NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.2

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.2 are provided here to help students understand the concept of similarity of triangles in a clear and structured way. These solutions are prepared by subject experts following the latest CBSE guidelines to support students in their board exam preparation.

Exercise 6.2 focuses on the criteria for similarity of triangles, including:

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.2

Triangles

Medium

Q.

Fill in the blanks using the correct word given in brackets:
(i) All circles are ­­­­­­­­_______. (congruent, similar)
(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)

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, E is a point on side CB produced of an isosceles triangle ABC}\\ \text{with AB}=\text{AC}\text{. If AD}\perp \text{BC and EF}\perp \text{AC, prove that }\Delta \text{ABD}\sim \Delta \text{ECF}\text{.}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, AD is a median of a triangle ABC and AM}\perp \text{BC. Prove that:}\\ {\text{(i) AC}}^{\text{2}}={\text{AD}}^{\text{2}}+\text{BC.DM}+{\left(\frac{\text{BC}}{2}\right)}^{\text{2}}\\ {\text{(ii) AB}}^{\text{2}}={\text{AD}}^{\text{2}}-\text{BC . DM}+{\left(\frac{\text{BC}}{2}\right)}^{\text{2}}\\ {\text{(iii) AC}}^{\text{2}}+{\text{AB}}^{\text{2}}={\text{2AD}}^{\text{2}}+\frac{1}{2}{\text{BC}}^{\text{2}}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, ABC is a triangle in which }\angle \text{ABC > 90° and AD}\perp \text{CB produced. Prove that}\\ {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2+2\mathrm{BC}.\mathrm{BD}.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, D is a point on hypotenuse AC of }\mathrm{\Delta }\text{ABC, DM}\perp \text{BC and DN}\perp \text{AB. Prove that:}\\ \text{(i) }{\mathrm{DM}}^2=\mathrm{DN}.\mathrm{MC}\text{ (ii) }{\mathrm{DN}}^2=\mathrm{DM}.\mathrm{AN}\end{array}$

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Triangles

Medium

Q.

$\text{In the following figure, PS is the bisector of }\angle \text{QPR of }\mathrm{\Delta }\text{PQR. Prove that }\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}.$

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Triangles

Medium

Q.

$\begin{array}{l}\text{The perpendicular from A on side BC of a }\mathrm{\Delta }\text{ABC intersects BC at D such that }\mathrm{DB}=3\mathrm{CD}\text{ (see the}\\ \text{following figure). Prove that }2{\mathrm{AB}}^2=2{\mathrm{AC}}^2+{\mathrm{BC}}^2.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, ABD is a triangle right angled\hspace{0.17em}at A and AC\hspace{0.33em}}\perp \mathrm{B}\text{D. }\\ \text{Show that}\\ \left(\text{i}\right){\text{ AB}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}}.\text{\hspace{0.33em}BD}\\ \left(\text{ii}\right){\text{ AC}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}.\hspace{0.33em}DC}\\ \left(\text{iii}\right){\text{ AD}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BD\hspace{0.33em}.\hspace{0.33em}CD}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and}\\ \text{median PM of }\mathrm{\Delta }\text{PQR (see the following figure). Show that }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively. }\\ \text{Prove that:}\\ \text{(i) }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}\\ \text{(ii) }\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}\end{array}$

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Triangles

Medium

Q.

In the following figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, altitudes AD and CE of }\mathrm{\Delta }\text{ABC intersect each other at the point P. Show that:}\\ \text{(i) }\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{CDP}\\ \text{(ii) }\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{CBE}\\ \text{(iii)}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{ADB}\\ \text{(iv)}\mathrm{\Delta }\text{PDC}~\mathrm{\Delta }\text{BEC}\end{array}$

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Triangles

Medium

Q.

$\text{In the following figure, if }\mathrm{\Delta }\text{ABE}\cong \mathrm{\Delta }\text{ACD, show that }\mathrm{\Delta }\text{ADE}~\mathrm{\Delta }\text{ABC.}$

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Triangles

Medium

Q.

$\text{In the following figure, }\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\text{ and }\angle \text{1}=\angle 2.\text{ Show that }\Delta \text{PQS}\sim \Delta \text{TQR}\text{.}$

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Triangles

Medium

Q.

State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

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Triangles

Medium

Q.

In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR.Show that BC ∥ QR.

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Triangles

Medium

Q.

$\text{In the following figure, }\mathrm{DE}\parallel \mathrm{OQ}\text{ and }\mathrm{DF}\parallel \mathrm{OR}.\text{\hspace{0.17em}Show that }\mathrm{EF}\parallel \mathrm{QR}.$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, }\mathrm{DE}\parallel \mathrm{AC}\text{ and }\mathrm{DF}\parallel \mathrm{AE}.\\ \text{Prove that }\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, if }\mathrm{LM}\parallel \mathrm{CB}\text{ and }\mathrm{LN}\parallel \mathrm{CD},\text{ prove that}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, two chords AB and CD of a circle intersect each other at the point P }\\ \text{(when produced) outside the circle. Prove that}\\ \text{(i) }\mathrm{\Delta }\text{PAC}~\mathrm{\Delta }\text{PDB (ii) }\mathrm{PA}\cdot \mathrm{PB}=\mathrm{PC}\cdot \mathrm{PD}\end{array}$

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• AA (Angle–Angle) Similarity Criterion

• SAS (Side–Angle–Side) Similarity Criterion

• SSS (Side–Side–Side) Similarity Criterion

Students learn how to prove that two triangles are similar using these conditions and apply the properties of similar triangles to solve problems involving proportional sides and corresponding angles.

By practicing this exercise, students strengthen their understanding of:

• Properties of similar triangles

• Ratio of corresponding sides

• Proof-based geometry questions

• Logical reasoning in geometry

The solutions are explained step-by-step to ensure conceptual clarity and exam-oriented preparation.

Q1. In Fig. 6.17 (i) and (ii), DE || BC.

Find EC in (i) and AD in (ii).

Reasoning:

By the Basic Proportionality Theorem (BPT) (Thales Theorem):
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

Two triangles are similar if:

1. Corresponding angles are equal.

2. Corresponding sides are proportional.

Solution:

(i)

Since DE || BC in ΔABC,

ΔABC ~ ΔADE

Therefore,

AD / DB = AE / EC

Given values:

1.5 / 3 = 1 / EC

Solving:

EC = 2 cm

(ii)

Similarly,

AD / DB = AE / EC

Given:

7.2 / 5.4 = AD / 1.8

Solving:

AD = 2.4 cm

Q2. E and F are points on sides PQ and PR of ΔPQR.

State whether EF || QR in each case:

(i)

PE = 3.9 cm
EQ = 3 cm
PF = 3.6 cm
FR = 2.4 cm

Check ratio:

PE / EQ = 3.9 / 3 = 1.3
PF / FR = 3.6 / 2.4 = 1.5

Since ratios are not equal,

EF is not parallel to QR.

(ii)

PE = 4 cm
EQ = 4.5 cm
PF = 8 cm
FR = 9 cm

PE / EQ = 4 / 4.5
PF / FR = 8 / 9

Both ratios are equal.

Therefore,

EF || QR

(iii)

PQ = 1.28 cm
PE = 0.18 cm

EQ = PQ − PE = 1.10 cm

PR = 2.56 cm
PF = 0.36 cm

FR = PR − PF = 2.20 cm

Now check ratio:

PE / EQ = 0.18 / 1.10
PF / FR = 0.36 / 2.20

Both simplify to:

9 / 55

Ratios are equal.

Therefore,

EF || QR

Q3. In Fig. 6.18, if LM || CB and LN || CD, prove that:

AM / AB = AN / AD

Solution:

In ΔABC,

LM || CB

By BPT:

AM / MB = AL / LC ..........(1)

In ΔACD,

LN || CD

AN / DN = AL / LC ..........(2)

From (1) and (2):

AM / MB = AN / DN

Rearranging and adding 1 on both sides:

AM / AB = AN / AD

Hence Proved.

Q4. In Fig. 6.19, DE || AC and DF || AE.

Prove that:

BF / FE = BE / EC

Solution:

In ΔABC,

DE || AC

BD / AD = BE / EC ..........(1)

In ΔABE,

DF || AE

BD / AD = BF / FE ..........(2)

From (1) and (2):

BE / EC = BF / FE

Hence Proved.

Q5. In Fig. 6.20, DE || OQ and DF || OR.

Show that EF || QR.

Solution:

In ΔPOQ,

DE || OQ

PE / EQ = PD / DO ..........(1)

In ΔPOR,

DF || OR

PF / FR = PD / DO ..........(2)

From (1) and (2):

PE / EQ = PF / FR

By Converse of BPT,

EF || QR

Q6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that

AB || PQ and AC || PR. Show that BC || QR.

Solution:

In ΔOPQ,

AB || PQ

OA / AP = OB / BQ ..........(1)

In ΔOPR,

AC || PR

OA / AP = OC / CR ..........(2)

From (1) and (2):

OB / BQ = OC / CR

By Converse of BPT:

BC || QR

Q7. Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Solution:

In ΔABC,
D is midpoint of AB

So,

AD = BD

If DE || BC, then by BPT:

AE / EC = AD / BD

Since AD = BD,

AE = EC

Hence, E is midpoint of AC.

Q8. Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Solution:

In ΔABC,

D is midpoint of AB → AD = BD

E is midpoint of AC → AE = EC

Therefore,

AD / BD = AE / EC

By Converse of BPT,

DE || BC

Q9. ABCD is a trapezium in which AB || DC.

Diagonals intersect at O. Show that:

AO / BO = CO / DO

Solution:

Construct XY || AB and XY || DC through O.

In ΔABC,

OY || AB

BY / CY = AO / OC ..........(1)

In ΔBCD,

OY || DC

BY / CY = BO / OD ..........(2)

From (1) and (2):

AO / OC = BO / OD

Hence Proved.

Q10. The diagonals of quadrilateral ABCD intersect at O such that:

AO / BO = CO / DO

Show that ABCD is a trapezium.

Solution:

Given:

AO / BO = CO / DO

Draw OE || AB.

Using BPT:

OA / OC = BE / CE

Comparing with given ratio:

OB / OD = BE / CE

Therefore,

OE || AB

Hence,

AB || CD

So ABCD is a trapezium.

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.2

Q1. What is the main focus of Exercise 6.2?

Answer:
Exercise 6.2 focuses on the similarity criteria of triangles and solving problems based on AA, SAS, and SSS similarity conditions.

Q2. What are the similarity criteria of triangles?

Answer:
Two triangles are similar if:

• Their corresponding angles are equal (AA similarity).

• Two sides are proportional and the included angle is equal (SAS similarity).

• All three sides are proportional (SSS similarity).

Q3. Why is similarity important in this chapter?

Answer:
Similarity helps in solving geometric problems involving proportional sides, height-distance problems, and proof-based questions, which are commonly asked in board exams.

Q4. How is Exercise 6.2 different from Exercise 6.1?

Answer:
Exercise 6.1 focuses mainly on the Basic Proportionality Theorem (BPT), while Exercise 6.2 deals specifically with the criteria for similarity of triangles.

Q5. How do NCERT Solutions help in exam preparation?

Answer:
These solutions provide clear, step-by-step explanations aligned with the CBSE syllabus, helping students improve their understanding and score better in examinations.