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Triangles formulas

28 formulas, 1 reference table and 31 common traps for CDS Mathematics Triangles, grouped by subtopic.

Full notes with worked examples

Angles of a Triangle

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Angle sum and the exterior angle

Angle sum and exterior angle

A+B+C=180∘,∠ACD=A+BA + B + C = 180^\circ, \qquad \angle ACD = A + B

Where two angle bisectors meet

Bisector angles

∠BIC=90∘+A2,∠BEC=90∘−A2\angle BIC = 90^\circ + \tfrac{A}{2}, \qquad \angle BEC = 90^\circ - \tfrac{A}{2}

Isosceles triangles and parallel lines

Base angles of an isosceles triangle

AB=AC  ⇒  ∠B=∠C=90∘−A2AB = AC \;\Rightarrow\; \angle B = \angle C = 90^\circ - \tfrac{A}{2}

Common traps

Opposite angles, not the adjacent one

The exterior angle at CC equals A+BA + B, the two angles AWAY from CC. It is 180∘180^\circ minus the interior angle at CC, not the sum including it.

Some questions answer themselves

A=B−CA = B - C with A+B+C=180∘A + B + C = 180^\circ forces B=90∘B = 90^\circ, so AA is acute whatever the statements say. In a data-sufficiency item, first check whether the stem alone already decides the answer.

Two angles form where the bisectors cross

Crossing lines make an angle and its supplement, say 115∘115^\circ and 65∘65^\circ. The angle ∠BIC\angle BIC faces the side BCBC and is always obtuse, so use the larger one; the smaller would make the third angle negative.

Equal angles face the equal sides

AB=ACAB = AC makes ∠B=∠C\angle B = \angle C, the angles at the ends of the base, not ∠A=∠B\angle A = \angle B. Name the angle opposite each equal side before you write it down.

Triangle Inequalities

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The triangle inequality

Triangle inequality

∣b−c∣<a<b+c|b - c| < a < b + c

The larger side faces the larger angle

Acute, right or obtuse

c2≶a2+b2(c the longest side)c^2 \lessgtr a^2 + b^2 \quad (c \text{ the longest side})

The medians against the perimeter

Sum of the medians

34(a+b+c)<ma+mb+mc<a+b+c\tfrac34 (a + b + c) < m_a + m_b + m_c < a + b + c

Common traps

Equal is not enough

4+5=94 + 5 = 9 gives a flat line, not a triangle. The inequality is strict.

When two options are both true

If a quantity is always positive, 'non-negative' is also true. CDS has printed both; mark the tighter description, 'positive', which is the one the setter means.

Compare the angles, not the picture

A figure that is not drawn to scale can make any side look the longest. Name the angles in the small triangle that holds the two sides, then compare the angles opposite them; the larger angle faces the longer side.

Read which way the inequality points

Statement questions print the true result reversed: 'the sum of two sides is less than twice the median'. The true fact is greater. Check any printed inequality on an equilateral triangle, where each median is 32\dfrac{\sqrt3}{2} of the side.

Congruence and Similarity

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The congruence rules

Rules that fix a triangle

SSS, SAS, ASA, AAS, RHS(not AAA, not SSA)\text{SSS},\ \text{SAS},\ \text{ASA},\ \text{AAS},\ \text{RHS} \qquad (\text{not AAA, not SSA})

Similar triangles and the shared-angle pattern

Shared-angle similarity

∠ADC=∠BAC  ⇒  AC2=DC⋅BC\angle ADC = \angle BAC \;\Rightarrow\; AC^2 = DC \cdot BC

Crossing lines between two poles

Height of the crossing point

h=aba+bh = \dfrac{ab}{a + b}

Common traps

The angle must be the included one

SAS needs the angle between the two named sides. A statement listing 'two sides and an angle' without saying which angle is not a valid rule.

Match vertices by the statement, not by the letters' positions

In △ABR∼△PQR\triangle ABR \sim \triangle PQR, BRBR matches QRQR and ARAR matches PRPR. Write the three pairs out before dividing, or the scale factor gets applied to the wrong side.

Not the average

Poles of 1212 and 2424 give 88, not 1818 (the average) and not anything that uses the distance between them. The crossing height is always less than the shorter pole.

Parallels, Midpoints and the Bisector Theorem

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A line parallel to one side

Basic proportionality

DE∥BC  ⇒  ADDB=AEEC,DEBC=ADABDE \parallel BC \;\Rightarrow\; \dfrac{AD}{DB} = \dfrac{AE}{EC}, \quad \dfrac{DE}{BC} = \dfrac{AD}{AB}

The midpoint theorem

Midpoint theorem

DE∥BC,DE=12BCDE \parallel BC, \quad DE = \tfrac12 BC

The angle-bisector theorem

Angle-bisector theorem

BDDC=ABAC\dfrac{BD}{DC} = \dfrac{AB}{AC}

Common traps

DE over BC is part over WHOLE

DEBC\dfrac{DE}{BC} equals ADAB\dfrac{AD}{AB}, not ADDB\dfrac{AD}{DB}. With AD:DB=1:4AD : DB = 1 : 4, DEDE is one-fifth of BCBC, not one-quarter.

Count every side once

When a triangle is cut into a middle triangle and three corner triangles, each side of the middle triangle belongs to exactly one corner triangle. So adding the corner perimeters counts the outer perimeter once and the middle perimeter once.

Pair each segment with the side that touches it

BDBD goes with ABAB — both end at BB — and DCDC with ACAC. Writing BDDC=ACAB\dfrac{BD}{DC} = \dfrac{AC}{AB} swaps the answer, and the swapped value is always one of the options.

Ratio of Areas of Triangles

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Areas of similar triangles

Area ratio of similar triangles

Area1Area2=(side1side2)2\dfrac{\text{Area}_1}{\text{Area}_2} = \left(\dfrac{\text{side}_1}{\text{side}_2}\right)^2

Joining the midpoints, again and again

Repeated midpoint triangles

Area after n steps=Area4 n\text{Area after } n \text{ steps} = \dfrac{\text{Area}}{4^{\,n}}

Same height, so compare the bases

Common height

[ABD][ADC]=BDDC\dfrac{[ABD]}{[ADC]} = \dfrac{BD}{DC}

Common traps

The length ratio is the square root

Cutting off one-third of the area makes DE=BC3DE = \dfrac{BC}{\sqrt3}, not BC3\dfrac{BC}{3}. And cutting a triangle into two EQUAL parts leaves the small triangle as half the whole, so k=12k = \dfrac{1}{\sqrt2}.

Count the steps, not the positions

The 4th and the 7th triangles are 33 steps apart, so the ratio is 43=644^3 = 64, not 444^4 or 474^7. Whether the original counts as the first does not matter.

The bases must lie on one line

The shared-height rule needs both bases on the same straight line (or on two parallel lines). Two triangles that share a vertex but have bases pointing different ways do not share a height.

Pythagoras Theorem

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Pythagorean triples

Triple generator

(m2−n2)2+(2mn)2=(m2+n2)2(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2

Perimeter and area together

Hypotenuse from perimeter and area

(a+b)2=c2+4Δ,c=P2−4Δ2P(a + b)^2 = c^2 + 4\Delta, \qquad c = \dfrac{P^2 - 4\Delta}{2P}

Ladders, poles and walks

Pythagoras

c2=a2+b2c^2 = a^2 + b^2

Common traps

A ratio fixes only the shape

'Sides in the ratio x:(x−1):(x−18)x : (x - 1) : (x - 18)' strictly allows any multiple. The paper means the sides ARE those expressions; solve for xx and reject any root that makes a side negative.

The squares of ALL three sides

a2+b2+c2=2c2a^2 + b^2 + c^2 = 2c^2, not c2c^2. A question giving 'the sum of the squares of the sides' wants c=sum÷2c = \sqrt{\text{sum} \div 2}.

Poles use the DIFFERENCE of heights

The vertical leg between two pole tips is h1−h2h_1 - h_2, not either height. That is also why 'heights differ by 1010 m' is enough information on its own.

The Altitude to the Hypotenuse

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The altitude is leg × leg ÷ hypotenuse

Altitude to the hypotenuse

p=abc,1p2=1a2+1b2p = \dfrac{ab}{c}, \qquad \dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}

The three similar triangles

Geometric-mean relations

p2=mn,AB2=m(m+n),AC2=n(m+n)p^2 = mn, \qquad AB^2 = m(m + n), \qquad AC^2 = n(m + n)

Altitudes of any triangle

Altitude from the area

ha=2Δa,a:b:c=1ha:1hb:1hch_a = \dfrac{2\Delta}{a}, \qquad a : b : c = \dfrac{1}{h_a} : \dfrac{1}{h_b} : \dfrac{1}{h_c}

Common traps

Find the right angle first

'Right-angled at BB' makes ACAC the hypotenuse. If the question gives BC=10BC = 10 and AC=12AC = 12 with the right angle at BB, the missing leg is 144−100\sqrt{144 - 100}, not 144+100\sqrt{144 + 100}.

A leg uses its own piece and the WHOLE hypotenuse

AB2=BD⋅BCAB^2 = BD\cdot BC, where BDBD is the piece touching BB and BCBC is the whole hypotenuse. Using BD⋅DCBD\cdot DC gives the altitude, not the leg, and using DCDC pairs the leg with the wrong piece.

Inverse, not direct

Altitudes 4:5:64 : 5 : 6 give sides 14:15:16=15:12:10\dfrac14 : \dfrac15 : \dfrac16 = 15 : 12 : 10, not 4:5:64 : 5 : 6. Then reorder to match the sides the question names.

Medians and Apollonius Theorem

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Apollonius' theorem

AB2+AC2=2(AD2+BD2)AB^2 + AC^2 = 2\left(AD^2 + BD^2\right)

Points on a leg of a right triangle

Two midpoints of the legs

AQ2+CP2=54 AC2AQ^2 + CP^2 = \tfrac54\,AC^2

Subtracting across an altitude

Across an altitude

AB2−AC2=BD2−DC2AB^2 - AC^2 = BD^2 - DC^2

Common traps

Half the base, not the base

Apollonius uses BD2=(BC2)2BD^2 = \left(\dfrac{BC}{2}\right)^2. Putting BC2BC^2 in its place is the usual slip, and it gives a median that is too short.

Measure from the right angle

AP2=AB2+BP2AP^2 = AB^2 + BP^2 uses the piece of the leg from the right angle BB to PP. With MM and NN trisecting BCBC, BNBN is two-thirds of BCBC, not one-third.

The general form needs a perpendicular

AB2−AC2=BD2−DC2AB^2 - AC^2 = BD^2 - DC^2 holds only when ADAD is the altitude. The isosceles form AB2−AD2=BD⋅DCAB^2 - AD^2 = BD\cdot DC is different: there DD can be any point of BCBC.

Centres of a Triangle

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Circumradius and inradius

Right-triangle radii

R=c2,r=a+b−c2R = \dfrac c2, \qquad r = \dfrac{a + b - c}{2}

Placing the triangle on axes

Circle in the corner of angle A

ρ=r 1−sin⁡(A/2)1+sin⁡(A/2)\rho = r\,\dfrac{1 - \sin(A/2)}{1 + \sin(A/2)}

The four centres and where they lie

CentreAcute triangleRight triangleObtuse triangle
Centroidinsideinsideinside
Incentreinsideinsideinside
Circumcentreinsidemidpoint of the hypotenuseoutside
Orthocentreinsideat the right-angle vertexoutside
In a right triangle the two legs are themselves altitudes, so they meet at the right angle.
The centroid and incentre never leave the triangle; the circumcentre and orthocentre do when an angle is obtuse.

Common traps

'On the triangle' is not 'inside'

The orthocentre of a right triangle is a vertex, and the circumcentre is the midpoint of a side: both lie ON the triangle. A statement saying 'inside' or 'outside' for a right triangle is false.

Halve it

a+b−ca + b - c is the incircle's DIAMETER in a right triangle. For 8,15,178, 15, 17 it is 66, and the radius is 33.

Half the angle

The corner circle's centre is on the bisector, so the distance from the vertex is ρsin⁡(A/2)\dfrac{\rho}{\sin(A/2)}, not ρsin⁡A\dfrac{\rho}{\sin A}. Get sin⁡A2\sin\dfrac A2 from cos⁡A\cos A with sin⁡2A2=1−cos⁡A2\sin^2\dfrac A2 = \dfrac{1 - \cos A}{2}.

Sine and Cosine Rules

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The sine rule

Sine rule

asin⁡A=bsin⁡B=csin⁡C=2R\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R

The cosine rule and ½ab sin C

Cosine rule and area

a2=b2+c2−2bccos⁡A,Δ=12bcsin⁡Aa^2 = b^2 + c^2 - 2bc\cos A, \qquad \Delta = \tfrac12 bc\sin A

Common traps

Sines, not angles

Angles in the ratio 1:2:31 : 2 : 3 give sides sin⁡30∘:sin⁡60∘:sin⁡90∘=1:3:2\sin 30^\circ : \sin 60^\circ : \sin 90^\circ = 1 : \sqrt3 : 2, not 1:2:31 : 2 : 3.

cos 120° is negative

At 120∘120^\circ the term −2bccos⁡A-2bc\cos A becomes +bc+bc, so the opposite side is longer than Pythagoras would give. Dropping the sign gives the 60∘60^\circ answer, which is always an option.

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