Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 6: Triangles
Chapter 6 Triangles is the highest-weighted chapter in the Geometry unit of CBSE Class 10 Mathematics. Students learn about similarity of triangles through Basic Proportionality Theorem (BPT), criteria for similarity (AA, SSS, SAS), the relationship between areas of similar triangles and the ratio of their corresponding sides, and the Pythagoras Theorem with its converse. This chapter carries 8 to 10 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 6 Triangles has 5 exercises for CBSE students (Exercise 6.1 to 6.5). Exercise 6.6 (optional) has been removed from the CBSE syllabus and will NOT appear in the 2027 board exam. However, solutions for Exercise 6.6 are provided below for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board) who may still include it in their curriculum.
Table of Contents
Exercise 6.1 - Similar Figures
3 questions
Exercise 6.2 - Basic Proportionality Theorem
5 questions
Exercise 6.3 - Criteria for Similarity of Triangles
5 questions
Exercise 6.4 - Areas of Similar Triangles
5 questions
Exercise 6.5 - Pythagoras Theorem
5 questions
Exercise 6.6 - Optional (For State Boards)
3 questions • This exercise has been removed from the CBSE 2026-
Previous Year Questions
16 questions (CBSE 2019-2026)
MCQ Practice
12 questions
6.1Exercise 6.1 - Similar Figures
Q1Fill in the blanks using the correct word given in brackets: (i) All circles are ____. (similar, congruent) (ii) All squares are ____. (similar, congruent) (iii) All ____ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ____ and (b) their corresponding sides are ____. (equal, proportional)
(i) All circles are similar.
(All circles have the same shape but can differ in size.)
(ii) All squares are similar.
(All squares have equal angles of 90 degrees and proportional sides.)
(iii) All equilateral triangles are similar.
(All equilateral triangles have angles of 60 degrees each, satisfying AA similarity.)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.
Q2Give two different examples of pairs of (i) similar figures (ii) non-similar figures.
(i) Similar figures:
- Any two circles (all circles are similar)
- Any two squares (all squares are similar)
(ii) Non-similar figures:
- A circle and a triangle (different shapes entirely)
- A rectangle and a rhombus (corresponding angles not equal in general)
Q3State whether the following quadrilaterals are similar or not: (i) Two rectangles (ii) Two rhombuses
(i) Two rectangles:
Not necessarily similar. All corresponding angles are 90 degrees (equal), but the corresponding sides may not be proportional. For example, a 2x3 rectangle and a 2x5 rectangle have equal angles but sides are not proportional (2/2 is not equal to 3/5).
(ii) Two rhombuses:
Not necessarily similar. All corresponding sides are proportional (since all sides of a rhombus are equal), but corresponding angles may not be equal. For example, a rhombus with angles 60 and 120 degrees is not similar to one with angles 80 and 100 degrees.
6.2Exercise 6.2 - Basic Proportionality Theorem
Q1In the figure, DE is parallel to BC. Find EC if AD = 1.5 cm, DB = 3 cm, and AE = 1 cm.
Since DE is parallel to BC, by the Basic Proportionality Theorem (BPT):
AD/DB = AE/EC
Substituting the given values:
1.5/3 = 1/EC
Cross-multiplying:
1.5 x EC = 3 x 1
EC = 3/1.5
EC = 2 cm
Q2E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF is parallel to QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
(i) PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 1.5
Since PE/EQ is not equal to PF/FR (1.3 is not equal to 1.5), by the converse of BPT, EF is NOT parallel to QR.
(ii) PE/QE = 4/4.5 = 8/9
PF/RF = 8/9
Since PE/QE = PF/RF = 8/9, by the converse of BPT, EF is parallel to QR.
Q6In the figure, A, B and C are points on OP, OQ and OR respectively such that AB is parallel to PQ and AC is parallel to PR. Show that BC is parallel to QR.
Given: AB is parallel to PQ and AC is parallel to PR.
To prove: BC is parallel to QR.
Proof:
In triangle OPQ, since AB is parallel to PQ:
By BPT: OA/AP = OB/BQ ... (i)
In triangle OPR, since AC is parallel to PR:
By BPT: OA/AP = OC/CR ... (ii)
From (i) and (ii):
OB/BQ = OC/CR
Now in triangle OQR, since OB/BQ = OC/CR, by the converse of BPT:
BC is parallel to QR. ∎
Q8Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Given: In triangle ABC, D is the mid-point of AB, and DE is parallel to BC, meeting AC at E.
To prove: E is the mid-point of AC (i.e., AE = EC).
Proof:
Since D is the mid-point of AB: AD = DB, so AD/DB = 1.
In triangle ABC, DE is parallel to BC.
By BPT: AD/DB = AE/EC
Substituting AD/DB = 1:
1 = AE/EC
AE = EC
Therefore, E is the mid-point of AC.
A line through the mid-point of one side, parallel to another side, bisects the third side. ∎
Q9ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Given: ABCD is a trapezium with AB parallel to DC. Diagonals AC and BD intersect at O.
To prove: AO/BO = CO/DO.
Construction: Draw a line EF through O parallel to AB (and hence parallel to DC), meeting AD at E.
In triangle ADC, EO is parallel to DC (since EF is parallel to DC).
By BPT: AE/ED = AO/OC ... (i)
In triangle ADB, EO is parallel to AB.
By BPT: DE/EA = DO/OB
So AE/ED = OB/DO ... (ii)
From (i) and (ii):
AO/OC = OB/DO
Cross-multiplying:
AO/BO = CO/DO (rearranging AO x DO = OB x OC) ∎
6.3Exercise 6.3 - Criteria for Similarity of Triangles
Q1State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you and also write the pairs of similar triangles in the symbolic form: (i) In triangles ABC and PQR: angle A = angle P = 60 degrees, angle B = angle Q = 80 degrees, angle C = angle R = 40 degrees.
(i) In triangle ABC and triangle PQR:
angle A = angle P = 60 degrees
angle B = angle Q = 80 degrees
angle C = angle R = 40 degrees
All three pairs of corresponding angles are equal.
By AA (Angle-Angle) similarity criterion:
Triangle ABC ~ Triangle PQR
(Note: When two angles of one triangle are equal to two angles of another triangle, the third angles are automatically equal since the angle sum is 180 degrees. So AA criterion is sufficient.)
Q4If the areas of two similar triangles are equal, prove that they are congruent.
Given: Triangle ABC ~ Triangle PQR and ar(ABC) = ar(PQR).
To prove: Triangle ABC is congruent to Triangle PQR.
Proof:
Since Triangle ABC ~ Triangle PQR:
ar(ABC)/ar(PQR) = (AB/PQ)^2 = (BC/QR)^2 = (CA/RP)^2
Given ar(ABC) = ar(PQR):
ar(ABC)/ar(PQR) = 1
Therefore:
(AB/PQ)^2 = 1, so AB = PQ
(BC/QR)^2 = 1, so BC = QR
(CA/RP)^2 = 1, so CA = RP
Since all three pairs of corresponding sides are equal:
Triangle ABC is congruent to Triangle PQR (by SSS congruence). ∎
Q6In the figure, QR/QS = QT/QR and angle 1 = angle 2. Show that triangle PQS ~ triangle TQR.
Given: QR/QS = QT/QR and angle 1 = angle 2.
To prove: Triangle PQS ~ Triangle TQR.
Proof:
angle 1 = angle 2 (given)
This means angle PQR = angle PRQ.
Therefore, triangle PQR is isosceles with PQ = PR ... (i)
Given: QR/QS = QT/QR
Rearranging: QS/QR = QR/QT
So QS × QT = QR^2 ... (ii)
In triangles PQS and TQR:
QR/QS = QT/QR (given) means the sides including angle Q are proportional.
angle PQS = angle TQR (common angle Q).
By SAS similarity criterion:
Triangle PQS ~ Triangle TQR. ∎
Q10CD and GH are respectively the bisectors of angle ACB and angle EGF such that D and H lie on sides AB and FE of triangles ABC and EFG respectively. If triangle ABC ~ triangle FEG, show that: (i) CD/GH = AC/FG (ii) Triangle DCB ~ Triangle HGE (iii) Triangle DCA ~ Triangle HGF.
Given: Triangle ABC ~ Triangle FEG. CD bisects angle ACB. GH bisects angle EGF.
(i) To prove: CD/GH = AC/FG
Since Triangle ABC ~ Triangle FEG:
angle ACB = angle FGE
So (1/2) angle ACB = (1/2) angle FGE
angle ACD = angle FGH ... (since CD and GH are bisectors)
Also, angle A = angle F (corresponding angles of similar triangles).
In triangles DCA and HGF:
angle ACD = angle FGH (proved above)
angle A = angle F (CPST)
By AA criterion: Triangle DCA ~ Triangle HGF
So CD/GH = AC/FG (corresponding sides of similar triangles). ∎
(ii) Triangle DCB ~ Triangle HGE:
angle DCB = angle HGE (half of equal angles ACB and FGE)
angle B = angle E (CPST of ABC ~ FEG)
By AA criterion: Triangle DCB ~ Triangle HGE. ∎
(iii) Triangle DCA ~ Triangle HGF:
Already proved in part (i) above.
angle ACD = angle FGH and angle A = angle F.
By AA criterion: Triangle DCA ~ Triangle HGF. ∎
Q12Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR. Show that triangle ABC ~ triangle PQR.
Given: AB/PQ = BC/QR = AD/PM. D is the midpoint of BC. M is the midpoint of QR.
To prove: Triangle ABC ~ Triangle PQR.
Proof:
Since D is the midpoint of BC: BD = BC/2.
Since M is the midpoint of QR: QM = QR/2.
Given: AB/PQ = BC/QR = AD/PM
Since BC/QR = (2BD)/(2QM) = BD/QM:
AB/PQ = BD/QM = AD/PM
In triangles ABD and PQM:
AB/PQ = BD/QM = AD/PM (proved above)
By SSS similarity criterion:
Triangle ABD ~ Triangle PQM
Therefore: angle ABD = angle PQM
i.e., angle ABC = angle PQR
Now in triangles ABC and PQR:
AB/PQ = BC/QR (given)
angle ABC = angle PQR (proved)
By SAS similarity criterion:
Triangle ABC ~ Triangle PQR. ∎
6.4Exercise 6.4 - Areas of Similar Triangles
Q1Let triangle ABC ~ triangle DEF and their areas be, respectively, 64 sq cm and 121 sq cm. If EF = 15.4 cm, find BC.
Since Triangle ABC ~ Triangle DEF:
ar(ABC)/ar(DEF) = (BC/EF)^2
Substituting:
64/121 = (BC/15.4)^2
Taking square root:
8/11 = BC/15.4
BC = 15.4 x (8/11)
BC = (15.4 x 8)/11
BC = 123.2/11
BC = 11.2 cm
Q3In the figure, the line segment XY is parallel to side AC of triangle ABC and it divides the triangle into two parts of equal areas. Find the ratio AX/AB.
Since XY is parallel to AC, by AA similarity:
Triangle BXY ~ Triangle BAC
Given: ar(BXY) = ar(trapezium AXYC)
Since ar(BXY) + ar(AXYC) = ar(BAC):
2 x ar(BXY) = ar(BAC)
ar(BXY)/ar(BAC) = 1/2
Using the area ratio theorem:
(BX/BA)^2 = 1/2
BX/BA = 1/sqrt(2)
Now AX = AB - BX:
AX/AB = 1 - BX/AB = 1 - 1/sqrt(2)
AX/AB = (sqrt(2) - 1)/sqrt(2)
Rationalising the denominator:
AX/AB = (2 - sqrt(2))/2
Q5D, E and F are respectively the mid-points of sides AB, BC and CA of triangle ABC. Find the ratio of the areas of triangles DEF and ABC.
Since D, E, F are mid-points of AB, BC, CA respectively:
By the Mid-Point Theorem, DE is parallel to AC and DE = AC/2.
Similarly, EF is parallel to AB and EF = AB/2.
And DF is parallel to BC and DF = BC/2.
In triangles DEF and ABC:
DE/AC = EF/AB = DF/BC = 1/2
By SSS similarity: Triangle DEF ~ Triangle ABC
(Note: The correspondence is D to A, E to B, F to C.)
Ratio of areas = (ratio of sides)^2 = (1/2)^2 = 1/4
ar(DEF)/ar(ABC) = 1/4
Alternatively, the four triangles DEF, ADF, BDE, EFC are all congruent and together make up triangle ABC. So ar(DEF) = ar(ABC)/4.
Q7Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Given: ABCD is a square with side a. An equilateral triangle ABE is described on side AB. An equilateral triangle ACF is described on diagonal AC.
To prove: ar(ABE) = (1/2) x ar(ACF).
Proof:
Side of square = a.
Diagonal of square = a x sqrt(2).
All equilateral triangles are similar (by AAA, since all angles are 60 degrees).
So Triangle ABE ~ Triangle ACF.
ar(ABE)/ar(ACF) = (AB/AC)^2
= (a / (a x sqrt(2)))^2
= (1/sqrt(2))^2
= 1/2
Therefore: ar(ABE) = (1/2) x ar(ACF).
The equilateral triangle on the side equals half the equilateral triangle on the diagonal. ∎
Q9Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio:
If sides of two similar triangles are in the ratio 4:9, then:
Ratio of areas = (ratio of corresponding sides)^2
= (4/9)^2
= 16/81
Ratio of areas = 16:81
6.5Exercise 6.5 - Pythagoras Theorem
Q1Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse: (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm
A triangle is right-angled if the square of the longest side equals the sum of squares of the other two sides.
(i) 7, 24, 25:
7^2 + 24^2 = 49 + 576 = 625
25^2 = 625
Since 7^2 + 24^2 = 25^2, it IS a right triangle. Hypotenuse = 25 cm.
(ii) 3, 8, 6:
3^2 + 6^2 = 9 + 36 = 45
8^2 = 64
Since 45 is not equal to 64, it is NOT a right triangle.
(iii) 50, 80, 100:
50^2 + 80^2 = 2500 + 6400 = 8900
100^2 = 10000
Since 8900 is not equal to 10000, it is NOT a right triangle.
(iv) 13, 12, 5:
5^2 + 12^2 = 25 + 144 = 169
13^2 = 169
Since 5^2 + 12^2 = 13^2, it IS a right triangle. Hypotenuse = 13 cm.
Q4ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC^2.
Given: Triangle ABC is isosceles and right-angled at C. So AC = BC (isosceles) and angle C = 90 degrees.
To prove: AB^2 = 2AC^2.
Proof:
In right triangle ABC, by Pythagoras theorem:
AB^2 = AC^2 + BC^2
Since triangle is isosceles with AC = BC:
AB^2 = AC^2 + AC^2
AB^2 = 2AC^2 ∎
Q6In an equilateral triangle ABC, D is a point on side BC such that BD = (1/3)BC. Prove that 9AD^2 = 7AB^2.
Given: ABC is equilateral with side a (AB = BC = CA = a). D is on BC with BD = a/3.
To prove: 9AD^2 = 7AB^2.
Construction: Draw AE perpendicular to BC. Since ABC is equilateral, E is the midpoint of BC.
So BE = a/2.
BD = a/3, so DE = BE - BD = a/2 - a/3 = a/6.
In right triangle AEB (Pythagoras theorem):
AE^2 = AB^2 - BE^2 = a^2 - (a/2)^2 = a^2 - a^2/4 = 3a^2/4
In right triangle AED (Pythagoras theorem):
AD^2 = AE^2 + DE^2
AD^2 = 3a^2/4 + (a/6)^2
AD^2 = 3a^2/4 + a^2/36
AD^2 = (27a^2 + a^2)/36
AD^2 = 28a^2/36
AD^2 = 7a^2/9
Multiplying both sides by 9:
9AD^2 = 7a^2 = 7AB^2
9AD^2 = 7AB^2 ∎
Q7In the given figure, ABC is a triangle in which angle ABC > 90 degrees and AD is perpendicular to CB produced. Prove that AC^2 = AB^2 + BC^2 + 2BC x BD.
Given: In triangle ABC, angle ABC > 90 degrees. AD is perpendicular to CB produced (D is on CB produced).
To prove: AC^2 = AB^2 + BC^2 + 2BC x BD.
Proof:
In right triangle ADC:
AC^2 = AD^2 + DC^2 (Pythagoras theorem)
AC^2 = AD^2 + (DB + BC)^2
AC^2 = AD^2 + DB^2 + BC^2 + 2BC x DB
AC^2 = (AD^2 + DB^2) + BC^2 + 2BC x BD
In right triangle ADB:
AD^2 + DB^2 = AB^2 (Pythagoras theorem)
Substituting:
AC^2 = AB^2 + BC^2 + 2BC x BD
AC^2 = AB^2 + BC^2 + 2BC x BD ∎
Q13D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^2 + BD^2 = AB^2 + DE^2.
Given: Triangle ABC with angle C = 90 degrees. D is on CA. E is on CB.
To prove: AE^2 + BD^2 = AB^2 + DE^2.
Proof:
In right triangle ACE (right angle at C):
AE^2 = AC^2 + CE^2 ... (i)
In right triangle BCD (right angle at C):
BD^2 = BC^2 + CD^2 ... (ii)
Adding (i) and (ii):
AE^2 + BD^2 = AC^2 + CE^2 + BC^2 + CD^2
AE^2 + BD^2 = (AC^2 + BC^2) + (CD^2 + CE^2) ... (iii)
In right triangle ABC (right angle at C):
AB^2 = AC^2 + BC^2 ... (iv)
In right triangle DCE (right angle at C):
DE^2 = CD^2 + CE^2 ... (v)
Substituting (iv) and (v) in (iii):
AE^2 + BD^2 = AB^2 + DE^2 ∎
6.6Exercise 6.6 - Optional (For State Boards)
This exercise has been removed from the CBSE 2026-27 syllabus. It is provided here for UP Board, Bihar Board, MP Board, and other state board students who still follow the older textbook.
Q1In the figure, PS is the bisector of angle QPR of triangle PQR. Prove that QS/SR = PQ/PR.
Given: In triangle PQR, PS bisects angle QPR.
To prove: QS/SR = PQ/PR.
Construction: Draw a line through R parallel to SP, meeting QP produced at T.
Since PS is parallel to RT and PR is a transversal:
angle SPR = angle PRT (alternate interior angles) ... (i)
Since PS is parallel to RT and QT is a transversal:
angle QPS = angle PTR (corresponding angles) ... (ii)
But angle QPS = angle SPR (PS is the bisector of angle QPR) ... (iii)
From (i), (ii), and (iii):
angle PRT = angle PTR
So triangle PTR is isosceles with PT = PR ... (iv)
In triangle QRT, PS is parallel to RT.
By BPT: QS/SR = QP/PT
From (iv), PT = PR:
QS/SR = QP/PR
QS/SR = PQ/PR (Angle Bisector Theorem) ∎
Q2In the figure, D is a point on hypotenuse AC of triangle ABC, right angled at B, such that BD is perpendicular to AC. If AD = 3.6 cm, CD = 6.4 cm, find AB, BC and BD.
Given: Triangle ABC, right angle at B. BD perpendicular to AC. AD = 3.6 cm, CD = 6.4 cm.
AC = AD + CD = 3.6 + 6.4 = 10 cm.
In right triangle ABC, BD is the altitude from the right angle to the hypotenuse.
Finding BD:
In triangles ADB and BDC:
angle ADB = angle BDC = 90 degrees
angle ABD = angle BCD (both complementary to angle DBC)
By AA: Triangle ADB ~ Triangle BDC
So AD/BD = BD/CD
BD^2 = AD x CD = 3.6 x 6.4 = 23.04
BD = 4.8 cm
Finding AB:
Triangle ADB ~ Triangle ABC (AA similarity, since angle A is common and angle ADB = angle ABC = 90 degrees)
AD/AB = AB/AC
AB^2 = AD x AC = 3.6 x 10 = 36
AB = 6 cm
Finding BC:
Triangle BDC ~ Triangle ABC (AA similarity)
CD/BC = BC/AC
BC^2 = CD x AC = 6.4 x 10 = 64
BC = 8 cm
Verification: AB^2 + BC^2 = 36 + 64 = 100 = AC^2 = 10^2. Correct.
Q7Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Given: ABCD is a rhombus with diagonals AC and BD intersecting at O.
To prove: AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2.
Proof:
Diagonals of a rhombus bisect each other at right angles.
So AO = CO = AC/2, BO = DO = BD/2, and angle AOB = 90 degrees.
In right triangle AOB:
AB^2 = AO^2 + OB^2 = (AC/2)^2 + (BD/2)^2 = AC^2/4 + BD^2/4
Similarly:
BC^2 = BO^2 + OC^2 = BD^2/4 + AC^2/4
CD^2 = CO^2 + OD^2 = AC^2/4 + BD^2/4
DA^2 = DO^2 + OA^2 = BD^2/4 + AC^2/4
Adding all four:
AB^2 + BC^2 + CD^2 + DA^2 = 4 x (AC^2/4 + BD^2/4)
= AC^2 + BD^2
AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2 ∎
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Ask AI TutorPrevious Year Board Exam Questions
In triangle ABC, DE is parallel to BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find DB and EC using Basic Proportionality Theorem.
Since DE is parallel to BC, by BPT: AD/DB = AE/EC. Also by AA similarity, Triangle ADE ~ Triangle ABC, so AD/AB = DE/BC. Thus AD/AB = 2/5, giving AB = 2.4 x 5/2 = 6 cm. So DB = AB - AD = 6 - 2.4 = 3.6 cm. Similarly, AE/AC = 2/5, giving AC = 3.2 x 5/2 = 8 cm. So EC = AC - AE = 8 - 3.2 = 4.8 cm. Therefore DB = 3.6 cm and EC = 4.8 cm.
Prove the Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In triangle ABC, DE parallel to BC where D is on AB and E is on AC. To prove: AD/DB = AE/EC. Construction: Join BE and CD. Draw EM perpendicular to AB and DN perpendicular to AC. Proof: ar(ADE) = (1/2) x AD x EM and ar(BDE) = (1/2) x DB x EM. So ar(ADE)/ar(BDE) = AD/DB ...(i). Similarly ar(ADE)/ar(CDE) = AE/EC ...(ii). Since triangles BDE and CDE are on the same base DE and between the same parallels DE and BC: ar(BDE) = ar(CDE) ...(iii). From (i), (ii), (iii): AD/DB = AE/EC. Hence proved.
Prove 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. Draw the necessary construction and write the complete proof using similarity of triangles.
Given: Right triangle ABC with angle B = 90 degrees. To prove: AC^2 = AB^2 + BC^2. Construction: Draw BD perpendicular to AC. Proof: In triangles ADB and ABC: angle A = angle A (common) and angle ADB = angle ABC = 90 degrees. By AA: Triangle ADB ~ Triangle ABC. So AD/AB = AB/AC, giving AB^2 = AD x AC ...(i). In triangles BDC and ABC: angle C = angle C (common) and angle BDC = angle ABC = 90 degrees. By AA: Triangle BDC ~ Triangle ABC. So CD/BC = BC/AC, giving BC^2 = CD x AC ...(ii). Adding (i) and (ii): AB^2 + BC^2 = AD x AC + CD x AC = AC(AD + CD) = AC x AC = AC^2. Hence proved.
In triangle ABC, angle B = 90 degrees and D is the mid-point of BC. Using the Pythagoras theorem, prove that AC^2 = AD^2 + 3CD^2. Show all steps of your working clearly.
Since D is mid-point of BC: BD = CD = BC/2, so BC = 2CD. In right triangle ABC: AC^2 = AB^2 + BC^2 ...(i). In right triangle ABD: AD^2 = AB^2 + BD^2, so AB^2 = AD^2 - BD^2 = AD^2 - CD^2 ...(ii). Substituting (ii) in (i): AC^2 = AD^2 - CD^2 + (2CD)^2 = AD^2 - CD^2 + 4CD^2 = AD^2 + 3CD^2. Hence proved.
Two triangles ABC and DBC are drawn on the same base BC. If the line AD intersects BC at point O, prove that the ratio of areas of triangle ABC and triangle DBC equals AO/DO using properties of similar triangles.
Draw AM perpendicular to BC and DN perpendicular to BC. ar(ABC)/ar(DBC) = (1/2 x BC x AM)/(1/2 x BC x DN) = AM/DN. In triangles AMO and DNO: angle AOM = angle DON (vertically opposite) and angle AMO = angle DNO = 90 degrees. By AA: Triangle AMO ~ Triangle DNO. So AM/DN = AO/DO. Therefore ar(ABC)/ar(DBC) = AO/DO. Hence proved.
If the areas of two similar triangles are in the ratio 25:64, what is the ratio of their corresponding sides? Use the theorem that the ratio of areas of similar triangles equals the square of the ratio of their corresponding sides.
For similar triangles, ratio of areas = (ratio of sides)^2. So (ratio of sides)^2 = 25/64. Taking square root: ratio of sides = 5/8. Ratio of corresponding sides = 5:8.
In the figure, DE is parallel to BC in triangle ABC. If AD/DB = 3/5, find the ratio of the area of triangle ADE to the area of triangle ABC using the theorem on areas of similar triangles.
Since DE parallel to BC, by AA criterion: Triangle ADE ~ Triangle ABC. AD/DB = 3/5, so AD/AB = AD/(AD+DB) = 3/(3+5) = 3/8. Using area ratio theorem: ar(ADE)/ar(ABC) = (AD/AB)^2 = (3/8)^2 = 9/64.
Prove that if in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar. This is known as the AA similarity criterion for triangles.
Given: Triangle ABC and Triangle DEF where angle A = angle D and angle B = angle E. To prove: AB/DE = BC/EF = CA/FD. Construction: Take points P on DE and Q on EF such that AB = DP and BC = EQ. Join PQ. Proof: In triangles ABC and DPQ: AB = DP, angle B = angle E, BC = EQ. By SAS congruence: Triangle ABC is congruent to Triangle DPQ. So angle DPQ = angle A = angle D. Since angle DPQ = angle D, PQ is parallel to DF (corresponding angles equal). By BPT in triangle DEF: DP/DE = EQ/EF. Since DP = AB and EQ = BC: AB/DE = BC/EF. Similarly AC/DF = AB/DE. Therefore AB/DE = BC/EF = CA/FD and hence Triangle ABC ~ Triangle DEF.
In triangle ABC, if AD is the median to side BC and E is the midpoint of median AD, then prove that the area of triangle BED is equal to one-fourth of the area of triangle ABC.
AD is the median of triangle ABC, so D is midpoint of BC. A median divides a triangle into two equal areas. So ar(ABD) = (1/2) ar(ABC) ...(i). E is the midpoint of AD. In triangle ABD, BE is the median (since E is midpoint of AD). So BE divides triangle ABD into two equal areas. ar(BED) = (1/2) ar(ABD) ...(ii). From (i) and (ii): ar(BED) = (1/2) x (1/2) ar(ABC) = (1/4) ar(ABC). Hence proved.
In triangle PQR, PD is perpendicular to QR. If PQ = 10 cm, QD = 8 cm and DR = 6 cm, find PR using the Pythagoras theorem applied to the two right triangles.
In right triangle PDQ: PD^2 = PQ^2 - QD^2 = 100 - 64 = 36, so PD = 6 cm. In right triangle PDR: PR^2 = PD^2 + DR^2 = 36 + 36 = 72. PR = 6 times sqrt(2) cm, which is approximately 8.49 cm.
In the figure, if triangle ABC ~ triangle DEF and their sides are of lengths (in cm) as marked along them, then find the lengths of the sides of each triangle.
Since Triangle ABC ~ Triangle DEF, corresponding sides are proportional. Using the given ratios: AB/DE = BC/EF = AC/DF. Let the unknown sides be expressed using the proportionality constant. Solving the equations using cross multiplication gives the specific side lengths. The key principle is that if triangles are similar with ratio k, then every pair of corresponding sides shares this same ratio k.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. The height of the tower is:
The pole and tower with their shadows form similar triangles (AA similarity, same sun angle). Height/Shadow ratio is constant. 6/4 = h/28. h = 6 x 28/4 = 42 m.
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first mentioned side is a right angle (converse of Pythagoras theorem).
Given: In triangle ABC, AC^2 = AB^2 + BC^2. To prove: angle B = 90 degrees. Construction: Construct triangle PQR with angle Q = 90 degrees, PQ = AB, QR = BC. Proof: In right triangle PQR: PR^2 = PQ^2 + QR^2 = AB^2 + BC^2 (by construction). But AC^2 = AB^2 + BC^2 (given). So PR^2 = AC^2, which gives PR = AC. In triangles ABC and PQR: AB = PQ, BC = QR, AC = PR. By SSS congruence: Triangle ABC is congruent to Triangle PQR. Therefore angle B = angle Q = 90 degrees. Hence proved.
In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1. Also, in triangle ABC, D and E are points on sides AB and AC such that DE is parallel to BC. If AD = 4 cm, DB = 6 cm, find AE/EC.
For the BPT part: Since DE is parallel to BC, by the Basic Proportionality Theorem: AD/DB = AE/EC. Given AD = 4 cm and DB = 6 cm, we get AE/EC = 4/6 = 2/3. This means AE and EC are in the ratio 2:3.
In triangle ABC, if angle A = angle C, and AB = 6 cm, BC = 8 cm, and AC = 6 cm, find whether triangle ABC is similar to triangle given, and prove that the triangles where DE is parallel to BC satisfy the Basic Proportionality Theorem.
Since angle A = angle C and AB = AC = 6 cm, triangle ABC is isosceles. For any line DE parallel to BC cutting AB at D and AC at E, by BPT: AD/DB = AE/EC. This follows directly from the theorem that a line parallel to one side of a triangle divides the other two sides in equal ratio. The proof uses equal area triangles on the same base between parallel lines.
If triangle ABC ~ triangle RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm, and PQ = 10 cm, then find QR.
Since Triangle ABC ~ Triangle RPQ, corresponding sides are proportional: AB/RP = BC/PQ = AC/QR. Substituting the given values: 3/6 = 5/10 = 6/QR. From 3/6 = 6/QR, cross-multiplying gives QR = 6 x 6/3 = 12 cm. The ratio of similarity is 1:2.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): If in two triangles, two angles of one triangle are respectively equal to two angles of the other triangle, then the two triangles are similar.
Reason (R): If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio.
Answer: (A)
Both are true. The assertion states the AA similarity criterion, and the reason correctly explains it. When two angles match, the third automatically matches (angle sum = 180 degrees), making corresponding sides proportional and thus the triangles similar.
AR2
Assertion (A): The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Reason (R): The areas of two similar triangles are in the ratio of the squares of corresponding sides.
Answer: (A)
Both are true and the reason explains the assertion. Since medians of similar triangles are in the same ratio as corresponding sides (can be proved using similarity), the ratio of areas equals the square of the ratio of medians, which equals the square of the ratio of sides.
AR3
Assertion (A): Two similar triangles with equal areas are congruent.
Reason (R): The ratio of areas of two similar triangles equals the square of the ratio of their corresponding sides.
Answer: (A)
Both true, and the reason explains the assertion. If ar(T1)/ar(T2) = 1 (equal areas), then (side1/side2)^2 = 1, so side1 = side2 for all corresponding sides. Equal corresponding sides means the triangles are congruent by SSS.
AR4
Assertion (A): If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
Reason (R): A line drawn parallel to one side of a triangle divides the other two sides proportionally.
Answer: (B)
Both are true. The assertion is the converse of BPT, and the reason is BPT itself. However, the reason (BPT) does not explain the assertion (converse of BPT). The converse requires a separate proof and is not a direct consequence of the original theorem.
AR5
Assertion (A): In an equilateral triangle ABC, if D is a point on BC such that BD = (1/3)BC, then 9AD^2 = 7AB^2.
Reason (R): In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Answer: (A)
Both are true, and the reason (Pythagoras theorem) is used in proving the assertion. The proof requires dropping a perpendicular from A to BC and applying Pythagoras theorem to the right triangles formed, giving AD^2 = 7AB^2/9.
AR6
Assertion (A): All congruent figures are similar, but all similar figures are not necessarily congruent.
Reason (R): Two polygons of the same number of sides are similar if all corresponding angles are equal and all corresponding sides are proportional.
Answer: (A)
Both true and the reason explains the assertion. Congruent figures have equal angles and equal sides (ratio 1:1), satisfying similarity conditions. But similar figures only need proportional sides, not equal sides. For example, two squares of different sizes are similar but not congruent.
AR7
Assertion (A): A triangle ABC with sides 6 cm, 10 cm and 8 cm is a right-angled triangle.
Reason (R): A triangle with sides a, b, c (where c is the largest) is right-angled if a^2 + b^2 = c^2.
Answer: (A)
Both true and the reason explains the assertion. Check: 6^2 + 8^2 = 36 + 64 = 100 = 10^2. Since the sum of squares of the two shorter sides equals the square of the longest side, by the converse of Pythagoras theorem, it is a right triangle with hypotenuse 10 cm.
Case Study Questions
Rohan is a class 10 student. He was doing a school project on shadows and heights. At a particular time of the day, he observed that a vertical lamp post of height 6 m casts a shadow of 4 m. At the same time, he measured the shadow of his school building and found it to be 28 m. He also noticed a tree nearby that cast a shadow of 8 m.
(i) What is the height of the school building? (1 mark)
The lamp post and building form similar triangles with the ground (same sun angle at the same time). Using AA similarity:
Height/Shadow = constant.
6/4 = h/28
h = 6 x 28/4 = 42 m.
The school building is 42 m tall.
(ii) What is the height of the tree? (1 mark)
Using the same ratio:
6/4 = h/8
h = 6 x 8/4 = 12 m.
The tree is 12 m tall.
(iii)(a) What is the ratio of the areas of the triangles formed by the lamp post and the school building with their shadows? (2 marks)
The triangles formed by the lamp post (height 6 m, shadow 4 m) and building (height 42 m, shadow 28 m) are similar.
Ratio of sides = 6/42 = 1/7.
Ratio of areas = (1/7)^2 = 1/49 or 1:49.
(iii)(b) If the shadow of the lamp post becomes 6 m at a different time, what will be the shadow of the tree at that time? (2 marks)
The ratio height/shadow remains constant for all objects at the same time.
New shadow of lamp post = 6 m. Height of lamp post = 6 m.
So height/shadow = 6/6 = 1.
Shadow of tree = height of tree/1 = 12 m.
A city council plans to build a triangular park ABC where angle B = 90 degrees, AB = 9 m and BC = 12 m. A straight path DE is to be built inside the park parallel to BC where D is on AB and E is on AC. The path DE divides the park into two regions, and the smaller triangular region ADE is to be a children's play area.
(i) Find the length of AC. (1 mark)
In right triangle ABC with angle B = 90 degrees:
AC^2 = AB^2 + BC^2 = 81 + 144 = 225.
AC = 15 m.
(ii) If AD = 3 m, find the length of DE. (1 mark)
Since DE is parallel to BC, Triangle ADE ~ Triangle ABC (AA similarity).
AD/AB = DE/BC
3/9 = DE/12
DE = 4 m.
(iii)(a) Find the ratio of the area of the children's play area (triangle ADE) to the area of the whole park (triangle ABC). (2 marks)
Since Triangle ADE ~ Triangle ABC with ratio AD/AB = 3/9 = 1/3:
ar(ADE)/ar(ABC) = (1/3)^2 = 1/9.
The play area is 1/9th of the total park area.
(iii)(b) If the council wants the play area to be exactly half of the total park area, how far from A should D be placed on AB? (2 marks)
For ar(ADE)/ar(ABC) = 1/2:
(AD/AB)^2 = 1/2
(AD/9)^2 = 1/2
AD^2 = 81/2 = 40.5
AD = 9/sqrt(2) = (9 x sqrt(2))/2, which is approximately 6.36 m.
MCQ Practice
MCQ1. In triangle ABC, DE is parallel to BC. If AD = 4 cm, DB = 5 cm, and AE = 8 cm, then EC is:
(A) 8 cm
(B) 10 cm
(C) 12 cm
(D) 14 cm
Answer: (B) - By BPT: AD/DB = AE/EC. So 4/5 = 8/EC. EC = 8 x 5/4 = 10 cm.
MCQ2. If triangle ABC ~ triangle DEF, AB = 4 cm, BC = 6 cm, EF = 9 cm, then DE is:
(A) 4 cm
(B) 6 cm
(C) 9 cm
(D) 8 cm
Answer: (B) - Since Triangle ABC ~ Triangle DEF: AB/DE = BC/EF. So 4/DE = 6/9. DE = 4 x 9/6 = 6 cm.
MCQ3. The areas of two similar triangles are in the ratio 16:81. The ratio of their corresponding sides is:
(A) 16:81
(B) 4:9
(C) 2:3
(D) 9:4
Answer: (B) - Ratio of sides = sqrt(ratio of areas) = sqrt(16/81) = 4/9. So the ratio is 4:9.
MCQ4. If in triangle ABC, AB = 6 cm, BC = 8 cm, AC = 10 cm, then angle B is:
(A) 30 degrees
(B) 45 degrees
(C) 60 degrees
(D) 90 degrees
Answer: (D) - AB^2 + BC^2 = 36 + 64 = 100 = AC^2 = 10^2. By converse of Pythagoras theorem, angle B = 90 degrees.
MCQ5. All equilateral triangles are:
(A) congruent
(B) similar
(C) both similar and congruent
(D) neither similar nor congruent
Answer: (B) - All equilateral triangles have angles 60-60-60, satisfying AA similarity. But they need not have equal sides, so they are similar but not necessarily congruent.
MCQ6. If triangle ABC ~ triangle PQR with AB/PQ = 1/3, then ar(ABC)/ar(PQR) is:
(A) 1/3
(B) 1/9
(C) 9
(D) 3
Answer: (B) - ar(ABC)/ar(PQR) = (AB/PQ)^2 = (1/3)^2 = 1/9.
MCQ7. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is:
(A) acute
(B) right angle
(C) obtuse
(D) cannot be determined
Answer: (B) - This is the converse of the Pythagoras Theorem. If c^2 = a^2 + b^2, the angle opposite to side c is a right angle (90 degrees).
MCQ8. A vertical stick 20 m long casts a shadow 10 m long. At the same time, a tower casts a shadow 50 m long. The height of the tower is:
(A) 50 m
(B) 100 m
(C) 120 m
(D) 200 m
Answer: (B) - Using similar triangles (same sun angle): 20/10 = h/50. h = 20 x 50/10 = 100 m.
MCQ9. If triangle ABC ~ triangle DEF, the perimeters of the two triangles are 30 cm and 20 cm respectively. If DE = 4 cm, then AB is:
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm
Answer: (A) - For similar triangles, ratio of perimeters = ratio of corresponding sides. So AB/DE = 30/20 = 3/2. AB = 4 x 3/2 = 6 cm.
MCQ10. Which of the following is NOT a similarity criterion for triangles?
(A) AA
(B) SSS
(C) SAS
(D) ASA
Answer: (D) - AA, SSS, and SAS are the three similarity criteria for triangles. ASA is a congruence criterion, not a similarity criterion. For similarity, we need proportional sides, not equal sides.
MCQ11. In triangle ABC, D and E are points on AB and AC respectively such that DE is parallel to BC. If AD = 2 cm, BD = 3 cm, then AE/AC is:
(A) 2/3
(B) 3/5
(C) 2/5
(D) 3/2
Answer: (C) - By BPT: AD/DB = AE/EC = 2/3. So AE/AC = AE/(AE+EC). If AE = 2k, EC = 3k, then AE/AC = 2k/5k = 2/5.
MCQ12. In a right triangle, the square of the hypotenuse is 169 sq cm. If one side is 5 cm, the other side is:
(A) 10 cm
(B) 12 cm
(C) 13 cm
(D) 14 cm
Answer: (B) - Hypotenuse^2 = 169, so hypotenuse = 13 cm. Other side = sqrt(169 - 25) = sqrt(144) = 12 cm.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
State the Basic Proportionality Theorem (Thales' Theorem).
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
If triangle ABC ~ triangle DEF with AB/DE = 2/3, what is the ratio of their areas?
Ratio of areas = (2/3)^2 = 4/9 or 4:9.
In triangle PQR, if PQ = 6 cm, QR = 8 cm, PR = 10 cm, is triangle PQR a right triangle?
Yes. PQ^2 + QR^2 = 36 + 64 = 100 = PR^2. By the converse of Pythagoras theorem, angle Q = 90 degrees.
Write the three criteria for similarity of triangles.
The three criteria are: AA (Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side) similarity criteria.
If a line divides two sides of a triangle in equal ratio, what can you say about the line?
By the converse of BPT, the line is parallel to the third side of the triangle.
2-3 Mark Questions
In triangle ABC, DE is parallel to BC. If AD = 3 cm, DB = 5 cm, AE = 4.5 cm, find EC.
By BPT: AD/DB = AE/EC. So 3/5 = 4.5/EC. EC = 4.5 x 5/3 = 7.5 cm.
The sides of two similar triangles are in the ratio 4:9. Find the ratio of (i) their areas (ii) their altitudes (iii) their medians.
(i) Ratio of areas = (4/9)^2 = 16/81. (ii) Ratio of altitudes = 4/9 (same as sides). (iii) Ratio of medians = 4/9 (same as sides).
In a right triangle ABC, right-angled at B, D is a point on AC such that BD is perpendicular to AC. If AD = 8 cm and DC = 2 cm, find BD.
BD^2 = AD x DC = 8 x 2 = 16. BD = 4 cm. (Using the property: altitude from right angle to hypotenuse creates similar triangles.)
In triangle ABC, angle B = 90 degrees and D is the midpoint of BC. Prove that AC^2 = AD^2 + 3CD^2.
BD = CD (D is midpoint), BC = 2CD. In right triangle ABC: AC^2 = AB^2 + BC^2. In right triangle ABD: AB^2 = AD^2 - BD^2 = AD^2 - CD^2. Substituting: AC^2 = AD^2 - CD^2 + 4CD^2 = AD^2 + 3CD^2.
3-4 Mark Questions
Prove the Basic Proportionality Theorem.
Given: DE parallel to BC in triangle ABC. To prove: AD/DB = AE/EC. Construction: Draw EM perpendicular to AB, DN perpendicular to AC. Join BE, CD. Proof: ar(ADE)/ar(BDE) = AD/DB (same height EM). ar(ADE)/ar(CDE) = AE/EC (same height DN). ar(BDE) = ar(CDE) (same base DE, between parallels DE and BC). Therefore AD/DB = AE/EC.
Prove the Pythagoras Theorem: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
Given: Triangle ABC with angle B = 90 degrees. Construction: Draw BD perpendicular to AC. In triangle ADB ~ triangle ABC (AA): AB^2 = AD x AC ...(i). In triangle BDC ~ triangle ABC (AA): BC^2 = CD x AC ...(ii). Adding: AB^2 + BC^2 = AC(AD + CD) = AC x AC = AC^2.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Let side = a. Altitude h bisects the base: base half = a/2. By Pythagoras: h^2 = a^2 - a^2/4 = 3a^2/4. So 4h^2 = 3a^2. Therefore 3(side)^2 = 4(altitude)^2.
Key Concepts
Important Formulas
Study Plan - Triangles
Similar Figures and Basic Proportionality Theorem
Read NCERT theory on similar figures and BPT. Understand the proof of BPT thoroughly. Solve Exercise 6.1 (all 3 questions) and Exercise 6.2 Q1-Q5.
BPT Applications and Converse
Complete Exercise 6.2 Q6-Q10. Focus on trapezium diagonal problems and proving lines parallel using converse of BPT.
Criteria for Similarity (AA, SSS, SAS)
Learn all three similarity criteria with examples. Solve Exercise 6.3 Q1-Q10. Practice identifying which criterion applies to each problem.
Areas of Similar Triangles and Pythagoras Theorem
Study the area ratio theorem. Solve Exercise 6.4. Begin Exercise 6.5 Q1-Q8. Focus on numerical problems using Pythagoras theorem.
Pythagoras Theorem Proofs and Previous Year Questions
Complete Exercise 6.5 Q9-Q17. Write the proof of Pythagoras theorem from memory. Solve all 16 PYQs from this page. Focus on proof-based questions.
Full Revision and Practice Test
Attempt all MCQs, Assertion-Reason, and Case Study questions under timed conditions. Review all important questions mark-wise. Revise common mistakes.
Common Mistakes to Avoid
- ✗Writing the similarity correspondence in wrong order. If Triangle ABC ~ Triangle PQR, then A corresponds to P, B to Q, C to R. The order of vertices matters for ratios.
- ✗Confusing similarity with congruence. Similar triangles have proportional sides (ratio can be any value), while congruent triangles have equal sides (ratio is exactly 1:1).
- ✗Forgetting to square the ratio when finding areas of similar triangles. The area ratio equals the SQUARE of the side ratio, not the side ratio itself.
- ✗In BPT problems, using AD/AB = AE/EC instead of AD/DB = AE/EC. The correct BPT statement divides the two sides in the same ratio (AD/DB = AE/EC), not segment to whole side.
- ✗Not drawing the correct construction line when proving BPT or Pythagoras theorem. In the BPT proof, you need perpendiculars EM and DN. In Pythagoras, you need BD perpendicular to AC.
- ✗Applying the converse of Pythagoras theorem incorrectly by not checking whether the LARGEST side's square equals the sum of squares of the other two. Always identify the longest side first.
Tips for Board Exam
- ✓BPT proof and Pythagoras theorem proof are guaranteed 5-mark questions. Memorise both proofs with constructions and practice writing them in under 7 minutes each.
- ✓For area ratio questions, always remember: ar(T1)/ar(T2) = (s1/s2)^2. This works for sides, altitudes, medians, and perimeters, but areas use the SQUARE of the ratio.
- ✓When identifying similar triangles, check AA first as it requires only two angle pairs. SSS and SAS need side calculations. Write the correspondence correctly in your answer.
- ✓In Pythagoras-based proofs like 9AD^2 = 7AB^2, always drop a perpendicular to create right triangles, then apply Pythagoras theorem to each right triangle separately.
- ✓For MCQs on similarity, verify your answer by checking if the ratio is consistent across all pairs of corresponding sides. A common trap is reversed correspondence.
Practice Triangles with AI Tutor
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Start Triangles - Choose Your ModeAbout Triangles - Class 10 Mathematics
Chapter 6 Triangles is the most important chapter in the Geometry unit of CBSE Class 10 Mathematics. It carries 8 to 10 marks in the board exam, making it the highest-weighted chapter in Geometry. The chapter builds on the concept of congruence learned in earlier classes and introduces the powerful idea of similarity, which extends geometric reasoning to figures that have the same shape but different sizes.
What Is Similarity?
Two figures are said to be similar if they have the same shape but not necessarily the same size. Congruent figures are a special case of similar figures where the size is also equal. For example, all circles are similar to each other, all squares are similar, and all equilateral triangles are similar. However, not all rectangles are similar (a 2 by 3 rectangle is not similar to a 2 by 5 rectangle even though all angles are 90 degrees) and not all triangles are similar.
For two polygons to be similar, two conditions must be satisfied: all corresponding angles must be equal, and all corresponding sides must be proportional. For triangles, these conditions are not independent. If the angles match, the sides are automatically proportional, and vice versa. This is what makes similarity criteria for triangles simpler than for general polygons.
Basic Proportionality Theorem
The Basic Proportionality Theorem, also known as Thales' Theorem or BPT, is the foundation of this chapter. It states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
In triangle ABC, if DE is parallel to BC where D lies on AB and E lies on AC, then AD/DB = AE/EC. The proof uses the concept of equal areas. By drawing perpendiculars from D and E and comparing the areas of triangles ADE, BDE, and CDE, we establish the proportionality. The key insight is that triangles BDE and CDE have equal areas because they share the same base DE and lie between the same parallel lines DE and BC.
The converse of BPT is equally important: if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. This converse is used to verify whether a given line segment is parallel to a side of the triangle.
BPT has several practical applications in the NCERT exercises. It is used to prove properties of trapeziums (diagonals of a trapezium divide each other proportionally), to establish that a line through the midpoint of one side parallel to another side bisects the third side (midpoint theorem), and to solve numerical problems involving finding unknown lengths.
Criteria for Similarity
Chapter 6 introduces three criteria for establishing similarity between triangles.
The AA (Angle-Angle) criterion states that if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. Since the sum of angles in any triangle is 180 degrees, when two pairs of angles match, the third pair automatically matches as well. This is the most commonly used criterion in board exam questions.
The SSS (Side-Side-Side) criterion states that if the corresponding sides of two triangles are proportional (in the same ratio), then the triangles are similar. For example, if AB/PQ = BC/QR = CA/RP, then triangle ABC is similar to triangle PQR.
The SAS (Side-Angle-Side) criterion states that if one angle of a triangle equals one angle of another triangle, and the sides including these angles are proportional, then the triangles are similar.
When writing the similarity relation, the order of vertices matters. If triangle ABC is similar to triangle PQR, it means A corresponds to P, B to Q, and C to R. This correspondence determines which sides are proportional: AB/PQ = BC/QR = CA/RP.
Areas of Similar Triangles
One of the most important theorems in this chapter states that the ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides. If triangle ABC is similar to triangle DEF with sides in ratio k, then the ratio of their areas is k squared.
This result extends to altitudes, medians, and angle bisectors as well. The ratio of corresponding altitudes, medians, or angle bisectors of similar triangles equals the ratio of their corresponding sides. But for areas, the ratio is always the square of the side ratio.
This theorem is frequently tested in board exams, both as numerical problems (given area ratio, find side ratio, or vice versa) and as proof-based questions (proving that the area of an equilateral triangle on the side of a square is half the area of the equilateral triangle on its diagonal).
Pythagoras Theorem
The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. If angle B equals 90 degrees in triangle ABC, then AC^2 = AB^2 + BC^2.
The proof of the Pythagoras Theorem given in NCERT uses the concept of similar triangles. By drawing the altitude BD from the right angle vertex B to the hypotenuse AC, two smaller triangles ADB and BDC are created. Both are similar to the original triangle ABC by the AA criterion. From triangle ADB similar to triangle ABC, we get AB^2 = AD times AC. From triangle BDC similar to triangle ABC, we get BC^2 = CD times AC. Adding these two equations gives AB^2 + BC^2 = AC times (AD + CD) = AC times AC = AC^2.
The converse of the Pythagoras Theorem states that if in a triangle the square of one side equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. The proof constructs a reference right triangle with two sides equal to the given triangle and uses SSS congruence to establish equality.
Applications of the Pythagoras Theorem in the NCERT exercises include verifying whether given triangles are right-angled, finding unknown sides, proving algebraic relationships between sides (such as 9AD^2 = 7AB^2 in an equilateral triangle), and extending to obtuse and acute angle triangles using the generalised Pythagoras results.
How Tested in Board Exams
Triangles questions in CBSE board exams follow predictable patterns. One MCQ worth 1 mark tests basic concepts like identifying similar triangles, finding area ratios, or applying the converse of Pythagoras theorem. One or two short answer questions worth 2 to 3 marks involve applying BPT to find unknown lengths, computing area ratios, or solving numerical Pythagoras problems. One long answer question worth 5 marks asks students to prove either the Basic Proportionality Theorem or the Pythagoras Theorem with full construction and logical steps.
Since 2023, CBSE has also included case study questions based on real-life applications of similarity and Pythagoras theorem, such as shadow-height problems, architectural designs, and land measurement scenarios. Assertion-Reason questions test conceptual understanding of similarity criteria and area theorems.
Tips for Full Marks
To score full marks from this chapter, students should focus on four areas. First, memorise the proofs of BPT and Pythagoras theorem with all construction steps. These are guaranteed long-answer questions worth 5 marks each. Second, practice identifying which similarity criterion (AA, SSS, or SAS) applies to a given problem and always write the triangle correspondence in the correct order. Third, remember that area ratios use the square of the side ratio. Fourth, for Pythagoras-based algebraic proofs, the standard technique is to drop a perpendicular to create right triangles and then apply the theorem to each.
Common mistakes include writing the similarity correspondence in wrong order, confusing similarity with congruence, forgetting to square the ratio for areas, and not identifying the longest side before applying the converse of Pythagoras theorem.
Connection to Other Chapters
Triangles connects to several other chapters in Class 10 Maths. The Pythagoras Theorem is directly used in Chapter 7 (Coordinate Geometry) through the distance formula, which is essentially Pythagoras applied on a coordinate plane. The concept of similarity appears in Chapter 9 (Some Applications of Trigonometry) where shadow and height problems use similar triangles. Chapter 10 (Circles) uses tangent properties that involve right triangles. The area concepts from this chapter connect to Chapter 11 (Areas Related to Circles). For students preparing for competitive exams, triangle similarity and Pythagoras theorem form the backbone of advanced geometry questions in Olympiads and JEE Foundation papers.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 6 Triangles for CBSE 2026-27?
For CBSE 2026-27, Chapter 6 Triangles has 5 exercises: Exercise 6.1 (3 questions on similar figures), Exercise 6.2 (10 questions on BPT), Exercise 6.3 (16 questions on similarity criteria), Exercise 6.4 (9 questions on areas of similar triangles), and Exercise 6.5 (17 questions on Pythagoras theorem). Exercise 6.6 (10 questions) is optional and removed from CBSE but included for state board students.
What is the Basic Proportionality Theorem (BPT) in Class 10 Maths?
The Basic Proportionality Theorem (also called Thales' Theorem) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In triangle ABC, if DE is parallel to BC, then AD/DB = AE/EC.
How many marks does Chapter 6 Triangles carry in the CBSE Class 10 board exam 2027?
Triangles is part of the Geometry unit (15 marks total) and carries approximately 8 to 10 marks. It is the highest-weighted chapter in Geometry. Typical distribution: 1 MCQ (1 mark) + 1-2 short answers (2-3 marks each) + 1 long proof (5 marks).
What are the three criteria for similarity of triangles in Chapter 6?
The three criteria are: (1) AA (Angle-Angle): two angles of one triangle equal two angles of another. (2) SSS (Side-Side-Side): all corresponding sides are proportional. (3) SAS (Side-Angle-Side): one pair of angles equal and the including sides proportional.
How to prove the Pythagoras Theorem for CBSE board exam?
In right triangle ABC with angle B = 90 degrees: (1) Draw BD perpendicular to AC. (2) Show Triangle ADB ~ Triangle ABC (AA) to get AB^2 = AD x AC. (3) Show Triangle BDC ~ Triangle ABC (AA) to get BC^2 = CD x AC. (4) Add: AB^2 + BC^2 = AC(AD + CD) = AC^2. This 5-mark proof requires the construction step and two applications of AA similarity.
What is the relationship between areas of similar triangles?
The ratio of areas of two similar triangles equals the square of the ratio of their corresponding sides. If Triangle ABC ~ Triangle PQR with sides in ratio a:b, then ar(ABC)/ar(PQR) = a^2/b^2. This also equals the square of the ratio of corresponding altitudes, medians, or perimeters.
Is Exercise 6.6 deleted from CBSE Class 10 Maths 2026-27?
Yes, Exercise 6.6 (which includes the Angle Bisector Theorem and related problems) has been removed from the CBSE 2026-27 syllabus and will not appear in the 2027 board exam. However, it may still be part of UP Board, Bihar Board, MP Board, and other state board syllabi.
What type of questions come from Chapter 6 Triangles in the CBSE board exam?
Five types: (1) MCQs on similarity, area ratios, BPT, Pythagoras - 1 mark. (2) Short answers applying BPT to find lengths - 2 marks. (3) Short proofs like AD^2 + 3CD^2 = AC^2 - 3 marks. (4) Long proofs of BPT or Pythagoras theorem - 5 marks. (5) Case study questions on shadows, heights, or park designs using similar triangles - 4 marks.