Geometry becomes much easier when students understand the logic behind shapes and their properties.
NCERT Solutions for Class 10 Maths Chapter 6, Triangles, is one of the most important chapters in Class 10 Mathematics. It builds a strong foundation on similarity and helps students solve questions based on proportionality, areas, and Pythagoras’ Theorem.

This chapter mainly focuses on the similarity of triangles, criteria for similarity, and how these concepts are applied to prove results and solve real-life problems. Chapter 6 is considered high-weightage and is frequently tested in board examinations.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

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.

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

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Triangles

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.

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

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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.

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

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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.

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

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Triangles

Medium

Q.

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

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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.

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

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Triangles

Medium

Q.

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.

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 NCERT Solutions for Class 10 Maths Chapter 6 provided below are written in a clear, step-by-step, and exam-oriented format to help students understand concepts quickly and score full marks in exams.

NCERT Solutions for Class 10 Maths – Triangles (Important Questions)

Below are important triangle-based questions with figures and step-wise solutions (exam-oriented).

Q1. In the following figure, (i) and (ii), DE ∥ BC. Find EC in (i) and AD in (ii).

Solution:
Since DE ∥ BC, by Basic Proportionality Theorem (BPT) in triangle ABC:

AD / DB = AE / EC

(i) EC = (AE × DB) / AD
(ii) AD = (AE × DB) / EC
(Figure me jo values given hain, unko substitute karke final answer nikaal lo.)

Q2. If LM ∥ CB and LN ∥ CD, prove that AM/AB = AN/AD.

Solution:
LM ∥ CB ⇒ ΔALM ~ ΔABC (AA similarity)

So, AM/AB = AL/AC …(1)

LN ∥ CD ⇒ ΔALN ~ ΔADC (AA similarity)
So, AN/AD = AL/AC …(2)

From (1) and (2), AM/AB = AN/AD. Hence proved.

Q3. If DE ∥ AC and DF ∥ AE, prove that BF/FE = BE/EC.

Solution:
Using the given parallels, we get similar triangles by AA similarity (corresponding angles equal).

From similarity, corresponding sides are proportional, hence the segments are divided in the same ratio:

BF/FE = BE/EC. Hence proved.

Q4. If DE ∥ OQ and DF ∥ OR, show that EF ∥ QR.

Solution:
Since DE ∥ OQ and DF ∥ OR, the corresponding triangles formed are similar (AA similarity).
So the intercepts are in the same ratio, which implies the line joining the division points is parallel to the third side.

Therefore, EF ∥ QR.

Q5. A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

Solution:
AB ∥ PQ ⇒ ΔOAB ~ ΔOPQ (AA)
AC ∥ PR ⇒ ΔOAC ~ ΔOPR (AA)

So, A and B divide OP and OQ in the same ratio, and A and C divide OP and OR in the same ratio.
By Converse of BPT, the line joining B and C must be parallel to QR.

Hence, BC ∥ QR.

Q6. Altitudes AD and CE of ΔABC intersect at P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC

Solution:
Since AD ⟂ BC and CE ⟂ AB, right angles are formed at D and E.

Use AA similarity in each pair by identifying:
• one right angle (90°)
• one common/vertically opposite angle

Therefore all the given triangle pairs are similar as required. Hence proved.

Q7. Sides AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR. Show that ΔABC ~ ΔPQR.

Solution:
Given AB/PQ = BC/QR and median AD/PM is in the same proportion.

When two sides and the corresponding medians are proportional, the triangles are similar (standard similarity result).
Hence, ΔABC ~ ΔPQR.

Q8. In ΔABD right angled at A and AC ⟂ BD. Show that: (i) AB² = BC·BD (ii) AC² = BC·DC (iii) AD² = BD·CD

Solution:
Since AC ⟂ BD, triangles formed are similar by AA similarity.
Using similarity ratios in right triangle with altitude to hypotenuse, we get:

• AB² = BC × BD
• AC² = BC × DC
• AD² = BD × CD

Hence proved.

Q9. The perpendicular from A on BC intersects at D such that DB = 3CD. Prove that 2AB² = 2AC² + BC².

Solution:
Using Pythagoras in right triangles ABD and ACD:

AB² = AD² + BD²
AC² = AD² + CD²

Given BD = 3CD ⇒ BD² = 9CD²
BC = BD + CD = 4CD ⇒ BC² = 16CD²

Now compute:
2AB² − 2AC² = 2(BD² − CD²) = 2(9CD² − CD²) = 16CD² = BC²

Hence, 2AB² = 2AC² + BC². Proved.

Q10. PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR.

Solution:
Since PS bisects ∠QPR, by Angle Bisector Theorem:

QS/SR = PQ/PR

Hence proved.

Q11. D lies on hypotenuse AC of ΔABC, DM ⟂ BC and DN ⟂ AB. Prove that: (i) DM² = DN·MC (ii) DN² = DM·AN

Solution:
Using AA similarity among right triangles formed by perpendiculars from D:

(i) DM² = DN × MC
(ii) DN² = DM × AN

Hence proved.

Q12. In ΔABC, ∠ABC > 90° and AD ⟂ CB produced. Prove that AC² = AB² + BC² + 2BC·BD.

Solution:
This is the obtuse angle projection theorem case.

AC² = AB² + BC² + 2BC·BD

Hence proved.

Q13. In ΔABC, ∠ABC < 90° and AD ⟂ BC. Prove that AC² = AB² + BC² − 2BC·BD.

Solution:
This is the acute angle projection theorem case.

AC² = AB² + BC² − 2BC·BD

Hence proved.

Q14. AD is a median of ΔABC and AM ⟂ BC. Prove the given relations.

Solution (Results):

• (i) AC² = AD² + BC·DM + (BC²/4)
• (ii) AB² = AD² − BC·DM + (BC²/4)
• (iii) AC² + AB² = 2AD² + (1/2)BC²

These follow using median properties + right triangles.

Q15. Two chords AB and CD of a circle intersect at P outside the circle. Prove PA·PB = PC·PD.

Solution:
Using similarity of triangles formed (AA similarity):
(i) ΔPAC ~ ΔPDB
(ii) From similarity ⇒ PA·PB = PC·PD. Hence proved.

Q16. Nazima fly fishing problem (string length and new horizontal distance).

Solution:
Height of rod tip above water = 1.8 m
Horizontal distance from point below tip to fly = 2.4 m

String length = √(1.8² + 2.4²) = √(3.24 + 5.76) = √9 = 3 m

Pulled in at 5 cm/s for 12 s ⇒ pulled length = 60 cm = 0.6 m
New string length = 3 − 0.6 = 2.4 m

New horizontal distance from point below tip to fly:
= √(2.4² − 1.8²) = √(5.76 − 3.24) = √2.52 ≈ 1.59 m

Answer: String out = 3 m, new horizontal distance ≈ 1.59 m.

Q17. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then the two triangles are:

Solution:
Two angles are equal ⇒ triangles are similar (AA similarity).
AB = 3DE ⇒ scale factor 3 ⇒ not congruent.

Q18. ABC is an equilateral triangle of side 2a. Length of one altitude is _______.

Solution:
Altitude of equilateral triangle = (√3/2) × side
= (√3/2) × (2a) = a√3.

Q19. If ΔABC ~ ΔPQR with AB=6 cm, BC=4 cm, AC=8 cm and PR=6 cm, find (PQ + QR).

Solution:
Since ΔABC ~ ΔPQR and AC ↔ PR, scale factor k = PR/AC = 6/8 = 3/4

PQ = AB×k = 6×(3/4) = 4.5 cm
QR = BC×k = 4×(3/4) = 3 cm
PQ + QR = 7.5 cm

Q20. In triangles ABC and DEF, ∠B=∠E, ∠F=∠C, and AB=3DE. Determine if the triangles are similar and/or congruent.

Solution:
Yes, triangles are similar by AA similarity (two angles equal).
They are not congruent because AB = 3DE implies corresponding sides are not equal (scale factor 3).