NCERT Solutions for Class 10 Maths Chapter 6: Triangles — Free PDF

Chapter 6 is one of the most important chapters for CBSE board exams. It covers the Basic Proportionality Theorem (BPT), criteria for similarity of triangles, areas of similar triangles, and the Pythagoras Theorem with its converse.

The chapter has 6 exercises covering:
- Exercise 6.1: Similar figures and basic concepts
- Exercise 6.2: BPT (Thales' Theorem) and its converse
- Exercise 6.3: AA, SSS, and SAS similarity criteria
- Exercise 6.4: Areas of similar triangles
- Exercise 6.5: Pythagoras Theorem and its converse
- Exercise 6.6: Mixed and challenging problems

This chapter typically carries 8-10 marks in the board exam, with theorem proofs being frequently asked long-answer questions. The proofs of BPT and the Pythagoras Theorem are both 5-mark questions that appear regularly. Students should not only memorise these proofs but also understand the reasoning behind each step, as examiners sometimes ask variations.

Similar Figures: Two figures are similar if they have the same shape but not necessarily the same size. All congruent figures are similar, but similar figures need not be congruent.

Similar Triangles: Two triangles are similar if:
- Their corresponding angles are equal, AND
- Their corresponding sides are in the same ratio (proportional)

If $\triangle ABC \sim \triangle DEF$, then $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$ and $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$.

BPT (Basic Proportionality Theorem / Thales' Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally:

Converse of BPT: If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

Similarity Criteria:
- AA (Angle-Angle): If two angles of one triangle are equal to two angles of another, the triangles are similar.
- SSS (Side-Side-Side): If all three pairs of corresponding sides are in the same ratio, the triangles are similar.
- SAS (Side-Angle-Side): If one pair of angles is equal and the including sides are proportional, the triangles are similar.

Theorem on Areas: If $\triangle ABC \sim \triangle DEF$, then:

The ratio of areas equals the square of the ratio of corresponding sides.

Pythagoras Theorem: In a right triangle with hypotenuse $c$ and legs $a$, $b$:

Converse of Pythagoras Theorem: If $c^2 = a^2 + b^2$ for the sides of a triangle, then the angle opposite the longest side $c$ is $90^\circ$.

**Problem 1: In $\triangle PQR$, $E$ and $F$ are points on $PQ$ and $PR$ respectively. $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm, $FR = 2.4$ cm. Is $EF \parallel QR$?**

Solution:

Since $\dfrac{PE}{EQ} \neq \dfrac{PF}{FR}$, by the converse of BPT, $EF$ is not parallel to $QR$.

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**Problem 2: In $\triangle ABC$, $DE \parallel BC$ with $AD = x$, $DB = x - 2$, $AE = x + 2$, $EC = x - 1$. Find $x$.**

Solution:
By BPT:

Cross-multiplying:

Verification: $AD = 4$, $DB = 2$, $AE = 6$, $EC = 3$. $\dfrac{4}{2} = \dfrac{6}{3} = 2$. ✓

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**Problem 3: $ABCD$ is a trapezium with $AB \parallel DC$. Diagonals intersect at $O$. If $\dfrac{AO}{OC} = \dfrac{BO}{OD}$, prove $AB \parallel DC$.**

Solution:
Draw a line through $O$ parallel to $AB$, meeting $AD$ at $E$.

In $\triangle DAB$, $EO \parallel AB$, so by BPT:

Given: $\dfrac{AO}{OC} = \dfrac{BO}{OD}$, which gives $\dfrac{DO}{OB} = \dfrac{CO}{OA}$ ... (2)

From (1) and (2): $\dfrac{DE}{EA} = \dfrac{CO}{OA}$.

In $\triangle ADC$, since $\dfrac{DE}{EA} = \dfrac{CO}{OA}$, by the converse of BPT, $EO \parallel DC$.

But $EO \parallel AB$, therefore $AB \parallel DC$. $\blacksquare$

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**Problem 4: In $\triangle ABC$, $DE \parallel BC$, $AD = 4$ cm, $AB = 12$ cm. If $AE = 3$ cm, find $AC$.**

Solution:
By BPT: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$

Answer: $AC = 9$ cm.

**Problem 1: $\triangle ABC \sim \triangle DEF$. The perimeters are $30$ cm and $18$ cm. If $BC = 9$ cm, find $EF$.**

Solution:
Since the triangles are similar, the ratio of perimeters equals the ratio of corresponding sides.

Answer: $EF = 5.4$ cm.

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**Problem 2: In $\triangle ABC$, $AD \perp BC$. Prove that $AB^2 + CD^2 = AC^2 + BD^2$.**

Solution:
In right $\triangle ADB$: $AB^2 = AD^2 + BD^2 \quad \cdots (1)$

In right $\triangle ADC$: $AC^2 = AD^2 + CD^2 \quad \cdots (2)$

From (1): $AD^2 = AB^2 - BD^2$.

Substituting in (2): $AC^2 = AB^2 - BD^2 + CD^2$.

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**Problem 3: In the figure, $\dfrac{AO}{OC} = \dfrac{BO}{OD} = \dfrac{1}{2}$. Show that $\triangle AOB \sim \triangle COD$.**

Solution:
Given: $\dfrac{AO}{OC} = \dfrac{BO}{OD} = \dfrac{1}{2}$.

$\angle AOB = \angle COD$ (vertically opposite angles).

In $\triangle AOB$ and $\triangle COD$:
- $\dfrac{AO}{CO} = \dfrac{BO}{DO} = \dfrac{1}{2}$ (given)
- $\angle AOB = \angle COD$ (vertically opposite)

By SAS similarity: $\triangle AOB \sim \triangle COD$. $\blacksquare$

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**Problem 4: $\triangle ABC$ and $\triangle DBC$ are on the same base $BC$. $AD$ and $BC$ intersect at $O$. Prove $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DBC)} = \dfrac{AO}{DO}$.**

Solution:
Draw $AM \perp BC$ and $DN \perp BC$.

$\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DBC)} = \dfrac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times BC \times DN} = \dfrac{AM}{DN}$

In $\triangle AOM$ and $\triangle DON$:
- $\angle AMO = \angle DNO = 90^\circ$
- $\angle AOM = \angle DON$ (vertically opposite)

By AA similarity: $\triangle AOM \sim \triangle DON$.

**Problem 1: The sides of two similar triangles are in the ratio $4:9$. Find the ratio of their areas.**

Solution:
By the area ratio theorem:

Answer: The ratio of areas is $16 : 81$.

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**Problem 2: $\triangle ABC \sim \triangle PQR$. $\text{ar}(\triangle ABC) = 64$ cm$^2$, $\text{ar}(\triangle PQR) = 121$ cm$^2$. If $QR = 15.4$ cm, find $BC$.**

Solution:

Answer: $BC = 11.2$ cm.

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**Problem 3: The areas of two similar triangles are $81$ cm$^2$ and $49$ cm$^2$. If the altitude of the larger triangle is $4.5$ cm, find the altitude of the smaller triangle.**

Solution:
For similar triangles, the ratio of altitudes equals the ratio of corresponding sides.

Answer: The altitude of the smaller triangle is $3.5$ cm.

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**Problem 4: Two similar triangles have areas in the ratio $25:36$. If a side of the first triangle is $10$ cm, find the corresponding side of the second.**

Solution:

Answer: $12$ cm.

**Problem 1: The sides of a triangle are $5$ cm, $12$ cm, and $13$ cm. Is it right-angled?**

Solution:
Check if the square of the longest side equals the sum of squares of the other two:

Since $13^2 = 5^2 + 12^2$, by the converse of the Pythagoras theorem, the triangle is right-angled with the right angle opposite the $13$ cm side.

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**Problem 2: A ladder $10$ m long reaches $8$ m high on a wall. Find the distance of the foot of the ladder from the wall.**

Solution:
Let the distance from the wall be $x$.

Answer: The foot of the ladder is $6$ m from the wall.

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**Problem 3: In $\triangle ABC$, $\angle B = 90^\circ$ and $D$ is the midpoint of $BC$. Prove $AC^2 = AD^2 + 3CD^2$.**

Solution:
Since $D$ is the midpoint of $BC$: $BD = CD = \dfrac{BC}{2}$, so $BC = 2CD$.

In right $\triangle ABC$:

In right $\triangle ABD$:

From (2): $AB^2 = AD^2 - CD^2$.

Substituting in (1):

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**Problem 4: $ABC$ is an isosceles triangle with $AB = AC = 13$ cm and $BC = 10$ cm. Find the altitude from $A$ to $BC$.**

Solution:
The altitude from $A$ meets $BC$ at $D$, the midpoint (by symmetry). So $BD = 5$ cm.

In right $\triangle ABD$:

Answer: The altitude is $12$ cm.

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**Problem 5: In a right triangle, the hypotenuse is $p$ cm. If the other two sides differ by $1$ cm, find them.**

Solution:
Let the sides be $a$ and $a + 1$.

Solving for specific values: if $p = 5$, then $25 = 2a^2 + 2a + 1 \implies 2a^2 + 2a - 24 = 0 \implies a^2 + a - 12 = 0 \implies (a+4)(a-3) = 0$. So $a = 3$ and the sides are $3$ cm and $4$ cm.

**Example 1: In $\triangle ABC$, $DE \parallel BC$ and $\dfrac{AD}{DB} = \dfrac{3}{5}$. If $AC = 4.8$ cm, find $AE$.**

Solution:
By BPT: $\dfrac{AD}{DB} = \dfrac{AE}{EC} = \dfrac{3}{5}$.

So $\dfrac{AE}{AC} = \dfrac{AE}{AE + EC} = \dfrac{3}{3+5} = \dfrac{3}{8}$.

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**Example 2: A vertical pole of length $6$ m casts a shadow $4$ m long. At the same time, a tower casts a shadow $28$ m long. Find the height of the tower.**

Solution:
The sun's rays create similar triangles (same angle of elevation).

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**Example 3: The diagonals of a quadrilateral $ABCD$ intersect at $O$ such that $\dfrac{AO}{OC} = \dfrac{BO}{OD}$. Show that $ABCD$ is a trapezium.**

Solution:
In $\triangle AOB$ and $\triangle COD$:
- $\dfrac{AO}{CO} = \dfrac{BO}{DO}$ (given)
- $\angle AOB = \angle COD$ (vertically opposite)

By SAS similarity: $\triangle AOB \sim \triangle COD$.

Therefore $\angle OAB = \angle OCD$ (corresponding angles of similar triangles).

These are alternate interior angles for lines $AB$ and $CD$ with transversal $AC$.

Since alternate interior angles are equal, $AB \parallel CD$.

Since one pair of opposite sides is parallel, $ABCD$ is a trapezium. $\blacksquare$

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**Example 4: In the given figure, $\triangle ABC$ and $\triangle AMP$ are right-angled at $B$ and $M$ respectively. Prove that $\dfrac{CA}{PA} = \dfrac{BC}{MP}$.**

Solution:
In $\triangle ABC$ and $\triangle AMP$:
- $\angle ABC = \angle AMP = 90^\circ$ (given)
- $\angle A$ is common

By AA similarity: $\triangle ABC \sim \triangle AMP$.

Corresponding sides are proportional:

Mistake 1: Writing the similarity correspondence incorrectly.
When you write $\triangle ABC \sim \triangle DEF$, the order of vertices matters. It means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Writing $\triangle ABC \sim \triangle FED$ would imply a completely different correspondence and give wrong proportions.

Mistake 2: Confusing similarity with congruence.
Similar triangles have equal angles and proportional sides but may differ in size. Congruent triangles are identical in both shape and size. SSS similarity requires ratios to be equal (not sides to be equal).

Mistake 3: Using the wrong ratio for area calculations.
The area ratio is the SQUARE of the side ratio, not the side ratio itself. If sides are in ratio $3:5$, areas are in ratio $9:25$, NOT $3:5$.

Mistake 4: Forgetting to state the similarity criterion.
In CBSE exams, you must explicitly state which criterion (AA, SSS, or SAS) you are using. Simply writing "triangles are similar" without stating the criterion will lose marks.

Mistake 5: Applying BPT when the line is not parallel.
BPT applies ONLY when the line is parallel to the third side. If the problem does not state or prove parallelism, you cannot directly use BPT.

Q1. In $\triangle ABC$, $DE \parallel BC$, $AD = 2$ cm, $DB = 6$ cm, $AE = 1.5$ cm. Find $EC$.

Answer: By BPT: $\dfrac{AD}{DB} = \dfrac{AE}{EC} \implies \dfrac{2}{6} = \dfrac{1.5}{EC} \implies EC = 4.5$ cm.

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Q2. $\triangle ABC \sim \triangle PQR$. If $AB = 6$, $PQ = 9$, and $\text{ar}(\triangle ABC) = 48$ cm$^2$, find $\text{ar}(\triangle PQR)$.

Answer: $\dfrac{48}{\text{ar}(\triangle PQR)} = \left(\dfrac{6}{9}\right)^2 = \dfrac{4}{9} \implies \text{ar}(\triangle PQR) = 108$ cm$^2$.

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Q3. A right triangle has legs $a = 9$ cm and $b = 40$ cm. Find the hypotenuse.

Answer: $c = \sqrt{81 + 1600} = \sqrt{1681} = 41$ cm.

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Q4. Two similar triangles have corresponding sides $5$ cm and $12$ cm. If the perimeter of the smaller triangle is $30$ cm, find the perimeter of the larger.

Answer: Ratio of perimeters $=$ ratio of sides $= \dfrac{5}{12}$. So perimeter of larger $= 30 \times \dfrac{12}{5} = 72$ cm.

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Q5. The sides of a triangle are $7$ cm, $24$ cm, and $25$ cm. Is it a right triangle?

Answer: $25^2 = 625$ and $7^2 + 24^2 = 49 + 576 = 625$. Since $25^2 = 7^2 + 24^2$, yes, it is a right triangle with the right angle opposite the $25$ cm side.

• BPT (Thales' Theorem): A line parallel to one side of a triangle divides the other two sides proportionally.
  - Three similarity criteria: AA (most commonly used), SSS (all side ratios equal), SAS (one angle + including sides proportional).
  - Area ratio of similar triangles equals the square of the corresponding side ratio.
  - Pythagoras Theorem: $c^2 = a^2 + b^2$ for a right triangle. Its converse verifies whether a triangle is right-angled.
  - Always write the vertex correspondence correctly when stating similarity.
  - BPT and Pythagoras proofs are high-value exam questions ($5$ marks each).
  - This is the highest-weightage geometry chapter — thorough preparation is essential.

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