PYQ Vault
Ebooks

AFCAT Numerical Ability · Formula sheet

Area and Perimeter formulas

19 formulas and 39 common traps for AFCAT Numerical Ability Area and Perimeter, grouped by subtopic.

Full notes with worked examples

Rectangles and Squares: Area, Perimeter, Diagonal and Cost

Learn this subtopic in the notes

Area, perimeter and diagonal of a rectangle and a square

Rectangle and square

A=lb,P=2(l+b),d=l2+b2;Asquare=a2=d22A = lb,\quad P = 2(l + b),\quad d = \sqrt{l^2 + b^2};\qquad A_{\text{square}} = a^2 = \dfrac{d^2}{2}

Finding the sides from two facts

Two facts, two sides

l+b=P2,2lb=(l+b)2−d2l + b = \dfrac{P}{2},\qquad 2lb = (l + b)^2 - d^2

Fencing, flooring, tiles and paths

Path outside a rectangle

path=(l+2w)(b+2w)−lb=2w(l+b)+4w2\text{path} = (l + 2w)(b + 2w) - lb = 2w(l + b) + 4w^2

Common traps

A square from its diagonal

A square with a diagonal of 6 cm has an area of 362=18\tfrac{36}{2} = 18 cm², not 36 cm². The diagonal is longer than the side by a factor of 2\sqrt 2.

Perimeter is a length

A 9 cm by 4 cm rectangle has area 36 cm² and perimeter 26 cm. Check the unit an option carries: cm² means an area was asked.

Halve the perimeter before splitting it

With a perimeter of 56 and a ratio of 4:34 : 3, split 28, not 56. Splitting the full perimeter doubles both sides and makes the area four times too big.

Shape is not size

'The length is three times the breadth' is true of a 3 by 1 rectangle and a 30 by 10 one. A statement like that needs a second fact about size before any area follows.

A path adds its width on both sides

A path 3 m wide round a 30 m lawn makes the outer length 36 m, not 33 m. The width is added at each end.

Same units before dividing

A 6 m by 4 m floor is 24 m², and a 40 cm tile is 0.16 m². Dividing 24 by 1,600 (the tile in cm²) gives nonsense.

Throw out the root that cannot fit

A quadratic for a path's width has two roots. A width that is more than half the plot's breadth would leave nothing inside, so it is rejected.

Triangles: Base and Height, Right Triangles and Heron's Formula

Learn this subtopic in the notes

Half the base times the height

Area of a triangle

A=12×b×h,h=2AbA = \tfrac12 \times b \times h,\qquad h = \dfrac{2A}{b}

Right triangles and Pythagoras

Pythagoras

c2=a2+b2,(a+b)2=(a−b)2+4abc^2 = a^2 + b^2,\qquad (a + b)^2 = (a - b)^2 + 4ab

Heron's formula and the isosceles triangle

Heron's formula

A=s(s−a)(s−b)(s−c),s=a+b+c2A = \sqrt{s(s - a)(s - b)(s - c)},\quad s = \dfrac{a + b + c}{2}

Common traps

Keep the half

A base of 14 and a height of 9 give 63, not 126. The product alone is the rectangle round the triangle.

The height is not the slant side

In a triangle with a 10 cm slant side and a 7 cm height on a 12 cm base, the area is 12×12×7=42\tfrac12 \times 12 \times 7 = 42. The slant side is not used.

The hypotenuse faces the right angle

With sides 9, 12 and 15, the legs are 9 and 12 and the area is 54. Taking 12 and 15 as the base and height gives 90, which is wrong: 15 is opposite the right angle.

Legs, not the hypotenuse, in the isosceles case

An isosceles right triangle with hypotenuse 424\sqrt 2 has legs of 4 and area 8. Squaring the hypotenuse and halving gives 16, twice too much.

s is half the perimeter

For sides 9, 10 and 17, s=18s = 18. Starting from 36 makes every bracket wrong and the root of a non-square.

Halve the base for the height

With equal sides 26 and base 20, the height uses half the base: 676−100\sqrt{676 - 100}. Using the whole base gives 276\sqrt{276}, which is not the height.

Rhombus, Trapezium, Parallelogram and Regular Polygons

Learn this subtopic in the notes

Rhombus and kite: working from the diagonals

Rhombus

A=12d1d2,side=(d12)2+(d22)2A = \tfrac12 d_1 d_2,\qquad \text{side} = \sqrt{\left(\tfrac{d_1}{2}\right)^2 + \left(\tfrac{d_2}{2}\right)^2}

Trapezium, parallelogram and any quadrilateral

Trapezium and quadrilateral

Atrap=12(a+b)h,Aquad=12d(h1+h2)A_{\text{trap}} = \tfrac12(a + b)h,\qquad A_{\text{quad}} = \tfrac12 d(h_1 + h_2)

Equilateral triangles and the regular hexagon

Equilateral triangle and hexagon

A△=34a2,Ahex=6×34a2=332a2A_{\triangle} = \dfrac{\sqrt 3}{4}a^2,\qquad A_{\text{hex}} = 6 \times \dfrac{\sqrt 3}{4}a^2 = \dfrac{3\sqrt 3}{2}a^2

Angles and perimeter of a regular polygon

Polygon angles

sum=(n−2)×180°,exterior=360°n\text{sum} = (n - 2) \times 180°,\qquad \text{exterior} = \dfrac{360°}{n}

Common traps

Halve the diagonals first

With diagonals 16 and 30 the side is 82+152=17\sqrt{8^2 + 15^2} = 17. Putting the whole diagonals into Pythagoras gives 34, twice the side.

Half the product

Diagonals 9 and 16 give an area of 72, not 144. The product of the diagonals is the rectangle drawn round the rhombus.

Each height goes with its own sides

Sides 20 and 16 with a height of 8 on the 20 cm sides give an area of 160. Multiplying 8 by 16 pairs the height with the wrong side and gives 128.

The slant side is not the height

In a trapezium the height is the perpendicular gap between the parallel sides. A slanting side of 10 cm tells you nothing until it is turned into that gap.

Height has a 2 under it, area a 4

For side 10 the height is 535\sqrt 3 and the area 25325\sqrt 3. Swapping the two denominators gives an area of 50350\sqrt 3, twice too big.

Six triangles, not three

A regular hexagon of side 10 holds six equilateral triangles of side 10, so its area is 1503150\sqrt 3. Counting three gives half the answer.

Sum or each

(n−2)×180°(n - 2) \times 180° is the total. For an octagon that is 1080°; each angle is 135°. Options often carry both.

Exterior angles always total 360°

Whatever the number of sides, the exterior angles of a polygon add to 360°. Going from an angle to the number of sides is quickest through the exterior angle.

Circles, Rings, Sectors and Grazing

Learn this subtopic in the notes

Circumference and area

Circle

C=2πr=πd,A=πr2C = 2\pi r = \pi d,\qquad A = \pi r^2

Rings, tracks and two circles

Area of a ring

A=π(R2−r2)=π(R+r)(R−r)A = \pi(R^2 - r^2) = \pi(R + r)(R - r)

Semicircles, arcs, sectors and clock hands

Arc and sector

arc=θ360×2πr,sector=θ360×πr2\text{arc} = \dfrac{\theta}{360} \times 2\pi r,\qquad \text{sector} = \dfrac{\theta}{360} \times \pi r^2

Animals tied with a rope

Grazed area

area=θ360×π (rope)2\text{area} = \dfrac{\theta}{360} \times \pi\,(\text{rope})^2

Common traps

Radius or diameter

A circumference of 88 gives a radius of 14 and a diameter of 28. Options usually carry both; read which one the question asks for.

Convert the distance first

A wheel's circumference is in cm and the road is often in km or m. 660 m is 66,000 cm; dividing 660 by 132 gives 5 turns instead of 500.

Difference of squares, not square of the difference

For radii 42 and 35 the ring is π(422−352)\pi(42^2 - 35^2), which is 1694 with 22/7. Squaring the width, π×72=154\pi \times 7^2 = 154, is far too small.

Width adds to the radius

A 7 m track round a circle of radius 14 m makes the outer radius 21 m and the outer diameter 42 m. Adding 7 to the diameter gives 35, which is wrong.

A semicircle's edge includes its diameter

The perimeter of a semicircle of radius 14 is 44+28=7244 + 28 = 72. The curved part alone, 44, is the arc, not the perimeter.

Minutes into degrees

In 20 minutes the minute hand turns 20×6°=120°20 \times 6° = 120°, a third of the circle. Using 20360\tfrac{20}{360} of the circle treats minutes as degrees.

Only the angle at the peg

At the corner of a square field the goat grazes a quadrant, 14\tfrac14 of the circle. Taking the whole circle counts grass on the far side of the fence.

Check the rope against the walls

The sector rule needs a rope shorter than the sides beside the peg. If the rope is longer than a side, the animal walks round that corner and grazes an extra sector there.

Combined Figures: Inscribed Shapes and Shaded Regions

Learn this subtopic in the notes

Figures inscribed in one another

Square and circle

circle in square: r=a2;square in circle: d=2r, A=2r2\text{circle in square: } r = \tfrac a2;\qquad \text{square in circle: } d = 2r,\ A = 2r^2

Shaded regions by subtraction

Shaded area

shaded=outer−removed,segment=sector−△\text{shaded} = \text{outer} - \text{removed},\qquad \text{segment} = \text{sector} - \triangle

Building a figure from pieces

Rectangle with semicircular ends

A=lb+π(b2)2,P=2l+πbA = lb + \pi\left(\tfrac b2\right)^2,\qquad P = 2l + \pi b

Common traps

Which length is shared

For a square inside a circle of radius 7, the diameter 14 is the square's diagonal, giving 98. Taking 14 as the side gives 196, the square outside the circle.

The circumradius is not half the side

An equilateral triangle of side 636\sqrt 3 has a circumradius of 6, from R=a3R = \tfrac{a}{\sqrt 3}. Half its side, 333\sqrt 3, is neither its circumradius nor its inradius.

A segment is not a sector

With radius 14 and a right angle at the centre, the sector is 154 but the segment is only 56. The triangle between the radii must come off.

Read what is shaded

In a square with quadrants at its corners, the shaded part may be the quadrants (154 for side 14) or what is left between them (42). Both numbers are usually among the options.

Two semicircles, one circle

A 50 m by 21 m ground with round ends adds one circle of radius 10.5, not two. Adding two full circles doubles the curved part.

Do not count the overlap twice

Splitting a T shape into a 6 by 2 bar and a stem, measure the stem from below the bar. Taking its full height through the bar counts the shared square twice.

Change in Dimensions: Scaling and Percentage Changes

Learn this subtopic in the notes

Scaling a figure and comparing areas

Scale factor k

P2P1=k,A2A1=k2\dfrac{P_2}{P_1} = k,\qquad \dfrac{A_2}{A_1} = k^2

Percentage changes in the sides

Two percentage changes

area change=a+b+ab100 %\text{area change} = a + b + \dfrac{ab}{100}\ \%

Common traps

Square the ratio for area

Radii 5:75 : 7 give areas 25:4925 : 49. Keeping 5:75 : 7 for the areas is the option most often wrongly picked.

Square roots on the way back

Areas 36:4936 : 49 mean sides 6:76 : 7. Giving 36:4936 : 49 for the sides, or squaring again, are the usual wrong options.

Percentages multiply, they do not add

Both sides of a rectangle up 20% raise the area by 44%, not 40%: 1.2×1.2=1.441.2 \times 1.2 = 1.44.

Undoing a rise is a smaller fall

After a 25% rise in the length, the breadth must fall 20%, not 25%: 1.25×0.8=11.25 \times 0.8 = 1.

More AFCAT Numerical Ability formula sheets