Mensuration 2D — SSC CGL Quantitative Aptitude Formulas and Traps

Rectangles, Squares and Other Quadrilaterals

Area and perimeter of a rectangle and a square

Rectangle and square

A = lb \qquad P = 2(l + b) \qquad A = a^2 \qquad P = 4a

$l, b$ length and breadth of the rectangle
$a$ side of the square

Diagonal of a rectangle and of a square

Diagonals

d = \sqrt{l^2 + b^2} \qquad d_{\text{square}} = a\sqrt{2} \qquad A_{\text{square}} = \frac{d^2}{2}

Tiles, paths, cut-outs and the cost of covering

Tiles and paths

N = \frac{\text{area to cover}}{\text{area of one tile}} \qquad A_{\text{path}} = (l + 2w)(b + 2w) - lb = 2w(l + b) + 4w^2

$w$ width of the path
$N$ number of tiles

Area of a parallelogram, a trapezium and a rhombus

Parallelogram, trapezium and rhombus

A = bh \qquad A = \frac{1}{2}(a + b)h \qquad A = \frac{1}{2}d_1 d_2

$a, b$ the two parallel sides of a trapezium
$h$ perpendicular height
$d_1, d_2$ diagonals of a rhombus

Common traps

Dividing a rectangle's perimeter by 4

Only a square has side

\frac{P}{4}

. For a rectangle,

\frac{P}{2}

is the sum

l + b

, and you still need a second fact to split it.

Multiplying lengths in different units

A length in metres and a breadth in centimetres must be changed to one unit before you multiply.

5\ \text{m} \times 40\ \text{cm}

is

5 \times 0.4 = 2\ \text{m}^2

, not 200.

Squaring the diagonal for a square's area

The square of the diagonal is twice the area. A square with diagonal 8 m has area

\frac{64}{2} = 32\ \text{m}^2

, not 64.

Adding the sides to get the diagonal

The diagonal of a rectangle is

\sqrt{l^2 + b^2}

, always shorter than

l + b

. A 6 by 8 rectangle has diagonal 10, not 14.

Adding the path width only once

A path on every side adds its width at both ends, so each side of the outer rectangle grows by

2w

, not

w

.

Rounding the tile count down

If the working gives 89.6 tiles, buy 90. A count of tiles, boxes or trips is always rounded up to cover the whole job.

Charging fencing by area

Fencing runs along the boundary, so its cost is perimeter times the rate per metre. Painting, paving and tiling cover the surface, so they use the area.

Using the slanting side as the height

A parallelogram's area uses the perpendicular height. A slanting side of 10 cm with a perpendicular height of 8 cm on a base of 12 cm gives

96\ \text{cm}^2

, not 120.

Forgetting the half in a trapezium

A trapezium's area is the AVERAGE of the parallel sides times the height. Adding the sides and multiplying by the height doubles the answer.

Areas of Triangles

Learn this subtopic in the notes

Half base times height, and two sides with the angle between them

Area of a triangle

A = \frac{1}{2}bh \qquad A = \frac{1}{2}ab\sin C

$b, h$ a base and the perpendicular height to it
$a, b, C$ two sides and the angle between them

Three sides: semi-perimeter, Heron's formula and the inradius

Heron's formula and the inradius

s = \frac{a + b + c}{2} \qquad A = \sqrt{s(s - a)(s - b)(s - c)} \qquad A = rs \qquad A_{\text{isosceles}} = \frac{b}{4}\sqrt{4a^2 - b^2}

$s$ semi-perimeter
$r$ radius of the inscribed circle

Area of an equilateral triangle and a regular hexagon

Equilateral triangle and regular hexagon

A = \frac{\sqrt{3}}{4}a^2 \qquad h = \frac{\sqrt{3}}{2}a \qquad A_{\text{hexagon}} = \frac{3\sqrt{3}}{2}a^2

$a$ side of the triangle or hexagon

Similar figures: scale the sides, square the area

Scaling of area

\frac{A_2}{A_1} = \left(\frac{s_2}{s_1}\right)^2 \qquad \text{area change} = 2x + \frac{x^2}{100}\ \%

$s_1, s_2$ matching sides of the two figures
$x$ percentage change in every side

Common traps

Pairing a base with the wrong height

The height must be perpendicular to the side you call the base. A slanting side is not a height unless the triangle has a right angle there.

Reading an area ratio as a base ratio

Area ratio is base ratio times height ratio. Areas

5 : 3

with heights

5 : 7

give bases

\frac{5}{3} \div \frac{5}{7} = 7 : 3

, not

5 : 3

.

Putting the perimeter into Heron's formula

Heron uses the SEMI-perimeter

s

. With the full perimeter every bracket is wrong and the answer is far too large.

Running Heron on a right triangle

Sides such as 7, 24, 25 or 15, 20, 25 form a right triangle. Half the product of the two shorter sides gives the area in one step.

Mixing up the height and the area

The height of an equilateral triangle is

\frac{\sqrt{3}}{2}a

, a length. The area is

\frac{\sqrt{3}}{4}a^2

. Writing

\frac{\sqrt{3}}{2}a^2

for the area doubles it.

Taking a hexagon as three triangles

Lines from the centre to the six corners cut a regular hexagon into SIX equilateral triangles, each with the hexagon's side.

Doubling the side doubles the area

Area scales as the square of the side. Doubling every side of a figure makes its area 4 times as large; tripling makes it 9 times.

A percentage change in area is not the side's change

A side up by 10% raises the area by 21%, not 10%. Square the factor

1.1

to get

1.21

.

Circumference, Wheels and Rings

Learn this subtopic in the notes

Circumference, wheel revolutions and fencing a circle

Circumference and revolutions

C = 2\pi r = \pi d \qquad N = \frac{\text{distance}}{C}

$r, d$ radius and diameter
$N$ number of revolutions

Area of a circle and how it scales with the radius

Area of a circle

A = \pi r^2 \qquad \frac{A_2}{A_1} = \left(\frac{r_2}{r_1}\right)^2 \qquad R^2 = r_1^2 + r_2^2

$R$ radius of the circle whose area is the sum of the other two

Area of a ring, a circular track or a path round a circle

Area of a ring

A = \pi(R^2 - r^2) = \pi(R - r)(R + r) \qquad A = \pi\left(\frac{c}{2}\right)^2

$R, r$ outer and inner radii
$c$ chord of the outer circle that touches the inner circle

Common traps

A whole-number ratio for circumference to diameter

Circumference divided by diameter is

\pi

for every circle, a little over 3. A ratio such as

3 : 5

would make the circumference shorter than the diameter, which no circle can have.

Dividing kilometres by centimetres

For revolutions, put the distance and the circumference in the same unit. A speed in km/h and a radius in cm must both become cm (and per minute, if revolutions per minute are asked).

Using a circumference ratio as an area ratio

Circumferences in the ratio

3 : 5

mean radii

3 : 5

and areas

9 : 25

. Square the length ratio before using it for area.

Adding radii to add areas

A circle with the area of two circles of radii 8 and 15 has

R = \sqrt{64 + 225} = 17

, not

8 + 15 = 23

. Radii add only when circumferences are added.

Squaring the width

A path 2 m wide round a circle of radius 10 m has area

\pi(12^2 - 10^2) = 44\pi

, not

\pi \times 2^2

. The width is a difference of radii, not a radius.

Swapping inner and outer radii

The outer radius is always the larger. For a path outside, add the width to the circle's radius; for a path inside, subtract it.

Arc Length and Sector Perimeter

Learn this subtopic in the notes

Arc length as a fraction of the circumference

Arc length

l = \frac{\theta}{360^\circ} \times 2\pi r \qquad C = \frac{360^\circ}{\theta} \times l

$\theta$ central angle in degrees
$l$ arc length

Perimeter of a sector

P = 2r + \frac{\theta}{360^\circ} \times 2\pi r \qquad P_{\text{semicircle}} = \pi r + 2r

Radians: arc length equals radius times angle

Radian measure

l = r\theta \qquad \pi \text{ radians} = 180^\circ \qquad \theta_{\text{rad}} = \theta^\circ \times \frac{\pi}{180^\circ}

$\theta$ central angle in radians in the first formula

Common traps

Doubling a diameter that is already a diameter

If the stem gives the diameter, the circumference is

\pi d

. Writing

2\pi d

doubles the arc.

Taking the minor arc for the major one

A chord cuts the circle into two arcs. The smaller uses

\theta

; the larger uses

360^\circ - \theta

. Read which one the stem asks for.

Leaving out the radii

The arc alone is not the perimeter of a sector. Add both straight edges,

2r

.

Counting the radius once

A sector has TWO straight edges, one along each radius. A semicircle's straight edge is the diameter, which is also

2r

.

Putting degrees into l = rθ

The rule

l = r\theta

needs

\theta

in radians. With

60^\circ

, first write

\frac{\pi}{3}

; otherwise the arc comes out 60 times the radius.

Taking the chord for the arc

2r\sin\frac{\theta}{2}

is the straight chord between the arc's ends. The curved arc is

r\theta

, which is always longer.

Sector Area

Learn this subtopic in the notes

Sector area as a fraction of the circle's area

Area of a sector from its angle

A = \frac{\theta}{360^\circ} \times \pi r^2

$\theta$ central angle in degrees

Sector area from its arc: half the arc times the radius

Area of a sector from its arc

A = \frac{1}{2}lr \qquad A = \frac{1}{2}r^2\theta

$l$ arc length
$\theta$ central angle in radians

Comparing two sectors given in degrees and in radians

Ratio of two sector areas

\frac{A_1}{A_2} = \frac{\theta_1 r_1^2}{\theta_2 r_2^2}

$\theta_1, \theta_2$ the two angles, in the same unit

Finding a sector from its perimeter

Sector perimeter and area together

P = 2r + l \qquad r(P - 2r) = 2A

Common traps

Using 2πr for an area

2\pi r

is the circumference, a length. A sector's AREA takes its fraction of

\pi r^2

.

Stopping at r squared

Working back from an area gives

r^2

. Take the square root before choosing an option:

r^2 = 225

means

r = 15

.

Putting degrees into half r squared theta

\frac{1}{2}r^2\theta

needs

\theta

in radians. For a

45^\circ

sector use

\frac{\pi}{4}

, or use the 360 fraction instead.

Dropping the half

Arc times radius is TWICE the sector's area, just as base times height is twice a triangle's area.

Comparing a degree number with a radian number

60

and

\frac{\pi}{3} \approx 1.05

look very different but are the same angle. Convert before comparing, or a

1 : 1

ratio looks like

57 : 1

.

Comparing by angle when the radii are unknown

Equal angles give equal areas only in circles of equal radius. If the stem does not say the radii are equal, the angles alone do not fix the ratio.

Keeping the wrong root

The perimeter-and-area equation is a quadratic with two positive roots. Both make a real sector; the stem's condition (a whole number, or the options) picks one.

Combined and Shaded Regions

Learn this subtopic in the notes

Area of a segment: sector minus triangle

Area of a segment

A_{\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

\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

r_{\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}}

$a$ side of the outer square or triangle
$r, R$ radius of the circle inside, radius of the circle outside

Common traps

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

\frac{\sqrt{3}}{4}r^2

is the triangle only at

60^\circ

(and, by the same sine,

120^\circ

). At

90^\circ

it is

\frac{r^2}{2}

; in general use

\frac{1}{2}r^2\sin\theta

.

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^\circ

.

Squaring the difference for a hole of the same shape

A hexagon of side 5 with a hexagonal hole of side 2 leaves

\frac{3\sqrt{3}}{2}(25 - 4)

, not

\frac{3\sqrt{3}}{2}(5 - 2)^2

.

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

\frac{a}{2\sqrt{3}}

and the circle through the corners has radius

\frac{a}{\sqrt{3}}

, exactly twice as large.

Mensuration 2D formulas

Rectangles, Squares and Other Quadrilaterals

Area and perimeter of a rectangle and a square

Diagonal of a rectangle and of a square

Tiles, paths, cut-outs and the cost of covering

Area of a parallelogram, a trapezium and a rhombus

Common traps

Areas of Triangles

Half base times height, and two sides with the angle between them

Three sides: semi-perimeter, Heron's formula and the inradius

Area of an equilateral triangle and a regular hexagon

Similar figures: scale the sides, square the area

Common traps

Circumference, Wheels and Rings

Circumference, wheel revolutions and fencing a circle

Area of a circle and how it scales with the radius

Area of a ring, a circular track or a path round a circle

Common traps

Arc Length and Sector Perimeter

Arc length as a fraction of the circumference

Radians: arc length equals radius times angle

Common traps

Sector Area

Sector area as a fraction of the circle's area

Sector area from its arc: half the arc times the radius

Comparing two sectors given in degrees and in radians

Finding a sector from its perimeter

Common traps

Combined and Shaded Regions

Area of a segment: sector minus triangle

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

Inscribed figures: a circle in a square or triangle, a square or hexagon in a circle

Common traps

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