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SSC CGL Maths · Triangles

Similar Triangles

Similar triangles have the same angles and their sides in one ratio k: every length scales by k and every area by k².

All Triangles notes

Why this matters

Twenty-two questions in the bank, 18 EASY and 4 MODERATE. Every one is the same two moves: find the scale factor k from one pair of matching lengths, then multiply a length by k or an area by k². Fifteen are about areas, and the most common wrong option is the answer you get by forgetting to square.

Concept 1 of 3: Similar triangles and the scale factor for lengths

A photo enlarged on a copier keeps every angle and stretches every length by the same amount. Two similar triangles are one triangle and its enlargement, so a single ratio describes the whole change.

Definition

  • Similar (written ∼) means the same angles, with every side multiplied by one number kk, the scale factor. Same shape, any size; k=1k = 1 is congruence.
  • Tests: AA (two pairs of equal angles are enough, the third follows), SSS (all three sides in one ratio), SAS (two sides in one ratio and the angle between them equal).
  • The letters pair in order, as for congruence: △ABC∼△PQR\triangle ABC \sim \triangle PQR gives ABPQ=BCQR=CARP=k\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = k.
  • Every length scales by kk: sides, perimeters, altitudes and medians.
  • A line parallel to one side cuts off a similar triangle: DE ∥ BC with D on AB and E on AC gives △ADE∼△ABC\triangle ADE \sim \triangle ABC, since the angles at D and B, and at E and C, are corresponding angles.
  • The shared-angle pattern: D on BC with ∠ADC=∠BAC\angle ADC = \angle BAC. Triangles ADC and BAC share ∠C\angle C, so they are similar by AA, and ACBC=DCAC\frac{AC}{BC} = \frac{DC}{AC}, that is AC2=BC×DCAC^2 = BC \times DC.
  • Fast route: find kk once, from the one pair where both lengths are known, then multiply.

Scale factor of similar triangles

k=ABPQ=BCQR=CARP=perimeter of ABCperimeter of PQRk = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{\text{perimeter of } ABC}{\text{perimeter of } PQR}
  • kthe scale factor from triangle PQR to triangle ABC

Worked example

△ABC∼△DEF\triangle ABC \sim \triangle DEF, with AB = 10 cm, BC = 14 cm, CA = 16 cm and DE = 15 cm. Find EF and the perimeter of triangle DEF.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q70Easy

Example 1 · Triangles · Similar Triangles

If △ABC∼△XYZ\triangle ABC \sim \triangle XYZ and AB=4 cm, BC=6 cm, XY=8 cm, what is the length of YZ?

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.
Drill 6 more on similar triangles and the scale factor for lengths

Concept 2 of 3: Areas of similar triangles: the square of the scale factor

Area is a length times a length. When both the base and the height are multiplied by k, the area is multiplied by k twice, which is k².

Definition

  • Ratio of areas =k2= k^2, the square of the ratio of any pair of matching lengths: sides, perimeters, altitudes or medians.
  • Going back: ratio of lengths =ratio of areas= \sqrt{\text{ratio of areas}}. Areas 49 : 64 give sides 7 : 8.
  • The rule holds for any two similar figures, not only triangles: similar squares, hexagons, or a map and the land.
  • Two areas with a known total: split the total in the ratio of the squares.
  • Fast route: area ratios in these questions are almost always ratios of perfect squares, so the square roots are whole numbers.

Areas of similar figures

Area1Area2=k2=(side1side2)2\frac{\text{Area}_1}{\text{Area}_2} = k^2 = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2

Worked example

Two similar triangles have matching sides of 6 cm and 9 cm. The area of the smaller is 28 cm². Find the area of the larger.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 1 · Q69Easy

Example 2 · Triangles · Similar Triangles

A right-angled triangle has a hypotenuse of 13 cm and one leg of 5 cm. A second right-angled triangle is similar to the first, and its hypotenuse is 39 cm. What is the area of the second triangle?

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.
Drill 10 more on areas of similar triangles: the square of the scale factor

Concept 3 of 3: A line parallel to one side: the areas it cuts off

A line across a triangle parallel to its base cuts off a smaller copy at the top. The copy's area follows from the square rule, and the strip underneath is whatever is left.

Definition

  • DE ∥ BC with D on AB and E on AC: △ADE∼△ABC\triangle ADE \sim \triangle ABC with k=ADAB=AEAC=DEBCk = \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}.
  • ar(ADE)ar(ABC)=(ADAB)2\frac{\text{ar}(ADE)}{\text{ar}(ABC)} = \left(\frac{AD}{AB}\right)^2. The strip DECB (a trapezium) is ar(ABC) − ar(ADE).
  • Use the whole side AB, not DB. If AD : DB = 2 : 3, then AD : AB = 2 : 5.
  • Working back from areas: if ar(ADE) : ar(DECB) = m:nm : n, then ar(ADE) : ar(ABC) = m:(m+n)m : (m + n); take the square root to get AD : AB, then subtract to get AD : DB.
  • If DE cuts the area into two equal halves, ADAB=12\frac{AD}{AB} = \frac{1}{\sqrt{2}}.
  • The diagonals of the strip: DC and BE meet at F. Then △DEF∼△CBF\triangle DEF \sim \triangle CBF (alternate angles), with k=DEBCk = \frac{DE}{BC}.

Area cut off by a parallel line

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

Worked example

In triangle ABC, DE ∥ BC with D on AB and E on AC, and AD : DB = 3 : 2. The area of triangle ABC is 75 cm². Find the area of triangle ADE and of the trapezium DECB.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 1 · Q70Moderate

Example 3 · Triangles · Similar Triangles

In △ABC\triangle ABC, a line segment DE is parallel to BC, with D on AB and E on AC. If the ratio of the area of △ADE\triangle ADE to the area of the trapezoid DECB is 4:21, what is the ratio of AD to DB?

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.
Drill 3 more on a line parallel to one side: the areas it cuts off

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Similar triangles and the scale factor for lengths

    Scale factor of similar triangles

    k=ABPQ=BCQR=CARP=perimeter of ABCperimeter of PQRk = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{\text{perimeter of } ABC}{\text{perimeter of } PQR}
  • Areas of similar triangles: the square of the scale factor

    Areas of similar figures

    Area1Area2=k2=(side1side2)2\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

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

Watch out for (8)

Test yourself on Triangles

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Triangles chapter testDrill all 22 questions