Geometry · Topic 4 of 7

🔁 Similarity & Congruence

2 exam question types, fully solved

Two triangles can match in two different ways. Similar triangles have the same shape but can be different sizes — like a photo and its enlargement. Congruent triangles are identical twins: same shape and same size. SSC loves both because, once you spot which case you are in, a single ratio or a single rule cracks the whole question.

Similarity — same shape, scaled size

Prove two triangles similar () with any ONE of these tests:

• AA: two pairs of angles equal (the third is then automatic).

• SSS: all three pairs of sides in the same ratio.

• SAS: two pairs of sides in the same ratio with the included angles equal.

Once similar, every pair of corresponding sides shares one ratio . Perimeters, medians and altitudes are all 1-D, so they use the same . But area is 2-D, so the ratio of areas is the square of the side ratio:

Basic Proportionality (Thales): a line parallel to one side cuts the other two sides in the same ratio — . Its special case, the Midpoint Theorem, gives a midsegment equal to the base.

Congruence — same shape AND same size

Prove two triangles congruent () with any ONE of these five tests:

• SSS — three sides equal.

• SAS — two sides and the included angle equal.

• ASA — two angles and the included side equal.

• AAS — two angles and a non-included side equal.

• RHS — right angle, hypotenuse and one side equal (right triangles only).

Watch out: is not a valid test. Once congruence is established, CPCT (Corresponding Parts of Congruent Triangles) lets you call every matching side and angle equal — just follow the vertex order. In we get and .

🔍Similarity LabSame shape, scaled size — sides ×k, areas ×k²

scale factor k

sides ×2 → perimeter ×2 → area ×4

Perimeter — 1-D, ratio = k

△ABC: 3 + 4 + 5 = 12

△DEF: 6 + 8 + 10 = 24

24 / 12 = 2 = k

Area — 2-D, ratio = k²

△ABC: ½ × 3 × 4 = 6

△DEF: ½ × 6 × 8 = 24

24 / 6 = 4 = k²

📌 Congruence = same shape and same size (k = 1, identical twins). Quick list: SSS · SAS · ASA · AAS · RHS. SSA is not a valid test. Once proved, CPCT makes every matching side and angle equal.

Two similar triangles have sides in the ratio , so their areas are in the ratio .

Q1Two similar triangles have corresponding sides in the ratio 2:3. What is the ratio of their areas?

Q2In triangle ABC, DE is parallel to BC with D on AB and E on AC. If AD = 3, DB = 6 and AE = 4, find EC.

Q3Two triangles are congruent if a right angle, the hypotenuse and one side match. This criterion is:

📚

Real exam questions — Similarity & Congruence

2 question types · 10 solved examples from real SSC papers

Similar means same shape (sides in a fixed ratio, areas in the square of that ratio); congruent means same shape AND size. Spot which one the figure gives you.

How to solve this type

Two triangles are SIMILAR (

\sim

) when they have the same shape but not necessarily the same size. Prove it with any one of: AA (two angles equal), SSS (all three sides in the same ratio) or SAS (two sides in the same ratio with the included angle equal). Once similar, every pair of corresponding sides is in the same ratio

k

, and so are the perimeters, medians and altitudes (all 1-D). But the ratio of AREAS is

k^2

\dfrac{\text{area}_1}{\text{area}_2}=\left(\dfrac{s_1}{s_2}\right)^2

. A line drawn parallel to one side (Basic Proportionality / Thales) cuts the other two sides in the same ratio:

\dfrac{AD}{DB}=\dfrac{AE}{EC}

The ratio of the areas of two similar triangles is $9:16$ . What is the ratio of their corresponding altitudes (heights)?

27:64

3:4

9:16

1:1

Two similar triangles have areas $64\,\text{cm}^2$ and $100\,\text{cm}^2$ . If a side of the smaller triangle is $8$ cm, the corresponding side of the larger triangle is:

A12 cmB9 cmC8 cmD10 cm

Two similar triangles have corresponding sides in the ratio $4:9$ . The ratio of their perimeters is:

2:3

4:9

16:81

8:27

In $\triangle ABC$ , $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ . If $AD=4$ , $DB=6$ and $AE=5$ , find $EC$ .

A7.5B5C8D6

In $\triangle ABC$ , $D$ and $E$ are the midpoints of $AB$ and $AC$ . If $DE=7$ cm, then $BC$ is:

A10 cmB21 cmC14 cmD7 cm