CDS Mathematics · Mensuration 2D

Combined & Shaded Regions

Areas of figures built from squares, circles, semicircles and triangles: write the shaded part as whole shapes added and taken away.

All Mensuration 2D notes CDS Mathematics strategy

Why this matters

The figure-heavy page of the chapter, and none of it is EASY. The work is bookkeeping rather than new formulas: decide which whole shapes the shading is made of, then add and subtract.

Concept 1 of 4: The whole minus the holes

Most shaded regions are one big shape with smaller shapes removed. Name the big shape, name what is cut out, and subtract. Four quarter-circles at the corners of a square make one full circle.

Definition

Square of side $2a$ minus its incircle: $(4 - \pi)a^2$ .
Circle through the corners of a square of side $a$ , minus the square: $\dfrac{(\pi - 2)a^2}{2}$ .
Circle round a rectangle: its diameter is the rectangle's diagonal.
A plate of uniform thickness: weight is proportional to area, so a cut-out removes the same fraction of weight.
A triangle with a sector cut at a vertex: the sector's angle is the triangle's angle at that vertex.

Shaded area

\text{shaded} = \text{whole} - \sum \text{removed pieces}

Worked example

From a square of side

14

cm, a quarter-circle of radius

7

cm is cut at each corner. Find the remaining area.

(\pi = \tfrac{22}{7})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q100Moderate

Example 1 · Mensuration 2D · Combined and Shaded Regions

In the figure given below, ABCD is a square of side 4 cm. Quadrants of a circle of diameter 2 cm are removed from the four corners and a circle of diameter 2 cm is also removed. What is the area of the shaded region ?

Read the hatching, not the words

The stem says 'the shaded region' and the figure decides what that is. Some papers shade the parts of a circle AND a triangle that do not overlap; others shade only the overlap. Decide from the figure before choosing a formula.

Drill 8 more on the whole minus the holes

Concept 2 of 4: Semicircles on a divided diameter

A diameter cut into equal parts, with semicircles drawn on some of the parts, produces figures made entirely of half-discs. Every area is a sum of

\dfrac12\pi r^2

terms, and every perimeter a sum of

\pi r

arcs.

Definition

Semicircle of radius $r$ : area $\dfrac{\pi r^2}{2}$ , arc $\pi r$ .
Semicircles on the two halves of a diameter $2R$ have radius $\dfrac R2$ : together they have half the area of the big semicircle, but the same arc length $\pi R$ .
A semicircle drawn on the same side as the shading is subtracted; on the other side, added.

Arcs on a split diameter

\pi r_1 + \pi r_2 + \cdots = \pi R \quad \text{when } r_1 + r_2 + \cdots = R

Worked example

AB

is a diameter of a circle of radius

8

cm, and

C

is the point on

AB

with

AC = 4

cm. Semicircles are drawn below

AB

AC

and

CB

. Find the area of the lower half-disc outside both small semicircles.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (I) 2020 — Elementary Mathematics · Q90Moderate

Example 2 · Mensuration 2D · Combined and Shaded Regions

Let

PQRS

be the diameter of a circle of radius 9 cm. The length

PQ

QR

and

RS

are equal. Semi-circle is drawn with

QS

as diameter (as shown in the given figure). What is the ratio of the shaded region to that of the unshaded region ?

Drill 5 more on semicircles on a divided diameter

Concept 3 of 4: Overlaps — lenses and crescents

Where two shapes overlap, the overlap is counted in both. Two quarter-circles drawn from opposite corners of a square overlap in a lens made of two segments. Semicircles on the sides of a right triangle leave crescents whose total is exactly the triangle.

Definition

Lens from quarter-circles centred at opposite corners of a square of side $a$ : two $90^\circ$ segments, $2\left(\dfrac{\pi a^2}{4} - \dfrac{a^2}{2}\right) = a^2\left(\dfrac{\pi}{2} - 1\right)$ .
Region covered by exactly one of two shapes: $A_1 + A_2 - 2\times\text{overlap}$ .
Crescents (lunes of Hippocrates): semicircles on the legs of a right triangle, minus the semicircle on the hypotenuse, plus the triangle, equal the triangle's area.

Lens from two quarter-circles

\text{lens} = a^2\left(\frac{\pi}{2} - 1\right)

Worked example

With opposite corners of a square of side

10

cm as centres, quarter-circles of radius

10

are drawn inside the square. Find the area common to both.

(\pi = 3.14)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — Elementary Mathematics · Q75Hard

Example 3 · Mensuration 2D · Combined and Shaded Regions

Considering two opposite vertices of a square of side 'a' as centres, two circular arcs are drawn within the square joining the other two vertices, thus forming two sectors. What is the common area in these two sectors ?

Half the lens is a single segment

A lens is two segments back to back. An option equal to half the right answer is usually the area of one segment.

Drill 6 more on overlaps — lenses and crescents

Concept 4 of 4: Figures built from pieces

A field or a figure made of rectangles, triangles and trapeziums is the sum of its pieces. The perimeter is only the OUTER boundary: sides that are glued together inside do not count.

Definition

Area of a composite figure $=$ sum of the areas of its pieces.
Perimeter $=$ only the outside edges; a shared side disappears from it.
Squares drawn outward on the sides of a triangle: total $=$ triangle $+$ the three squares.

Composite figure

\text{Area} = \sum \text{pieces}, \qquad \text{Perimeter} = \text{outer edges only}

Worked example

An equilateral triangle is drawn outward on one side of a square of side

6

cm. Find the perimeter and the area of the whole figure.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (II) 2017 — Elementary Mathematics · Q59Hard

Example 4 · Mensuration 2D · Combined and Shaded Regions

An isosceles triangle is drawn outside on one of the sides of a square as base in such a way that the perimeter of the complete figure is

\frac{7}{6}

times the perimeter of the original square. What is the ratio of area of the triangle to the area of the original square ?

Drill 2 more on figures built from pieces

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 (4)

The whole minus the holes
Shaded area
$\text{shaded} = \text{whole} - \sum \text{removed pieces}$
Semicircles on a divided diameter
Arcs on a split diameter
$\pi r_1 + \pi r_2 + \cdots = \pi R \quad \text{when } r_1 + r_2 + \cdots = R$
Overlaps — lenses and crescents
Lens from two quarter-circles
$\text{lens} = a^2\left(\frac{\pi}{2} - 1\right)$
Figures built from pieces
Composite figure
$\text{Area} = \sum \text{pieces}, \qquad \text{Perimeter} = \text{outer edges only}$

Watch out for (2)

Read the hatching, not the words
→ The whole minus the holes
Half the lens is a single segment
→ Overlaps — lenses and crescents

Test yourself on Mensuration 2D

20 past CDS questions from this chapter, timed at 24 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 test Drill all 25 questions