Chapter 6: Triangles

VVI questions — the ones the board asked most (20)

1. Q1Short answer2 marksAsked: 2020, 2023

   ABC is an isosceles triangle whose angle C is right angle. Prove that AB^2 = 2AC^2.

   

   Medha Ma’am

   AskMedha’s AI teacher

   Hear it from Medha Ma’am · 60 s

   Model answer · Pythagoras theorem, isosceles right triangle

   1. 1.In triangle ABC, angle C = 90°, so by the Pythagoras Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Here AB is the hypotenuse, so AB2 = AC2 + BC2.1 mark
   2. 2.Since triangle ABC is an isosceles triangle with angle C = 90°, the two sides containing the right angle are equal, so AC = BC. Putting BC = AC in AB2 = AC2 + BC2, we get AB2 = AC2 + AC2 = 2AC2. Hence AB2 = 2AC2 is proved.1 mark

   Diagram: Draw right triangle ABC with the right angle at C; label the vertices A, B, C, mark angle C as 90°, and mark sides AC and BC as equal.

   • Write 'Pythagoras Theorem' by name and state AB2 = AC2 + BC2 using the hypotenuse AB.
   • State clearly that AC = BC because the triangle is isosceles with angle C = 90°, before substituting.

   Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme

2. Q2Long answer5 marksAsked: 2025

   If one angle of a triangle is equal to one angle of the other triangle and the sides included between these angles are proportional then prove that the triangles are similar.

   

   Medha Ma’am

   AskMedha’s AI teacher

   Hear it from Medha Ma’am · 160 s

   Model answer · SAS similarity criterion proof

   1. 1.Given: in ΔABC and ΔDEF, ∠A = ∠D and AB/DE = AC/DF. To prove: ΔABC ~ ΔDEF.1 mark
   2. 2.Construction: take P on DE and Q on DF such that DP = AB and DQ = AC; join PQ.1 mark
   3. 3.In ΔABC and ΔDPQ, AB = DP, ∠A = ∠D, AC = DQ, so ΔABC ≅ ΔDPQ by SAS congruence; hence ∠B = ∠DPQ and ∠C = ∠DQP.1 mark
   4. 4.Since DP/DE = DQ/DF, by the converse of the Basic Proportionality Theorem PQ ∥ EF, so the corresponding angles ∠DPQ = ∠E and ∠DQP = ∠F.1 mark
   5. 5.Thus ∠A = ∠D, ∠B = ∠E and ∠C = ∠F, so by the AAA (AA) similarity criterion ΔABC ~ ΔDEF.1 mark

   Diagram: Draw triangle ABC with P on AB and Q on AC such that AP = DE and AQ = DF, join PQ parallel to BC; draw a separate triangle DEF with vertices D, E, F labelled, corresponding to A, P, Q.

   Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme

3. Q3Long answer5 marksAsked: 2026

   State and prove Pythagoras' theorem.

   

   Medha Ma’am

   AskMedha’s AI teacher

   Hear it from Medha Ma’am · 100 s

   Model answer · Pythagoras theorem, proof

   1. 1.Pythagoras Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.1 mark
   2. 2.Given a right triangle ABC right angled at B.1 mark
   3. 3.To prove that AC2 = AB2 + BC2.0.5 marks
   4. 4.Construction: Let us draw BD ⊥ AC.0.5 marks
   5. 5.Proof: In triangle ADB and triangle ABC, triangle ADB ~ triangle ABC, since if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other. So, AD/AB = AB/AC (sides are proportional), or AD . AC = AB2 ... (1).1 mark
   6. 6.Also, triangle BDC ~ triangle ABC (by the same theorem of similarity). So, CD/BC = BC/AC, or CD . AC = BC2 ... (2).0.5 marks
   7. 7.Adding (1) and (2), AD . AC + CD . AC = AB2 + BC2, or AC (AD + CD) = AB2 + BC2, or AC . AC = AB2 + BC2, or AC2 = AB2 + BC2. This proves the Pythagoras Theorem.0.5 marks

   Diagram: Draw right triangle ABC with the right angle at B; draw BD perpendicular from B to the hypotenuse AC, meeting it at D; label vertices A, B, C and foot of perpendicular D, with the right angle marked at B.

   • Write the word 'Given', 'To prove' and 'Construction' as separate headings before stating them.
   • State clearly that triangle ADB ~ triangle ABC and triangle BDC ~ triangle ABC, using the similarity theorem about the perpendicular from the right angle to the hypotenuse.
   • End with the line 'AC2 = AB2 + BC2' after adding the two proportional relations, to show the final result.

   Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme

4. Q4Long answer5 marksAsked: 2026

   Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

   How the model answer begins

   1. Given two triangles ABC and PQR such that triangle ABC is similar to triangle PQR.

   The full answer from Medha Ma’am →
5. Q5Long answer5 marksAsked: 2026

   Prove that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on each side of the perpendicular are similar to each other and to the original triangle.

   How the model answer begins

   1. We are given a right triangle ABC, right angled at B. Let BD be the perpendicular drawn from the vertex of the right angle B to the hypotenuse AC, so that BD ⊥ AC.

   The full answer from Medha Ma’am →
6. Q6Long answer5 marksAsked: 2022

   Prove that if a line is drawn parallel to one side of a triangle, the other two sides are divided in the same ratio.

   How the model answer begins

   1. We are given a triangle ABC in which a line parallel to side BC intersects the other two sides AB and AC at D and E respectively, that is, DE || BC.

   The full answer from Medha Ma’am →
7. Q7Long answer5 marksAsked: 2021

   Prove that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

   How the model answer begins

   1. We are given a triangle ABC in which a line DE intersects sides AB and AC at points D and E respectively such that AD/DB = AE/EC.

   The full answer from Medha Ma’am →
8. Q8Short answer2 marksAsked: 2025

   E is a point on the extended part of the side AD of a parallelogram ABCD and BE intersects CD at F; then prove that triangle ABE ~ triangle CFB.

   How the model answer begins

   1. In parallelogram ABCD, AB is parallel to CD, and angle A is equal to angle C because opposite angles of a parallelogram are equal. Since F lies on CD, angle C is the same as angle FCB, so angle A is equal to angle FCB.

   The full answer from Medha Ma’am →
9. Q9Short answer2 marksAsked: 2025

   ABC is an isosceles right triangle with angle C as right angle. Prove that AB^2 = 2AC^2.

   How the model answer begins

   1. According to the Pythagoras Theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In triangle ABC, angle C is a right angle, so AB is the hypotenuse.…

   The full answer from Medha Ma’am →
10. Q10Short answer2 marksAsked: 2025

    E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC, prove that triangle ABD ~ triangle ECF.

    How the model answer begins

    1. Since AB = AC, ∠ABC = ∠ACB. E lies on CB produced, so ∠ABD is the same angle as ∠ABC and ∠ECF the same as ∠ACB; hence ∠ABD = ∠ECF.

    The full answer from Medha Ma’am →
11. Q11Short answer2 marksAsked: 2025

    Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Then prove that triangle ABC ~ triangle PQR.

    How the model answer begins

    1. (The question's 'PR' is read as QR — the median PM is drawn to QR.) AD and PM are medians, so BC = 2BD and QR = 2QM. From AB/PQ = BC/QR = AD/PM, AB/PQ = BD/QM = AD/PM, so ΔABD ~ ΔPQM by SSS similarity and ∠B = ∠Q.

    The full answer from Medha Ma’am →
12. Q12Short answer2 marksAsked: 2025

    triangle ABC and triangle DEF are similar and their areas are 9 cm^2 and 64 cm^2 respectively. If DE = 5.1 cm then find AB.

    How the model answer begins

    1. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Since triangle ABC and triangle DEF are similar, AB and DE are corresponding sides, so ar(ABC)/ar(DEF) =…

    The full answer from Medha Ma’am →
13. Q13Short answer2 marksAsked: 2025

    In a triangle PQR, two points S and T are on the sides PQ and PR respectively such that PS/SQ = PT/TR and angle PST = angle PRQ, then prove that deltaPQR is an isosceles triangle.

    How the model answer begins

    1. It is given that PS/SQ = PT/TR. So, by the converse of the Basic Proportionality Theorem (Thales Theorem), ST is parallel to QR.

    The full answer from Medha Ma’am →
14. Q14Short answer2 marksAsked: 2025

    In a triangle PQR, two points S and T are on the sides PQ and PR respectively such that PS/SQ = PT/TR and angle PST = angle PRQ, then prove that triangle PQR is an isosceles triangle.

    How the model answer begins

    1. It is given that PS/SQ = PT/TR. By the converse of the Basic Proportionality Theorem (Thales theorem), if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.…

    The full answer from Medha Ma’am →
15. Q15Short answer2 marksAsked: 2025

    E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC, prove that triangle ABD similar to triangle ECF.

    How the model answer begins

    1. Since AB = AC, ∠ABC = ∠ACB (angles opposite equal sides); as E lies on CB produced, ∠ABD = ∠ECF.

    The full answer from Medha Ma’am →
16. Q16Short answer2 marksAsked: 2025

    E is a point on side CB produced of an isosceles ΔABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

    How the model answer begins

    1. In triangle ABC, AB = AC, since ABC is an isosceles triangle. So, angle B = angle C, because angles opposite to equal sides of a triangle are equal.

    The full answer from Medha Ma’am →
17. Q17Short answer2 marksAsked: 2025

    Sides AB and BC and median AD of a ΔABC are respectively proportional to sides PQ and PR and median PM of another ΔPQR. Then prove that ΔABC ~ ΔPQR.

    How the model answer begins

    1. (The question's 'PR' is read as QR — the median PM is drawn to QR.) AD and PM are medians, so BC = 2BD and QR = 2QM. From AB/PQ = BC/QR = AD/PM, AB/PQ = BD/QM = AD/PM, so ΔABD ~ ΔPQM by SSS similarity and ∠B = ∠Q.

    The full answer from Medha Ma’am →
18. Q18Short answer2 marksAsked: 2025

    ΔABC and ΔDEF are similar and their areas are 9 cm^2 and 64 cm^2 respectively. If DE = 5.1 cm then find AB.

    How the model answer begins

    1. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Since ∆ABC ~ ∆DEF, ar(ABC)/ar(DEF) = (AB/DE)^2, where AB and DE are corresponding sides of the two…

    The full answer from Medha Ma’am →
19. Q19Short answer2 marksAsked: 2026

    Write the side-angle-side criterion of the similarity of two triangles and also represent with figure.

    How the model answer begins

    1. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar. This criterion is referred to as the SAS (Side-Angle-Side)…

    The full answer from Medha Ma’am →
20. Q20Short answer2 marksAsked: 2026

    ΔABC ~ ΔPQR and their areas are 64 cm^2 and 121 cm^2 respectively. If QR = 15.4 cm then find the value of BC.

    How the model answer begins

    1. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Since ΔABC ~ ΔPQR, their areas and corresponding sides BC and QR are related as Area(ΔABC)/Area(ΔPQR) =…

    The full answer from Medha Ma’am →

20 more questions of this chapter, each with its model answer, in the AskMedha app.

Objective questions with the board’s key (10)

1. 1. Bihar Board 2026

   If the lengths of diagonals of a rhombus are 10 cm and 24 cm respectively then each side of the rhombus is

   • (A) 5 cm
   • (B) 12 cm
   • (C) 13 cm ✓
   • (D) 17 cm

   Answer: (C)

2. 2. Bihar Board 2026

   If the diagonals of the parallelogram are equal and bisect each other at right angle then the quadrilateral is

   • (A) rectangle
   • (B) square ✓
   • (C) trapezium
   • (D) rhombus

   Answer: (B)

3. 3. Bihar Board 2026

   What relation holds between the diagonals of rhombus?

   • (A) equal
   • (B) intersect at right angle
   • (C) bisect at right angle ✓
   • (D) none of these

   Answer: (C)

4. 4. Bihar Board 2026

   All the triangles of which of the following types are similar ?

   • (A) Scalene triangle
   • (B) Isosceles triangle
   • (C) Equilateral triangle ✓
   • (D) None of these

   Answer: (C)

5. 5. Bihar Board 2026

   Two polygons having the same number of sides are similar. Then the relation among their corresponding sides is

   • (A) equal
   • (B) directly proportional ✓
   • (C) inversely proportional
   • (D) none of these

   Answer: (B)

6. 6. Bihar Board 2026

   Ratio of the areas of two similar triangles is 4 : 9 ; then the ratio of their corresponding sides is

   • (A) 3 : 2
   • (B) 9 : 4
   • (C) 4 : 9
   • (D) 2 : 3 ✓

   Answer: (D)

7. 7. Bihar Board 2026

   If delta ABC ~ delta DEF and AB/DE = 3/5 then FD/CA =

   • (A) 9 : 25
   • (B) 25 : 9
   • (C) 3 : 5
   • (D) 5 : 3 ✓

   Answer: (D)

8. 8. Bihar Board 2026

   delta ABC ~ delta PQR and their perimeters are 32 cm and 24 cm respectively. If AB = 10 cm then PQ =

   • (A) 7 cm
   • (B) 40/3 cm
   • (C) 15/2 cm ✓
   • (D) 15 cm

   Answer: (C)

9. 9. Bihar Board 2026

   Length of the altitude of an equilateral triangle having side 2a cm is

   • (A) a*sqrt(3)/2 cm
   • (B) a*sqrt(2)/3 cm
   • (C) a*sqrt(2) cm
   • (D) a*sqrt(3) cm ✓

   Answer: (D)

10. 10. Bihar Board 2026

    Which of the following options is not a Pythagorean triplet ?

    • (A) 12, 35, 37
    • (B) 28, 45, 53
    • (C) 16, 63, 65
    • (D) 17, 61, 69 ✓

    Answer: (D)

Questions parents and students ask

How many marks does "Triangles" carry in Bihar Board Class 10 Maths?

In the Bihar Board papers of 2021–2026, this chapter carried about 8 marks and about 11 questions a year; by marks it ranks 4 in Maths.

Which are the most important (VVI) questions of "Triangles"?

The 20 questions above are from the board’s real papers, each with the years it was asked; the most-asked come first with their full model answers. All 40 written questions of this chapter, with answers, are in the AskMedha app.

Do the answers follow the board’s marking scheme?

BSEB does not publish a marking scheme. Each answer is written point by point on a structure derived from past papers and checked by a second, independent model; it is never called the board’s official scheme.