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CDS Mathematics · Formula sheet

Mensuration 2D formulas

31 formulas and 20 common traps for CDS Mathematics Mensuration 2D, grouped by subtopic.

Full notes with worked examples

Areas of Triangles

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Base × height, or two sides and the angle between them

Two sides and the included angle

Area=12bh=12 absin⁡C\text{Area} = \tfrac12 bh = \tfrac12\,ab\sin C

Heron's formula — three sides only

Heron's formula

Area=s(s−a)(s−b)(s−c),s=a+b+c2\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \tfrac{a+b+c}{2}

A right triangle from its perimeter

Area from the sum of the legs

Area=ab2=(a+b)2−c24\text{Area} = \frac{ab}{2} = \frac{(a+b)^2 - c^2}{4}

The equilateral triangle

Equilateral triangle, side a

h=32a,Area=34a2=h23h = \tfrac{\sqrt3}{2}a, \qquad \text{Area} = \tfrac{\sqrt3}{4}a^2 = \tfrac{h^2}{\sqrt3}

Scale the sides, square the area

Similar figures

A1A2=(s1s2)2\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2

Recover the sides first

Sides from altitudes

a:b:c=1ha:1hb:1hca : b : c = \frac{1}{h_a} : \frac{1}{h_b} : \frac{1}{h_c}

Common traps

The angle must be the one between the two sides

12absin⁡C\dfrac12 ab\sin C needs CC to sit between aa and bb. If the stem gives two angles and one side, find the third angle first and use the side opposite it, as in Area=c2sin⁡Asin⁡B2sin⁡C\text{Area} = \dfrac{c^2\sin A\sin B}{2\sin C}.

Two sides plus the perimeter is three sides

A stem that gives two sides and the perimeter has given all three: subtract to get the third, then test for a right triangle before reaching for Heron.

Don't solve for the legs unless you must

Solving the quadratic for aa and bb works but costs a minute. The identity 2ab=(a+b)2−c22ab = (a + b)^2 - c^2 gives the area in one line.

Altitudes and sides go the opposite way

The longest side has the shortest altitude. Altitudes 3:5:63 : 5 : 6 give sides 13:15:16=10:6:5\dfrac13 : \dfrac15 : \dfrac16 = 10 : 6 : 5, not 3:5:63 : 5 : 6.

Rectangles, Squares & Other Quadrilaterals

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Rectangles and squares as a pair of equations

Square from its diagonal

Area=s2=d22\text{Area} = s^2 = \frac{d^2}{2}

Borders, paths and percentage changes

Path of width w outside

Path=(l+2w)(b+2w)−lb\text{Path} = (l + 2w)(b + 2w) - lb

The rhombus — half the product of the diagonals

Rhombus

Area=12d1d2,4a2=d12+d22\text{Area} = \tfrac12 d_1 d_2, \qquad 4a^2 = d_1^2 + d_2^2

Parallelogram and trapezium

Apar=absin⁡θ,Atrap=12(p+q) hA_{\text{par}} = ab\sin\theta, \qquad A_{\text{trap}} = \tfrac12 (p + q)\,h

Cut it into triangles

Quadrilateral split along a diagonal

[ABCD]=[ABC]+[ACD][ABCD] = [ABC] + [ACD]

Common traps

The diagonal is an option on purpose

When a stem gives a square's diagonal (or the product of both diagonals), the diagonal itself is usually one of the options. Read the question again: it asks for the side.

Width counts twice

A 11 m path on every side adds 22 m to the length and 22 m to the breadth. Adding 11 m to each gives the wrong outer rectangle.

An unequal-leg trapezium needs two right triangles

When the slant sides differ, drop both perpendiculars. The two overhangs add to the difference of the parallel sides, and each leg gives its own Pythagoras equation with the same height. Subtract the two equations to find the overhangs.

Equal Perimeters & Re-bent Wires

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The wire keeps its length

Length is conserved

perimeter of the old shape=perimeter of the new shape\text{perimeter of the old shape} = \text{perimeter of the new shape}

Same perimeter: the rounder shape wins

Equal perimeter

circlesquare=4π,trianglesquare=439\frac{\text{circle}}{\text{square}} = \frac{4}{\pi}, \qquad \frac{\text{triangle}}{\text{square}} = \frac{4\sqrt3}{9}

Same perimeter: the square beats every rectangle

Square minus rectangle

s2−lb=(l−b2)2s^2 - lb = \left(\frac{l-b}{2}\right)^2

Common traps

A semicircle's boundary includes the diameter

A wire bent into a semicircle forms the arc AND the straight diameter: πr+2r\pi r + 2r. Using πr\pi r alone gives a larger radius. When no printed option matches either reading, 'None of the above' is the answer the paper wants.

Equal perimeter and equal area flip the answer

Equal perimeters: the circle has the MORE area. Equal areas: the circle has the LESS perimeter. Both are the same fact, so read which quantity the stem fixes before choosing.

Circumference, Wheels & Rings

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Circumference and wheel revolutions

Wheel revolutions

n=distance2πrn = \frac{\text{distance}}{2\pi r}

Area grows as the square of the radius

Circle

A=πr2,A1A2=(r1r2)2A = \pi r^2, \qquad \frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2

The ring between two circles

Ring from a tangent chord

Ring=π(R2−r2)=π(chord2)2\text{Ring} = \pi(R^2 - r^2) = \pi\left(\tfrac{\text{chord}}{2}\right)^2

Common traps

Radius or diameter?

Stems switch between the two. Revolutions use the circumference πd=2πr\pi d = 2\pi r; dividing by πr\pi r doubles the answer and that doubled value is usually an option.

Equal-area rings get thinner outward

Split a disc into rings of equal area and the radii go as 1,2,3,…\sqrt1, \sqrt2, \sqrt3, \ldots. The ratio of neighbouring radii, 1+1m\sqrt{1 + \tfrac1m}, falls as you move out.

"Cannot be determined" is the bait

The ring's area needs only R2−r2R^2 - r^2, not RR and rr separately. When an option says the data are insufficient, the tangent-chord fact is usually what makes it wrong.

Arcs, Sectors & Segments

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Arc length — radius times angle in radians

Arc length

l=rθ=θ∘360∘×2πrl = r\theta = \frac{\theta^\circ}{360^\circ}\times 2\pi r

Sector area — a slice of the disc

Sector

A=θ360∘ πr2=12rlA = \frac{\theta}{360^\circ}\,\pi r^2 = \tfrac12 r l

The segment — sector minus triangle

Minor segment

Segment=12r2(θ−sin⁡θ)\text{Segment} = \tfrac12 r^2(\theta - \sin\theta)

Common traps

Never put degrees into rθ

l=rθl = r\theta needs radians. Multiplying the radius by 3030 instead of π6\dfrac{\pi}{6} is the commonest slip, and the options rarely include anything that wild, so a strange result is the warning.

Sector is not segment

The sector includes the triangle; the segment does not. The sector's value is almost always among the options for a segment question.

Inscribed & Circumscribed Figures

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Square in a circle, circle in a square

In a circle of radius r

square=2r2,equilateral triangle=334r2\text{square} = 2r^2, \qquad \text{equilateral triangle} = \tfrac{3\sqrt3}{4}r^2

A triangle's incircle and circumcircle

Inradius and circumradius

r=Δs,R=abc4Δr = \frac{\Delta}{s}, \qquad R = \frac{abc}{4\Delta}

Figures in a semicircle or a quarter circle

Square in a semicircle

s2=45r2s^2 = \tfrac45 r^2

A square or rectangle wedged into a corner

Square in the right angle

s=aba+bs = \frac{ab}{a + b}

Regular polygons — hexagon and octagon

Hexagon and octagon

Ahex=332a2,octagon side=a(2−1)A_{\text{hex}} = \tfrac{3\sqrt3}{2}a^2, \qquad \text{octagon side} = a(\sqrt2 - 1)

Common traps

The inradius and the circumradius swap easily

For an equilateral triangle a3\dfrac{a}{\sqrt3} is the circumradius and a23\dfrac{a}{2\sqrt3} the inradius. Check with R=2rR = 2r: the circle through the corners is the bigger one.

Side or perimeter?

These stems often ask for the perimeter or the area of the square, and the side itself is an option. Finish the question.

Touching Circles

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Join the centres

Distance between centres

O1O2=r1+r2  (outside),O1O2=R−r  (inside)O_1O_2 = r_1 + r_2 \;(\text{outside}), \qquad O_1O_2 = R - r \;(\text{inside})

The gap between touching circles

Three equal touching circles

gap=34(2r)2−3⋅60∘360∘πr2=r22(23−π)\text{gap} = \frac{\sqrt3}{4}(2r)^2 - 3\cdot\frac{60^\circ}{360^\circ}\pi r^2 = \frac{r^2}{2}(2\sqrt3 - \pi)

Common traps

Reject the root that doesn't fit the figure

The touching condition is a quadratic, and one root is usually impossible: a circle wider than the rectangle, or one that would overlap the other. Check the root against the figure before choosing.

The sectors add to half a circle, not a whole one

Three 60∘60^\circ sectors make 180∘180^\circ, half a circle. Subtracting a whole circle gives a negative gap, and an option built on it is usually printed.

Combined & Shaded Regions

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The whole minus the holes

Shaded area

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

Semicircles on a divided diameter

Arcs on a split diameter

πr1+πr2+⋯=πRwhen r1+r2+⋯=R\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

lens=a2(π2−1)\text{lens} = a^2\left(\frac{\pi}{2} - 1\right)

Figures built from pieces

Composite figure

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

Common traps

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.

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.

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