\theta.
Drill 6 more on area of a segment: sector minus triangle

Concept 2 of 3: Shaded areas: the whole minus the holes, or the sum of the pieces

Most shaded figures are a simple outline with simple shapes cut out of it, or a few simple shapes placed side by side. Name each piece, find its area with a rule you already know, then subtract or add.

Definition

  • Whole minus holes: shaded area == outer shape −- every piece removed (a pond in a field, a hole in a nut).
  • Sum of pieces: a figure built from shapes side by side is the sum of their areas. A share of the total is that piece ÷\div the total ×100\times 100.
  • Equal circles at the corners: the sectors drawn at the three corners of any triangle have angles adding to 180∘180^\circ, so together they make half a circle. At the four corners of a square or rectangle they add to 360∘360^\circ, a full circle.
  • A hole of the same shape: a regular hexagon of side aa with a hexagonal hole of side bb leaves 332(a2−b2)\frac{3\sqrt{3}}{2}(a^2 - b^2), the difference of the squares of the sides, not the square of the difference.
  • Fast route: never find the corner sectors one by one; add their angles first.

Corner sectors of equal radius

three corners of a triangle: 12πr2four corners of a rectangle: πr2\text{three corners of a triangle: } \frac{1}{2}\pi r^2 \qquad \text{four corners of a rectangle: } \pi r^2

Worked example

A square of side 14 cm has a quarter circle of radius 7 cm cut from each corner. Find the area that is left. (Take π=227\pi = \frac{22}{7}.)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 23 Sep 2025, Shift 1 · Q58Easy

Example 2 · Mensuration 2D · Combined and Shaded Regions

A rectangular park is 60 m by 40 m. A semicircular pond of diameter 40 m is inside. What percent of the park's area does the pond occupy?

Counting each corner sector as a full circle

Three equal circles drawn at a triangle's corners each put only a sector inside it. Together those sectors are half of one circle, because the triangle's angles add to 180∘180^\circ.

Squaring the difference for a hole of the same shape

A hexagon of side 5 with a hexagonal hole of side 2 leaves 332(25−4)\frac{3\sqrt{3}}{2}(25 - 4), not 332(5−2)2\frac{3\sqrt{3}}{2}(5 - 2)^2.
Drill 4 more on shaded areas: the whole minus the holes, or the sum of the pieces

Concept 3 of 3: Inscribed figures: a circle in a square or triangle, a square or hexagon in a circle

When one figure fits exactly inside another, they touch at a few points, and those points fix one size from the other. Find the one length they share (a diameter, a diagonal, a radius) and every area follows.

Definition

  • Circle in a square of side aa: the diameter equals the side, so r=a2r = \frac{a}{2}. The corners left over have area a2−πa24a^2 - \frac{\pi a^2}{4}.
  • Square in a circle of radius rr: the square's diagonal is the diameter, so its area is (2r)22=2r2\frac{(2r)^2}{2} = 2r^2.
  • Circle in an equilateral triangle of side aa: from area == inradius ×\times semi-perimeter, r=a23r = \frac{a}{2\sqrt{3}}. The circle through the corners has radius R=a3=2rR = \frac{a}{\sqrt{3}} = 2r.
  • Regular hexagon in a circle of radius RR: its side equals RR, so its area is 332R2\frac{3\sqrt{3}}{2}R^2, about 83%83\% of the circle.
  • Fast route: an inscribed figure is always smaller than the one around it. An option larger than the outer figure, or a leftover area that is negative, is wrong. Always safe for elimination.

Inscribed figures

rin square=a2Asquare in circle=2r2rin equilateral=a23Requilateral=a3r_{\text{in square}} = \frac{a}{2} \qquad A_{\text{square in circle}} = 2r^2 \qquad r_{\text{in equilateral}} = \frac{a}{2\sqrt{3}} \qquad R_{\text{equilateral}} = \frac{a}{\sqrt{3}}
  • aside of the outer square or triangle
  • r, Rradius of the circle inside, radius of the circle outside

Worked example

A circle is inscribed in an equilateral triangle of side 12312\sqrt{3} cm. Find the radius of the circle and the area of the triangle outside it.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 3 · Q59Moderate

Example 3 · Mensuration 2D · Combined and Shaded Regions

A regular hexagon is inscribed inside a circle of radius 14 cm. Find the area of the hexagon.

Taking a square's side as the diameter of the circle round it

For a square INSIDE a circle, the diagonal is the diameter. The side equals the diameter only when the circle is inside the square.

Mixing up a triangle's two circles

In an equilateral triangle the circle inside has radius a23\frac{a}{2\sqrt{3}} and the circle through the corners has radius a3\frac{a}{\sqrt{3}}, exactly twice as large.
Drill 3 more on inscribed figures: a circle in a square or triangle, a square or hexagon in a circle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Area of a segment: sector minus triangle

    Area of a segment

    Asegment=θ360∘πr2−12r2sin⁡θA_{\text{segment}} = \frac{\theta}{360^\circ} \pi r^2 - \frac{1}{2}r^2\sin\theta
  • Shaded areas: the whole minus the holes, or the sum of the pieces

    Corner sectors of equal radius

    three corners of a triangle: 12πr2four corners of a rectangle: πr2\text{three corners of a triangle: } \frac{1}{2}\pi r^2 \qquad \text{four corners of a rectangle: } \pi r^2
  • Inscribed figures: a circle in a square or triangle, a square or hexagon in a circle

    Inscribed figures

    rin square=a2Asquare in circle=2r2rin equilateral=a23Requilateral=a3r_{\text{in square}} = \frac{a}{2} \qquad A_{\text{square in circle}} = 2r^2 \qquad r_{\text{in equilateral}} = \frac{a}{2\sqrt{3}} \qquad R_{\text{equilateral}} = \frac{a}{\sqrt{3}}

Watch out for (6)

Test yourself on Mensuration 2D

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Mensuration 2D chapter testDrill all 16 questions