SSC CGL Quantitative Aptitude · Formula sheet

Triangles formulas

19 formulas, 1 reference table and 49 common traps for SSC CGL Quantitative Aptitude Triangles, grouped by subtopic.

Angles of a Triangle

Angle sum of a triangle, with ratios and equations

Angle sum

\angle A + \angle B + \angle C = 180^\circ \qquad \text{one part} = \frac{180^\circ}{a + b + c}

$a, b, c$ the numbers in the ratio of the angles

The exterior angle of a triangle

Exterior angle

\angle ACD = \angle A + \angle B = 180^\circ - \angle ACB

$D$ a point on BC extended beyond C

Isosceles and equilateral triangles

Base angle of an isosceles triangle

\text{base angle} = \frac{180^\circ - \text{apex}}{2}

Common traps

Dividing 180° by one part of the ratio

The size of one part is 180° divided by the sum of all the parts. Only then multiply by the part the question asks for.

Allowing two obtuse or right angles

Two angles of 90° or more already reach 180°, so a triangle has at most one right or obtuse angle.

Giving the difference in parts, not degrees

With angles in the ratio 2 : 5 : 11 the difference between the largest and smallest is 9 parts of 10°, which is 90°, not 9°.

Adding the interior angle beside the exterior one

The exterior angle at C equals ∠A + ∠B, the two angles away from C. The interior angle at C is its supplement, not part of the sum.

Answering with the exterior angle as an interior one

An exterior angle of 120° is not an angle of the triangle. The interior angle at that corner is 60°, and the largest interior angle has to be found from the three interior angles.

Putting the equal angles at the wrong corners

The equal angles are opposite the equal sides. With AC = BC the equal angles are ∠A and ∠B, and ∠C is the apex.

Halving the apex to get a base angle

Each base angle is half of what is left after the apex: (180° − apex) ÷ 2. An apex of 80° gives base angles of 50°, not 40°.

Sides of a Triangle

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Triangle inequality: the third side lies between the difference and the sum

Range of the third side

|a - b| < x < a + b \qquad \text{each side} < \frac{a + b + c}{2}

$a, b$ the two known sides
$x$ the third side

Pythagoras theorem, triples and the acute or obtuse test

Pythagoras and the equilateral height

c^2 = a^2 + b^2 \qquad h = \frac{\sqrt{3}}{2}a \qquad PQ^2 - QS^2 = PR^2 - SR^2

$c$ the hypotenuse
$h$ the height of an equilateral triangle of side a
$S$ the foot of the perpendicular from P to QR

The altitude to the hypotenuse of a right triangle

Altitude to the hypotenuse

BD = \frac{AB \times BC}{AC} \qquad AB^2 = AD \times AC \qquad BD^2 = AD \times DC

$B$ the right angle
$D$ the foot of the altitude on the hypotenuse AC

Common traps

Including the ends of the range

A third side equal to the sum or to the difference of the other two flattens the triangle into a line, so the range uses strict signs at both ends.

Counting from the difference itself

With sides 4 and 10 the third side runs from 7 to 13, not from 6 to 14. That gives 7 whole-number values, which is 2 × 4 − 1.

Repeating the short side of an isosceles triangle

With sides 4 cm and 9 cm, the triangle 4, 4, 9 fails because 4 + 4 is less than 9. The repeated side must be the 9 cm one.

Testing with a side that is not the longest

The acute or obtuse test compares the square of the longest side with the sum of the squares of the other two. Any other pairing tells you nothing.

Using the whole base in an isosceles triangle

The perpendicular from the apex splits the base into two equal halves. Put half the base into Pythagoras, then double the result if the base is asked.

Taking the radius as the hypotenuse

A right triangle with its corners on a circle has the diameter as its hypotenuse. A circle of radius 5 cm gives a hypotenuse of 10 cm.

Pairing a leg with the far piece

AB² = AD × AC uses the piece AD that touches A. Using DC instead gives BC², the other leg.

Congruent Triangles

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Reading the matching corners from a congruence statement

Matching parts

\triangle ABC \cong \triangle PQR \;\Rightarrow\; AB = PQ,\; BC = QR,\; CA = RP,\; \angle A = \angle P,\; \angle B = \angle Q,\; \angle C = \angle R

Common traps

Taking equal angles as congruence

Three equal angles fix only the shape of a triangle. A small and a large equilateral triangle have the same angles and are not congruent.

Believing congruent triangles can differ in area

Congruent triangles are exact copies, so any option saying their areas or perimeters differ is false.

Matching corners by alphabet or by position

In △ABC ≅ △FDE, A matches F, B matches D and C matches E. The pairing comes only from the order the letters are written in.

Trusting how the figure is drawn

Two congruent triangles may be drawn turned or flipped. Read the matching sides from the statement, then check the figure against it.

Accepting two sides and a non-included angle

Two sides with an angle that is not between them (SSA) do not fix a triangle. The only exception is a right angle, which is the RHS rule.

Calling a right-angle case SAS

With a right angle, the hypotenuse and a leg equal, the right angle is not between the two given sides. The rule is RHS, not SAS.

Missing the shared side or the crossing angles

A side common to both triangles, or two vertically opposite angles, are equal parts that the question does not state. Count them.

Similar Triangles

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Similar triangles and the scale factor for lengths

Scale factor of similar triangles

k = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{\text{perimeter of } ABC}{\text{perimeter of } PQR}

$k$ the scale factor from triangle PQR to triangle ABC

Areas of similar triangles: the square of the scale factor

Areas of similar figures

\frac{\text{Area}_1}{\text{Area}_2} = k^2 = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2

A line parallel to one side: the areas it cuts off

Area cut off by a parallel line

\frac{\text{ar}(ADE)}{\text{ar}(ABC)} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{DE}{BC}\right)^2

Common traps

Reading similar as equal

Similar triangles have equal angles, not equal sides. Their sides are equal only when the scale factor is 1, and then they are congruent.

SAS similarity with the wrong angle

For SAS similarity the equal angle must lie between the two sides that are in proportion. An equal angle anywhere else proves nothing.

Turning the ratio over halfway

Decide which triangle goes on top and keep it there for every pair, or the scale factor flips from k to 1/k.

Using k for areas

Sides in the ratio 2 : 3 give areas in the ratio 4 : 9. Leaving the ratio as 2 : 3 is the most common wrong option.

Squaring on the way back

From an area ratio to a side ratio, take the square root. Areas 36 : 49 mean sides 6 : 7, not 1296 : 2401.

Squaring the perimeter ratio for perimeters

Perimeters, altitudes and medians are lengths, so they scale by k. Only areas scale by k².

Squaring AD : DB

The similar triangles are ADE and ABC, so the ratio to square is AD : AB, the part over the whole side.

Treating the strip as the whole triangle

When the question gives the small triangle against the trapezium, add the two to get the whole triangle before taking a square root.

Parallels, Midpoints and the Bisector Theorem

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Basic proportionality theorem: a line parallel to one side

Basic proportionality theorem

\frac{AD}{DB} = \frac{AE}{EC} \qquad \frac{DE}{BC} = \frac{AD}{AB}

$D, E$ points on AB and AC with DE parallel to BC

The midpoint theorem and the quarter triangle

Midpoint theorem

XY = \frac{1}{2}BC \qquad \text{ar}(XYZ) = \frac{1}{4}\,\text{ar}(ABC)

$X, Y$ the midpoints of AB and AC
XYZthe triangle joining the midpoints of all three sides

The angle bisector theorem

Angle bisector theorem

\frac{BD}{DC} = \frac{AB}{AC} \qquad BD = \frac{AB}{AB + AC} \times BC

$D$ the point where the bisector of angle A meets BC

Common traps

Using AD : DB for the parallel segment

DE and BC are sides of the similar triangles ADE and ABC, so DE : BC = AD : AB. With AD : DB = 1 : 2, DE is a third of BC, not a half.

Reading 5AE = 3EC as AE : EC = 5 : 3

If five of AE equal three of EC, then AE is the shorter piece: AE : EC = 3 : 5.

Taking the midpoint triangle as half the area

The midpoint triangle has half the side lengths, so it has a quarter of the area. Half is the ratio of perimeters.

Calling the midpoint triangle congruent to the whole

The midpoint triangle is congruent to the three corner triangles beside it, and only similar to the original triangle.

Pairing each piece with the far side

BD touches B, so it goes with AB; DC goes with AC. Swapping them gives the right numbers in the wrong order, and that answer is usually printed.

Putting the third side into the ratio

The ratio uses only the two sides that meet at the bisected angle. The side being cut is used afterwards, to share it out.

Medians and Apollonius Theorem

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The centroid divides each median in the ratio 2 : 1

Centroid on a median

AG = \frac{2}{3}AD \qquad GD = \frac{1}{3}AD

$AD$ the median from A to the midpoint D of BC
$G$ the centroid

A median halves the area, and the centroid makes equal parts

Areas made by medians

\text{ar}(ABD) = \frac{1}{2}\,\text{ar}(ABC) \qquad \text{ar}(GBC) = \frac{1}{3}\,\text{ar}(ABC)

Apollonius theorem: the length of a median

Apollonius theorem

AB^2 + AC^2 = 2(AD^2 + BD^2) \qquad AD = \frac{1}{2}\sqrt{2AB^2 + 2AC^2 - BC^2}

$D$ the midpoint of BC, so BD is half of BC

Common traps

Reading 2 : 1 from the wrong end

The longer part of a median is the one at the corner. GD, at the midpoint end, is one third of the median.

Using whole medians when two medians cross at 90°

The right angle is at the centroid, so the right triangle's legs are the two-thirds parts AG and BG, not the full medians.

Taking the centroid triangles as quarters

GAB, GBC and GCA are each a third of the triangle. The quarter belongs to the triangle that joins the midpoints of the sides.

Thinking a median quarters the area

A median halves one side and keeps the height, so it halves the area.

Dropping the 2 in Apollonius

The theorem is AB² + AC² = 2AD² + 2BD². Writing AD² + BD² on the right halves the answer's square.

Using the whole side where the half is needed

BD in the theorem is half of BC, because D is the midpoint. Put BC itself in only through the form with 2AB² + 2AC² − BC².

Centres of a Triangle

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The angle at the incentre between two angle bisectors

Angle at the incentre

\angle BIC = 90^\circ + \frac{\angle A}{2}

$I$ the incentre, where the bisectors of angles B and C meet

Inradius: the area divided by the semi-perimeter

Inradius

r = \frac{\text{Area}}{s} \qquad s = \frac{a + b + c}{2} \qquad r_{\text{right}} = \frac{a + b - c}{2}

$s$ the semi-perimeter
$c$ the hypotenuse, in the right-triangle form

Circumradius of a right, an equilateral and any triangle

Circumradius

R = \frac{abc}{4 \times \text{Area}} \qquad R_{\text{right}} = \frac{\text{hypotenuse}}{2} \qquad R_{\text{equilateral}} = \frac{a}{\sqrt{3}}

$a, b, c$ the three sides

The four centres of a triangle and where each one lies

Centre	Lines that meet there	Acute triangle	Right triangle	Obtuse triangle
Centroid	the three medians	inside	inside	inside
Incentre	the three angle bisectors	inside	inside	inside
Circumcentre	the perpendicular bisectors of the sides	inside	at the midpoint of the hypotenuse	outside, beyond the longest side The circumradius of a right triangle is half its hypotenuse.
Orthocentre	the three altitudes	inside	at the corner with the right angle	outside, beyond the obtuse corner The two legs of a right triangle are altitudes of each other, so they meet at the right angle.

Centroid and incentre never leave the triangle; circumcentre and orthocentre do.

Common traps

Swapping the circumcentre and the orthocentre of a right triangle

In a right triangle the circumcentre is the midpoint of the hypotenuse and the orthocentre is the right-angle corner.

Making the incentre equidistant from the corners

The incentre is the same distance from the three sides. The point the same distance from the three corners is the circumcentre.

Adding the whole of the opposite angle

The angle at the incentre is 90° plus half the opposite angle. With ∠A = 60° it is 120°, not 150°.

Using the wrong angle of the triangle

The formula for ∠BIC uses ∠A, the corner whose bisector is not involved. Angles B and C have already been halved inside triangle BIC.

Dividing the area by the full perimeter

The inradius is the area divided by the semi-perimeter. Dividing by the whole perimeter gives half the radius.

Adding the hypotenuse in the right-triangle form

For a right triangle r = (a + b − c) ÷ 2: the two legs minus the hypotenuse. Adding all three sides gives the semi-perimeter instead.

Halving a leg instead of the hypotenuse

The circumradius of a right triangle is half the hypotenuse, the longest side. Half of a leg is not a radius of anything here.

Swapping R and r in an equilateral triangle

The circumradius a ÷ √3 is twice the inradius a ÷ (2√3). The circle through the corners is the bigger one.

Triangles formulas

Angles of a Triangle

Angle sum of a triangle, with ratios and equations

The exterior angle of a triangle

Isosceles and equilateral triangles

Common traps

Sides of a Triangle

Triangle inequality: the third side lies between the difference and the sum

Pythagoras theorem, triples and the acute or obtuse test

The altitude to the hypotenuse of a right triangle

Common traps

Congruent Triangles

Reading the matching corners from a congruence statement

Common traps

Similar Triangles

Similar triangles and the scale factor for lengths

Areas of similar triangles: the square of the scale factor

A line parallel to one side: the areas it cuts off

Common traps

Parallels, Midpoints and the Bisector Theorem

Basic proportionality theorem: a line parallel to one side

The midpoint theorem and the quarter triangle

The angle bisector theorem

Common traps

Medians and Apollonius Theorem

The centroid divides each median in the ratio 2 : 1

A median halves the area, and the centroid makes equal parts

Apollonius theorem: the length of a median

Common traps

Centres of a Triangle

The angle at the incentre between two angle bisectors

Inradius: the area divided by the semi-perimeter

Circumradius of a right, an equilateral and any triangle

The four centres of a triangle and where each one lies

Common traps

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