The shape is the same.
All circles are similar.
Class 10 · Maths · Chapter 6 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Triangles
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1. Read — Similar figures · basic proportionality · AAA/SSS/SAS · areas · Pythagoras, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure shows one triangle with a line parallel to the base and the intercept ratio AD/DB = AE/EC.
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समरूप आकृतियाँ · NCERT 6.2 · Exercise 6.1
Figures of the same shape are called similar. The size may differ. All circles are similar. All squares are similar. All equilateral triangles are similar.
Congruent figures have both the same shape and the same size. Every congruent pair is similar. Not every similar pair is congruent. A triangle and a square are not similar, because the shape is different. Exercise 6.1 is this recognition.
Two polygons with the same number of sides are similar only when corresponding angles are equal and corresponding sides are in the same ratio. Angles alone are not enough: a square and a long rectangle both have angles of 90°, but the sides are not proportional. A side ratio without the angles is not enough either.
Question: Two squares have sides 4 cm and 6 cm. Are they similar? Are they congruent? Write the ratio.
Formula: The ratio of corresponding sides = smaller / larger.
Substitution: 4/6 = 2/3.
Answer: Both are squares, so they are similar. The size differs, so they are not congruent. The exact ratio is 2:3.
In a BSEB answer write the sentence “similar but not congruent” with an example.
On CBSE do not call a square and a rectangle similar just because the angles match.
The shape is the same.
All circles are similar.
False — the size may differ.
4/6.
2:3.
The size is equal too.
Congruent implies similar. The ratio is 1:1.
All three angles are 60°, so the shape is the same and they are similar. They are congruent only when the sides are equal, that is, the ratio is 1:1.
आधारभूत समानुपातिकता — थेल्स · NCERT 6.3 · Exercise 6.2 · AD/DB = AE/EC
In triangle ABC, if DE is parallel to BC, with D on AB and E on AC, then AD/DB = AE/EC. Also AD/AB = AE/AC.
This is the basic proportionality theorem. Exercise 6.2 of the current NCERT is on it. The line must cut the sides at distinct points. Do not place D on B.
Use the ratio only when the figure says DE || BC. AD and DB are pieces of one side. AE and EC are pieces of the other. Writing AD/AE = DB/EC scrambles the order, unless that form really is equivalent. Keep the unit centimetre.
Question: DE || BC, AD = 3 cm, DB = 6 cm, AE = 4 cm. Find EC.
Formula: AD/DB = AE/EC.
Substitution: 3/6 = 4/EC. 1/2 = 4/EC. EC = 8.
Answer: EC = 8 cm. Check: AD/AB = 3/9 = 1/3 and AE/AC = 4/12 = 1/3.
In a BSEB answer keep the formula line and the substitution apart. Do not jump to 8 cm.
On CBSE write the form that is given. Do not swap AD/DB with AD/AB.
3/6 = 4/EC.
EC = 8 cm.
True — this is the other form of Thales.
The lower piece of the other side.
EC.
AD/DB = AE/EC. 2/3 = 4/EC. EC = 6 cm. AC = AE + EC = 10 cm. Check: AD/AB = 2/5 and AE/AC = 4/10 = 2/5.
थेल्स का विलोम · Equal ratios mean the line is parallel
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side. If AD/DB = AE/EC, then DE || BC.
This is the converse. The direct theorem assumes parallel and gives the ratio. The converse assumes the ratio and gives parallel. If the ratios differ, the line is not taken as parallel.
AD = 2 cm, DB = 4 cm, AE = 3 cm, EC = 6 cm. 2/4 = 1/2 and 3/6 = 1/2. They are equal, so DE || BC. If EC were 5 cm, then 3/5 ≠ 1/2, and you would not call them parallel.
Question: AD = 2 cm, DB = 4 cm, AE = 3 cm, EC = 6 cm. Is DE || BC?
Formula: Converse — if AD/DB = AE/EC then DE || BC.
Substitution: AD/DB = 2/4 = 1/2. AE/EC = 3/6 = 1/2.
Answer: Both ratios are 1/2, so DE is parallel to BC.
In a BSEB answer write both fractions in lowest terms, then the equal sign.
On CBSE, “almost equal” is not accepted. 1/2 and 3/5 are not equal.
The converse.
DE is parallel to BC.
False — the ratios are not equal.
Lowest terms.
1/2.
The converse starts from the ratio.
The direct theorem assumes parallel and gives the ratio.
True — the ratio is given, and parallel is to be proved.
AD/DB = 4/6 = 2/3. AE/EC = 6/9 = 2/3. By the converse, DE || BC.
कसौटी AAA, SSS और SAS · NCERT 6.4 · Exercise 6.3 · ΔABC ~ ΔDEF
Two triangles are similar if corresponding angles are equal. The third angle is fixed by 180°, so AA is enough. This is called the AAA criterion.
SSS: the three corresponding sides are in one ratio. SAS: one angle is equal and both arms of that angle are in one ratio. ΔABC ~ ΔDEF means A↔D, B↔E, C↔F. The current exercise 6.3 is these criteria. SAS congruence asks for equal lengths; SAS similarity asks for a ratio.
If the sides are 3, 4, 5 and 6, 8, 10, the ratio is 1:2. They are similar by SSS. They are not congruent, because the lengths are not equal. If both triangles have angles 50°, 60° and 70°, they are similar by AA, without measuring sides.
Question: ΔABC ~ ΔPQR, AB = 4 cm, PQ = 6 cm. Which side corresponds to AB, and what is the ratio?
Formula: The order is A↔P, B↔Q, C↔R, so AB/PQ.
Substitution: AB/PQ = 4/6 = 2/3.
Answer: AB corresponds to PQ. The exact ratio is 2:3. Do not divide by QR.
In a BSEB answer write both the name of the criterion and the correspondence. Only the ~ symbol is incomplete.
On CBSE do not write SAS similarity as SAS congruence. One asks for a ratio, the other for equal lengths.
The angles are equal.
AAA, or AA.
False — the ratio is 1:2, so they are similar, not congruent.
The first two letters.
PQ.
The angle is included.
One equal angle and its arms in one ratio.
A↔D and B↔E, so DE corresponds to AB. AB/DE = 5/15 = 1/3. The exact ratio is 1:3.
समरूप त्रिभुजों के क्षेत्रफल · Older Bihar exercise 6.4 · (side)²
If ΔABC ~ ΔDEF, then ar(ABC)/ar(DEF) = (AB/DE)². The square is of the side, not of the area.
This is theorem 6.6 and exercise 6.4 of the older Bihar book. The current NCERT chapter 6 stops its exercises at the criteria in 6.3. Write area in square centimetres. If the side ratio is 2:3, the area ratio is 4:9.
If the areas are 16 cm² and 36 cm², the ratio is 16/36 = 4/9. The side ratio is √(4/9) = 2/3. If the smaller corresponding side is 4 cm, the larger is 4 × 3/2 = 6 cm. Before the square root, see that the triangles are given as similar.
Question: ΔABC ~ ΔDEF, the areas are 16 cm² and 36 cm². AB = 4 cm and AB corresponds to DE. Find DE.
Formula: ar(ABC)/ar(DEF) = (AB/DE)².
Substitution: 16/36 = (4/DE)². 4/9 = (4/DE)². 2/3 = 4/DE.
Answer: DE = 6 cm. Check: (4/6)² = 4/9 and 16/36 = 4/9.
In a BSEB answer write the square line on its own. Do not call 16/36 the side ratio.
Current CBSE exercise 6.3 does not ask this result. Exercise 6.4 of the older book does.
10-second revision
Take squares.
4:9.
False — it is the ratio of the squares of the sides.
The square root of 4/9.
2:3.
4 × 3/2.
6 cm.
True — the side ratio is 1, so the size is equal too.
ar1/ar2 = (3/5)² = 9/25. 18/ar2 = 9/25. ar2 = 18 × 25/9 = 50. The larger area is 50 cm².
पाइथागोरस और उसका विलोम · Older Bihar exercise 6.5 · a² + b² = c²
In a right triangle, hypotenuse² = base² + perpendicular². Converse: if the square of one side equals the sum of the squares of the other two, the angle opposite the longest side is a right angle.
Exercise 6.5 of the older Bihar book is this. If a perpendicular is drawn from the right-angled vertex to the hypotenuse, the two small triangles are similar to the whole triangle. The book reaches Pythagoras from that similarity. The current NCERT chapter 6 goes to the summary after the criteria.
Among 6 cm, 8 cm and 10 cm, the longest is 10 cm. Compare 6² + 8² with 10². If they match, the angle is a right angle. The same check works for 5, 12, 13: 25 + 144 = 169. The unit is centimetre on a side and square centimetre on a square.
Question: The legs of a right triangle are 6 cm and 8 cm. Find the hypotenuse.
Formula: c² = a² + b².
Substitution: c² = 6² + 8² = 36 + 64 = 100.
Answer: c = 10 cm, because a length is positive. Check: 36 + 64 = 100.
In exercise 6.5 of the older BSEB book, the line that adds the squares should be visible.
The current CBSE exercises of chapter 6 do not ask Pythagoras separately. Write a similarity criterion there.
10-second revision
36 + 64 = 100.
10 cm.
True — 25 + 144 = 169 = 13².
36 + 64.
100. The hypotenuse is 10 cm.
7² + 24² = 49 + 576 = 625 = 25². By the converse, the angle opposite 25 cm is a right angle.
Pick a type. The 30 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
One shape.
All squares are similar.
3/6 = 4/EC.
8 cm.
The converse.
DE is parallel to BC.
The order of the letters.
PQ.
Squares.
4:9.
36 + 64.
10 cm.
The ratio is 1:2.
Similar by SSS, not congruent because the size differs. Both are also right-angled.
All squares are similar.
Similar, ratio 2:3. Not congruent.
2:3.
6 cm.
The sum of squares.
25 + 144 = 169 = 13².
A rectangle and a square.
Angles alone are not enough. A side ratio is needed too.
The sum of angles.
The third angle is fixed, so AAA and AA are the same statement.
True — the shape is the same.
False — the size may differ.
True — Thales.
False — the ratios must be equal.
False — DE corresponds to AB.
True.
True — 36 + 64 = 100.
False — the sides are not proportional.
EC.
The lower piece.
2:3.
4/6.
4:9.
Squares.
100.
The square of the hypotenuse.
Angle E.
The second letter.
2/3.
4/6 in lowest terms.
169.
13².
1:1.
Equal size.
Thales is the ratio, AAA is the angles, areas use the square, and Pythagoras is the square of the hypotenuse.
6.1 is similar figures, 6.2 is Thales, 6.3 is the criteria. Pythagoras is 6.5 of the older book.
Assertion (A): If DE || BC and AD = 3 cm, DB = 6 cm, AE = 4 cm, then EC = 8 cm.
Reason (R): AD/DB = AE/EC.
Both are true and R is the basis of the calculation.
Assertion (A): All circles are similar.
Reason (R): The ratio of the areas of similar triangles equals the square of the sides.
Both are true, but R does not explain why circles are similar.
Assertion (A): Triangles with sides 3, 4, 5 and 6, 8, 10 are similar.
Reason (R): They are congruent as well.
A is true, by SSS and the ratio 1:2. R is false.
Assertion (A): 6 cm, 8 cm and 11 cm are the sides of a right triangle.
Reason (R): In the converse, the square of the longest side must equal the sum of the squares of the other two.
A is false, 36 + 64 = 100 ≠ 121. R is true.
Assertion (A): Squares of 4 cm and 6 cm are similar but not congruent.
Reason (R): The shape is the same and the size ratio is 2:3, not 1:1.
Both are true and R explains A.
The squares are similar. Equal circles are also congruent. A triangle and a square have different shapes. 3-4-5 and 6-8-10 are similar.
Angles are AAA, three sides are SSS, the included angle is SAS, and the hypotenuse is Pythagoras.
Figures of the same shape are similar; the size need not be equal.
If DE || BC, then AD/DB = AE/EC.
ar(ABC)/ar(DEF) = (AB/DE)² when ΔABC ~ ΔDEF.
In a right triangle, hypotenuse² = the sum of the squares of the other two sides.
AD/DB = AE/EC. 5/10 = AE/8. 1/2 = AE/8. AE = 4 cm.
AB/DE = BC/EF. 6/9 = 8/EF. 2/3 = 8/EF. EF = 12 cm.
(4/6)² = 16/36 = 4/9. 24/ar = 4/9. ar = 24 × 9/4 = 54. The larger area is 54 cm².
c² = 81 + 144 = 225. c = 15 cm.
AD/DB = AE/EC. 4/4 = 5/EC. EC = 5 cm. AB = 8 cm, AC = 10 cm. AD/AB = 4/8 = 1/2 and AE/AC = 5/10 = 1/2. Both forms give the same ratio.
10/5 = 24/12 = 26/13 = 2. Similar by SSS, ratio 1:2. 25 + 144 = 169 = 13², so the first is right-angled, and so is the second, because 100 + 576 = 676 = 26². The area ratio is (1/2)² = 1:4.
The sun makes the same angle, so the triangles are similar by AA. Height over shadow is equal: h/6 = 2/1.5. h = 6 × 2/1.5 = 6 × 4/3 = 8. The pole is 8 m high.
This model set is for practice. It is not a question from any year’s annual examination.
The size is the same too.
Circles of equal radius are congruent.
2/3 = 4/EC.
EC = 6 cm.
225.
The hypotenuse is 15 cm.
AD/DB = 3/5. AE/EC = 4/6 = 2/3. 3/5 ≠ 2/3, so DE is not parallel to BC.
17² = 8² + b². 289 = 64 + b². b² = 225. b = 15 cm.
Assertion (A): All equilateral triangles are similar.
Reason (R): Their sides are always equal, so they are congruent.
A is true, the angles are 60°. R is false, the sides may differ.
These are competency-based practice questions. They are not a copy of any year’s paper.
Letter order A-D, B-E.
DE corresponds to AB. 4/8 = 1/2. EF does not correspond.
h/9 = 1.5/1.
h = 9 × 1.5 = 13.5 m. AA similarity.
Assertion (A): Similar triangles of areas 16 cm² and 36 cm² have corresponding sides in the ratio 2:3.
Reason (R): The ratio of the areas is the square of the ratio of the sides.
Both are true. 16/36 = 4/9 = (2/3)².
Assertion (A): Any pair of rectangles is similar.
Reason (R): For polygons, a ratio of corresponding sides is needed along with the corresponding angles.
A is false. A square and a long rectangle are not similar. R is true.
It is a right angle. c² = 8² + 15² = 64 + 225 = 289. c = 17 m.
Exercise 6.3 of the current NCERT asks the similarity criteria AAA, SSS and SAS. Areas are exercise 6.4 of the older Bihar book, and Pythagoras is 6.5.
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What you learned
| What | Keep this |
|---|---|
| Similar | same shape, size may differ |
| Congruent | same shape and same size |
| Thales | AD/DB = AE/EC if DE || BC |
| AAA / AA | equal corresponding angles |
| Areas | ar ratio = (side ratio)² |
| Pythagoras | a² + b² = c² |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.