NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.3

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.3 are provided here to help students master the concept of similar triangles and their properties. These solutions are prepared by subject experts according to the latest CBSE syllabus and guidelines to ensure accurate and step-by-step explanations.

Exercise 6.3 mainly focuses on the properties of similar triangles, including:

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.3

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.

$\begin{array}{l}\text{D, E and F are respectively the mid-points of sides AB, BC and CA of }\mathrm{\Delta }\text{ABC. Find the ratio of the areas}\\ \text{of }\mathrm{\Delta }\text{DEF and }\mathrm{\Delta }\text{ABC.}\end{array}$

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Triangles

Easy

Q.

If $$\rm \triangle {ABC} \sim \triangle {PQR}\ {\rm in\ which\ }\ {AB}=6 {~\rm cm},\ {BC}=4 {~\rm cm}, {AC\ }=8 {~\rm cm}$$ and PR = 6 cm, then find the length of (PQ + QR)

[CBSE - 2025]

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Triangles

Medium

Q.

ABC is an equilateral triangle of side 2a, then length of one of its altitude is _______ .

[CBSE - 2020]

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Triangles

Medium

Q.

In triangles $$\rm ABC\ {\rm and}\ {DEF}, \angle {B}=\angle {E}, \angle {F}=\angle {C}\ {\rm and}\ {AB}=3 {DE}$$​.Then, the two triangles are:

[CBSE - 2025]

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Triangles

Medium

Q.

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the following figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

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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.

$\begin{array}{l}\text{ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles}\\ \text{ABC and BDE is}\\ \text{(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the}\\ \text{equilateral triangle described on one of its diagonals.}\end{array}$

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Triangles

Medium

Q.

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

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Triangles

Medium

Q.

If the areas of two similar triangles are equal, prove that they are congruent.

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Triangles

Medium

Q.

$\text{In the following figure, }\Delta \text{ODC}\sim \Delta \text{OBA, }\angle \text{BOC}=125°\text{and }\angle \text{CDO}=70°.\text{ Find }\angle \text{DOC, }\angle \text{DCO and }\angle \text{OAB}\text{.}$

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Triangles

Medium

Q.

In the following figure, ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O, show that ar(ΔABC) (ΔDBC) = AO DO.

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Triangles

Medium

Q.

$\begin{array}{l}\text{If AD and PM are medians of triangles ABC and PQR, respectively where }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR, prove that}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}.\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Sides AB and AC and median AD of a triangle ABC }\\ \text{are respectively proportional to sides PQ and PR and }\\ \text{median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR}\end{array}$

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Triangles

Medium

Q.

$\text{D is a point on the side BC of a triangle ABC such that }\angle \text{ADC}=\angle {\text{BAC. Show that CA}}^{\text{2}}=\text{CB}·\text{CD.}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{CD and GH are respectively the bisectors of }\mathrm{\Delta }\text{ACB and }\mathrm{\Delta }\text{EGF such that D and H lie on sides AB and FE}\\ \text{of }\mathrm{\Delta }\text{ABC and }\mathrm{\Delta }\text{EFG respectively. If }\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{FEG, show that:}\\ \text{(i) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCB}~\mathrm{\Delta }\text{HGE}\\ \text{(iii)\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{HGF}\end{array}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD}\\ \text{at F. Show that }\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{CFB.}\end{array}$

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Triangles

Medium

Q.

$\text{S and T are points on sides PR and QR of }\mathrm{\Delta }\text{PQR such that }\angle \text{P}=\angle \text{RTS. Show that }\mathrm{\Delta }\text{RPQ}~\mathrm{\Delta }\text{RTS.}$

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Triangles

Medium

Q.

$\begin{array}{l}\text{Diagonals AC and BD of a trapezium ABCD with AB}\parallel \text{DC intersect each other at the point O. Using}\\ \text{a similarity criterion for two triangles, show that }\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}.\end{array}$

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Triangles

Medium

Q.

In triangles $ABC$ and $DEF$, it is given that $\angle B = \angle E$, $\angle F = \angle C$, and $AB = 3DE$. Determine if the two triangles are similar and/or congruent, and justify your reasoning.

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• The ratio of corresponding sides

• The ratio of areas of similar triangles

• Relationship between sides and areas

• Applications of similarity in problem-solving

Students learn that:

• The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

• If two triangles are similar, their corresponding angles are equal and corresponding sides are proportional.

By practicing Exercise 6.3, students improve their:

• Conceptual understanding of similarity

• Logical reasoning and proof-writing skills

• Problem-solving techniques for board exams

These NCERT solutions are explained in a simple and exam-oriented manner to help students score high marks.

Q1. State which pairs of triangles in Fig. 6.34 are similar.

Write the similarity criterion used and write the pairs in symbolic form.

(1)

Reasoning:
If all corresponding angles of two triangles are equal, then the triangles are similar.
This is called AAA (Angle–Angle–Angle) similarity criterion.

Solution:
In triangles ΔABC and ΔPQR:

∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°

All corresponding angles are equal.

By AAA criterion:
ΔABC ~ ΔPQR

(2)

Reasoning:
If corresponding sides of two triangles are proportional, the triangles are similar.
This is called SSS (Side–Side–Side) similarity criterion.

Solution:
In triangles ΔABC and ΔQRP:

AB/QR = 2/4 = 1/2
BC/PR = 2.5/5 = 1/2
AC/PQ = 3/6 = 1/2

All sides are proportional.

By SSS criterion:
ΔABC ~ ΔQPR

(3)

Reasoning:
For SSS similarity, all three corresponding sides must be in the same ratio.

Solution:
In triangles ΔLMP and ΔFED:

LM/FE = 2.7/5
MP/ED = 2/4 = 1/2
LP/FD = 3/6 = 1/2

Ratios are not all equal.

Hence, triangles are not similar:
ΔLMP is not similar to ΔFED

(4)

Reasoning:
If one corresponding angle is equal and the sides including that angle are proportional, the triangles are similar.
This is called SAS (Side–Angle–Side) similarity criterion.

Solution:
In triangles ΔNML and ΔPQR:

NM/PQ = 2.5/5 = 1/2
ML/QR = 5/10 = 1/2
∠M = ∠Q = 70°

Sides including the angle are proportional and included angle is equal.

By SAS criterion:
ΔNML ~ ΔPQR

(5)

Reasoning:
SAS applies only when the equal angle is included between the proportional sides.

Solution:
In triangles ΔABC and ΔDFE:

AB/DF = 2.5/5 = 1/2
BC/EF = 3/6 = 1/2

But the angle given is ∠A = ∠F = 80°, while the proportional sides AB and BC include ∠B (not ∠A).
So SAS condition is not satisfied.

Hence, triangles are not similar:
ΔABC is not similar to ΔDFE

(6)

Reasoning:
If all corresponding angles are equal, triangles are similar (AAA).
(Alternate method: AA similarity also works.)

Solution:
In ΔDEF:

∠D = 70°
∠E = 80°
∠F = 30° (Angle sum property)

In ΔPQR:

∠P = 70°
∠Q = 80°
∠R = 30°

All corresponding angles are equal.

By AAA criterion:
ΔDEF ~ ΔPQR

Alternate Method (AA):
If two corresponding angles are equal, triangles are similar.
So ΔDEF ~ ΔPQR by AA criterion as well.

Q2. In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°.

Find ∠DOC, ∠DCO and ∠OAB.

Solution:

Since ∠DOC and ∠COB form a linear pair:

∠DOC = 180° − 125°
∠DOC = 55°

In ΔODC (angle sum property):

∠DCO = 180° − (∠DOC + ∠CDO)
∠DCO = 180° − (55° + 70°)
∠DCO = 55°

Given ΔODC ~ ΔOBA, so corresponding angles are equal:

∠DCO = ∠OAB

Therefore:

∠OAB = 55°

Final Answers:
∠DOC = 55°
∠DCO = 55°
∠OAB = 55°

Q3. Diagonals AC and BD of trapezium ABCD (AB || DC) intersect at O.

Show that: OA/OB = OC/OD

Reasoning:
If two angles of one triangle are equal to two angles of another triangle, triangles are similar (AA).

Solution:
Consider ΔAOB and ΔCOD:

∠AOB = ∠COD (vertically opposite angles)
∠BAO = ∠DCO (alternate interior angles as AB || DC)

So, ΔAOB ~ ΔCOD (AA criterion)

Hence, corresponding sides are proportional:

OA/OB = OC/OD

Q4. In Fig. 6.36, QR/QT = QS/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Solution:
Given ∠1 = ∠2 and sides around that angle are proportional.

By SAS similarity, ΔPQS ~ ΔTQR

Q5. S and T are points on PR and QR of ΔPQR such that ∠PRT = ∠RTS.

Show that ΔRPQ ~ ΔRTS.

Solution:
Two angles are equal and one angle is common → AA criterion

Therefore, ΔRPQ ~ ΔRTS

Q6. In Fig. 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Solution:
From congruence:

AE = AD
AB = AC

Also ∠DAE = ∠BAC (common angle)

By SAS similarity:

ΔADE ~ ΔABC

Q7. In Fig. 6.38, altitudes AD and CE of ΔABC intersect at P. Show that:

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

Solution (All by AA similarity):

(i) ∠AEP = ∠CDP = 90° and ∠APE = ∠CPD (vertically opposite)
⇒ ΔAEP ~ ΔCDP

(ii) ∠ADB = ∠CEB = 90° and one common angle
⇒ ΔABD ~ ΔCBE

(iii) Both have 90° and one common angle
⇒ ΔAEP ~ ΔADB

(iv) Both have 90° and one common angle
⇒ ΔPDC ~ ΔBEC

Q8. E is a point on side AD produced of parallelogram ABCD and BE intersects CD at F.

Show that ΔABE ~ ΔCFB.

Solution:
∠BAE = ∠FCB (opposite angles of parallelogram)
∠AEB = ∠FBC (alternate angles as AE || BC)

By AA criterion:
ΔABE ~ ΔCFB

Q9. In Fig. 6.39, ABC and AMP are right triangles (right angled at B and M). Prove:

(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP

Solution:
(i) Both have a right angle and one common angle → AA similarity
⇒ ΔABC ~ ΔAMP

(ii) In similar triangles, corresponding sides are proportional:
CA/PA = BC/MP

Q10. CD and GH bisect ∠ACB and ∠EGF. If ΔABC ~ ΔFEG, show that:

(i) CD/AC = GH/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Solution:
Since ΔABC ~ ΔFEG, corresponding angles equal.
Angle bisectors create equal halves, so triangles formed are similar by AA.
Hence required relations are proved.

Q11. In Fig. 6.40, E is on CB produced of isosceles ΔABC (AB = AC). If AD ⟂ BC and EF ⟂ AC, prove ΔABD ~ ΔECF.

Solution:
∠ADB = ∠EFC = 90°
and ∠ABD = ∠ECF (angles in isosceles triangle)
By AA similarity: ΔABD ~ ΔECF

Q12. AB, BC and median AD of ΔABC are proportional to PQ, QR and median PM of ΔPQR.

Show that ΔABC ~ ΔPQR.

Solution:
Given proportional corresponding sides and medians, use SAS similarity through median triangles.
Hence ΔABC ~ ΔPQR.

Q13. D is a point on BC of ΔABC such that ∠ADC = ∠BAC. Show that CA² = CB × CD.

Solution:
ΔABC ~ ΔDAC (AA criterion)

So, CA/CD = CB/CA
⇒ CA² = CB × CD

Q14. AB, AC and median AD are proportional to PQ, PR and median PM. Show that ΔABC ~ ΔPQR.

Solution:
Using construction + median properties and SSS/SAS similarity, triangles become similar.
Hence ΔABC ~ ΔPQR.

Q15. A pole 6 m casts shadow 4 m. Tower casts shadow 28 m. Find tower height.

Solution:
Similar triangles formed by sun rays:

6/4 = H/28
H = (6 × 28) / 4 = 42 m

Height of tower = 42 m

Q16. If AD and PM are medians of similar triangles ΔABC ~ ΔPQR, prove that:

AB/AD = PQ/PM

Solution:
Since triangles are similar, corresponding sides are proportional.
Medians of similar triangles are also proportional in the same ratio.
Hence AB/AD = PQ/PM.

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.3

Q1. What is the main topic of Exercise 6.3?

Answer:
Exercise 6.3 focuses on the relationship between the sides and areas of similar triangles and solving problems based on these properties.

Q2. What is the important formula used in this exercise?

Answer:
If two triangles are similar, then:

$$\frac{\text{Area of } \triangle_1}{\text{Area of } \triangle_2} = \left( \frac{\text{Corresponding side of } \triangle_1}{\text{Corresponding side of } \triangle_2} \right)^2$$

Area of △2​Area of △1​​=(Corresponding side of △2​Corresponding side of △1​​)2

Q3. How do you find the ratio of areas of similar triangles?

Answer:
First, find the ratio of corresponding sides. Then square that ratio to get the ratio of their areas.

Q4. Why is Exercise 6.3 important for board exams?

Answer:
Questions based on similarity and area ratios are frequently asked in CBSE board exams. Understanding these concepts helps in solving proof-based and application-based questions efficiently.

Q5. How do NCERT Solutions help students?

Answer:
NCERT Solutions provide clear, step-by-step explanations aligned with CBSE guidelines, making it easier for students to understand and revise the chapter effectively.