NCERT Solutions for Class 10 Maths Chapter 6: Triangles
1. SEO Strategy Introduction & Chapter Master Overview
Chapter 6, Triangles, is arguably the most conceptually rigorous and high-scoring geometry unit in the CBSE Class 10 Mathematics curriculum. In the annual CBSE Board Examinations, geometry carries a substantial weightage of 15 marks, of which Triangles alone directly accounts for 6 to 8 marks. The questions span from 1-mark objective questions (testing similarity conditions) to 2-mark and 3-mark analytical problems, as well as mandatory 4-mark Case Study questions or 5-mark long-answer geometric proofs.
The fundamental core of this chapter is the transition from Congruence (studied in Class 9) to Similarity. Two geometric figures are defined as congruent if they possess identical shape and identical size. In contrast, two geometric figures are defined as similar ($\sim$) if they possess the same shape, but not necessarily the same size. For two polygons (and specifically triangles) with the same number of sides to be similar, two non-negotiable conditions must be simultaneously fulfilled:
- All corresponding angles must be strictly equal ($\angle A = \angle D, \angle B = \angle E, \angle C = \angle F$).
- All corresponding sides must be in the same ratio (or proportion) ($\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$).
A major focus of the CBSE Class 10 rationalized syllabus centers on the Basic Proportionality Theorem (BPT / Thales Theorem) and the three fundamental Similarity Criteria: Angle-Angle-Angle / Angle-Angle (AAA / AA), Side-Side-Side (SSS), and Side-Angle-Side (SAS).
Under official CBSE evaluation guidelines, geometric proofs demand a rigid structural presentation:
- Given: Formal mathematical specification of the known geometric configuration and labeled vertices.
- To Prove: Precise mathematical statement of the target equality, ratio, or similarity relation.
- Construction (if needed): Auxiliary lines, perpendiculars, or extensions explicitly defined with dashed lines.
- Proof: Step-by-step logical deduction where every single equation or statement is paired with an explicit geometric axiom, theorem, or property in parentheses.
Plaintext
+----------------------------------------------------------------------------------------------------+
| MASTER SUMMARY: TRIANGLES |
+-----------------------------------+-----------------------------------+----------------------------+
| Concept / Theorem | Mathematical Statement / Rule | Key Examination Notes |
+-----------------------------------+-----------------------------------+----------------------------+
| Similar Polygons Condition | (i) Corresponding angles equal | Both conditions must hold |
| | (ii) Corresponding sides in ratio | simultaneously |
+-----------------------------------+-----------------------------------+----------------------------+
| Basic Proportionality Theorem | If DE || BC in Triangle ABC, | Most tested theorem proof; |
| (BPT / Thales Theorem) | then AD / DB = AE / EC | Standard 3/5 mark question |
+-----------------------------------+-----------------------------------+----------------------------+
| Converse of BPT | If AD / DB = AE / EC in | Used to prove lines |
| | Triangle ABC, then DE || BC | are parallel |
+-----------------------------------+-----------------------------------+----------------------------+
| AA (Angle-Angle) Similarity | If 2 angles of one triangle equal | Most frequently applied |
| Criterion | 2 angles of another, then Tri 1 ~ | similarity shortcut |
| | Tri 2 | |
+-----------------------------------+-----------------------------------+----------------------------+
| SSS (Side-Side-Side) Similarity | If AB/DE = BC/EF = AC/DF, | Ratio of all 3 sides must |
| Criterion | then Tri ABC ~ Tri DEF | be identical constant |
+-----------------------------------+-----------------------------------+----------------------------+
| SAS (Side-Angle-Side) Similarity | If AB/DE = AC/DF and | Angle must be strictly the |
| Criterion | angle A = angle D, then Tri ~ Tri | INCLUDED angle |
+-----------------------------------+-----------------------------------+----------------------------+
| Symbolic Order of Vertices | Tri ABC ~ Tri DEF implies | Vertex correspondence must |
| | A <-> D, B <-> E, C <-> F | strictly match |
+-----------------------------------+-----------------------------------+----------------------------+
2. Exercise 6.1 Solutions (Page No. 122-123)
Question 1. 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) (Page No. 122)
Answer:
(i) All circles are similar.
Reason: Every circle has the exact same circular shape regardless of its radius $r$. Circles are only congruent if their radii are identical, but all circles are universally similar.
(ii) All squares are similar.
Reason: In every square, all four interior angles are strictly $90^\circ$, and all four sides are equal, ensuring that the ratio of corresponding sides between any two squares is always constant.
(iii) All equilateral triangles are similar.
Reason: Every equilateral triangle has interior angles fixed at $60^\circ, 60^\circ, 60^\circ$, satisfying the AAA similarity criterion unconditionally. Isosceles triangles do not necessarily have equal corresponding angles.
(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.
Reason: This represents the fundamental formal definition of similarity for rectilinear geometric figures.
Question 2. Give two different examples of pair of:
(i) similar figures
(ii) non-similar figures (Page No. 122)
Answer:
(i) Pair of Similar Figures:
- Any two equilateral triangles: For instance, $\triangle ABC$ of side $3\text{ cm}$ and $\triangle PQR$ of side $7\text{ cm}$. All corresponding interior angles are $60^\circ$, and the ratio of corresponding sides is uniformly $\frac{3}{7}$.
- Any two regular hexagons: Two regular hexagons of side lengths $2\text{ cm}$ and $5\text{ cm}$ respectively. All interior angles are identically $120^\circ$, and their sides maintain a uniform ratio of $\frac{2}{5}$.
(ii) Pair of Non-Similar Figures:
- A square and a rectangle: Although corresponding angles are all equal ($90^\circ$), the ratio of corresponding adjacent sides is not equal.
- An equilateral triangle and a scalene right-angled triangle: Their corresponding interior angles are unequal ($60^\circ, 60^\circ, 60^\circ$ versus $90^\circ, 60^\circ, 30^\circ$).
Question 3. State whether the following quadrilaterals are similar or not:
A rhombus $PQRS$ with sides of length $1.5\text{ cm}$ each and non-right angles, and a square $ABCD$ with sides of length $3.0\text{ cm}$ each and interior angles equal to $90^\circ$. (Page No. 123)
Answer:
For two quadrilaterals to be similar, two conditions must be satisfied simultaneously:
- Their corresponding sides must be proportional.
- Their corresponding angles must be equal.
Step 1: Check side proportionality:
$$\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RS}{CD} = \frac{SP}{DA} = \frac{1.5}{3.0} = \frac{1}{2}$$
The ratio of corresponding sides is constant and equal to $\frac{1}{2}$.
Step 2: Check angle equality:
In rhombus $PQRS$, the interior angles are not right angles ($\angle P \neq 90^\circ, \angle Q \neq 90^\circ$). In square $ABCD$, all interior angles are strictly right angles ($\angle A = \angle B = \angle C = \angle D = 90^\circ$).
$$\angle P \neq \angle A, \quad \angle Q \neq \angle B$$
Conclusion:
Although the corresponding sides are proportional, the corresponding angles are not equal. Therefore, quadrilateral $PQRS$ and quadrilateral $ABCD$ are not similar.
3. Theorem Master Proof: Basic Proportionality Theorem (Thales Theorem)
Theorem 6.1 (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. [CBSE STANDARD FREQUENT 5-MARKER]
Plaintext
A
/ \
/ \
/ \
D-------E (DE || BC)
/ \ / \
/ \ / \
/ \ / \
B-------C-------
Given:
A triangle $\triangle ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively (i.e., $DE \parallel BC$).
To Prove:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Construction:
Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
Proof:
Recall that the area of a triangle is given by:
$$\text{Area}(\triangle) = \frac{1}{2} \times \text{base} \times \text{height}$$
Consider $\triangle ADE$ with base $AD$ and altitude $EN$:
$$\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN \quad \text{— (1)}$$
Consider $\triangle BDE$ with base $DB$ and altitude $EN$ (since $EN$ is perpendicular to the entire line $AB$):
$$\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN \quad \text{— (2)}$$
Dividing equation (1) by equation (2):
$$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{— (3)}$$
Similarly, taking base $AE$ for $\triangle ADE$ with altitude $DM$:
$$\text{ar}(\triangle ADE) = \frac{1}{2} \times AE \times DM \quad \text{— (4)}$$
For $\triangle CDE$ with base $EC$ and altitude $DM$:
$$\text{ar}(\triangle CDE) = \frac{1}{2} \times EC \times DM \quad \text{— (5)}$$
Dividing equation (4) by equation (5):
$$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{— (6)}$$
Now, observe that $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallel lines $DE$ and $BC$.
By standard geometric properties:
$$\text{ar}(\triangle BDE) = \text{ar}(\triangle CDE) \quad \text{— (7)}$$
Substituting equation (7) into equation (3) and comparing with equation (6):
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Hence Proved.
4. Exercise 6.2 Solutions (Page No. 128-130)
Question 1. In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $EC$ in (i) and $AD$ in (ii).
(i) In $\triangle ABC$, $AD = 1.5\text{ cm}, DB = 3\text{ cm}, AE = 1\text{ cm}$.
(ii) In $\triangle ABC$, $DB = 7.2\text{ cm}, AE = 1.8\text{ cm}, EC = 5.4\text{ cm}$. (Page No. 128)
Answer:
(i) In $\triangle ABC$, since $DE \parallel BC$, by the Basic Proportionality Theorem (BPT):
$$\begin{aligned} \frac{AD}{DB} &= \frac{AE}{EC} \\ \frac{1.5}{3} &= \frac{1}{EC} \\ \frac{1}{2} &= \frac{1}{EC} \\ EC &= 2\text{ cm} \end{aligned}$$
Thus, $EC = 2\text{ cm}$.
(ii) In $\triangle ABC$, since $DE \parallel BC$, by the Basic Proportionality Theorem (BPT):
$$\begin{aligned} \frac{AD}{DB} &= \frac{AE}{EC} \\ \frac{AD}{7.2} &= \frac{1.8}{5.4} \\ \frac{AD}{7.2} &= \frac{1}{3} \\ AD &= \frac{7.2}{3} = 2.4\text{ cm} \end{aligned}$$
Thus, $AD = 2.4\text{ cm}$.
Question 2. $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$:
(i) $PE = 3.9\text{ cm}, EQ = 3\text{ cm}, PF = 3.6\text{ cm}$ and $FR = 2.4\text{ cm}$
(ii) $PE = 4\text{ cm}, QE = 4.5\text{ cm}, PF = 8\text{ cm}$ and $RF = 9\text{ cm}$
(iii) $PQ = 1.28\text{ cm}, PR = 2.56\text{ cm}, PE = 0.18\text{ cm}$ and $PF = 0.36\text{ cm}$ (Page No. 128)
Answer:
By the Converse of Basic Proportionality Theorem, if $\frac{PE}{EQ} = \frac{PF}{FR}$, then $EF \parallel QR$.
(i) Computing individual ratios:
$$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$$
$$\frac{PF}{FR} = \frac{3.6}{2.4} = \frac{3}{2} = 1.5$$
Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, $EF$ is not parallel to $QR$.
(ii) Computing individual ratios:
$$\frac{PE}{QE} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$$
$$\frac{PF}{RF} = \frac{8}{9}$$
Since $\frac{PE}{QE} = \frac{PF}{RF} = \frac{8}{9}$, by the Converse of BPT, $EF \parallel QR$.
(iii) Calculating segment lengths:
$$EQ = PQ – PE = 1.28 – 0.18 = 1.10\text{ cm}$$
$$FR = PR – PF = 2.56 – 0.36 = 2.20\text{ cm}$$
Computing ratios:
$$\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$$
$$\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$$
Since $\frac{PE}{EQ} = \frac{PF}{FR}$, by the Converse of BPT, $EF \parallel QR$.
Question 3. In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$. (Page No. 128)
Answer:
Given: In quadrilateral $ABCD$, $LM \parallel CB$ and $LN \parallel CD$.
To Prove: $\frac{AM}{AB} = \frac{AN}{AD}$.
Proof:
- In $\triangle ABC$, since $LM \parallel CB$, applying BPT gives:$$\frac{AM}{MB} = \frac{AL}{LC}$$Inverting and adding $1$ to both sides (or using the corollary form $\frac{\text{part}}{\text{whole}}$):$$\frac{AM}{AM + MB} = \frac{AL}{AL + LC} \implies \frac{AM}{AB} = \frac{AL}{AC} \quad \text{— (1)}$$
- In $\triangle ADC$, since $LN \parallel CD$, applying BPT gives:$$\frac{AN}{ND} = \frac{AL}{LC} \implies \frac{AN}{AD} = \frac{AL}{AC} \quad \text{— (2)}$$
Comparing equations (1) and (2):
$$\frac{AM}{AB} = \frac{AN}{AD}$$
Hence Proved.
Question 4. In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$. (Page No. 128)
Answer:
Given: In $\triangle ABC$, $DE \parallel AC$ and $DF \parallel AE$.
To Prove: $\frac{BF}{FE} = \frac{BE}{EC}$.
Proof:
- In $\triangle BAE$, since $DF \parallel AE$, applying BPT gives:$$\frac{BF}{FE} = \frac{BD}{DA} \quad \text{— (1)}$$
- In $\triangle BAC$, since $DE \parallel AC$, applying BPT gives:$$\frac{BE}{EC} = \frac{BD}{DA} \quad \text{— (2)}$$
From equations (1) and (2), both ratios equal $\frac{BD}{DA}$:
$$\frac{BF}{FE} = \frac{BE}{EC}$$
Hence Proved.
Question 5. In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$. (Page No. 129)
Answer:
Given: In $\triangle PQR$ with interior point $O$, $DE \parallel OQ$ and $DF \parallel OR$.
To Prove: $EF \parallel QR$.
Proof:
- In $\triangle PQO$, since $DE \parallel OQ$, by Basic Proportionality Theorem:$$\frac{PE}{EQ} = \frac{PD}{DO} \quad \text{— (1)}$$
- In $\triangle POR$, since $DF \parallel OR$, by Basic Proportionality Theorem:$$\frac{PF}{FR} = \frac{PD}{DO} \quad \text{— (2)}$$
- Equating (1) and (2):$$\frac{PE}{EQ} = \frac{PF}{FR}$$
- In $\triangle PQR$, since the line segment $EF$ divides sides $PQ$ and $PR$ in the same ratio ($\frac{PE}{EQ} = \frac{PF}{FR}$), by the Converse of BPT:$$EF \parallel QR$$Hence Proved.
Question 6. In Fig. 6.21, $A, B$ and $C$ are points on $OP, OQ$ and $OR$ respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that $BC \parallel QR$. (Page No. 129)
Answer:
Given: Points $A, B, C$ on rays $OP, OQ, OR$ respectively with $AB \parallel PQ$ and $AC \parallel PR$.
To Prove: $BC \parallel QR$.
Proof:
- In $\triangle OPQ$, since $AB \parallel PQ$, by BPT:$$\frac{OA}{AP} = \frac{OB}{BQ} \quad \text{— (1)}$$
- In $\triangle OPR$, since $AC \parallel PR$, by BPT:$$\frac{OA}{AP} = \frac{OC}{CR} \quad \text{— (2)}$$
- From equations (1) and (2):$$\frac{OB}{BQ} = \frac{OC}{CR}$$
- In $\triangle OQR$, since the line segment $BC$ divides sides $OQ$ and $OR$ proportionally, by the Converse of Basic Proportionality Theorem:$$BC \parallel QR$$Hence Proved.
Question 7. Using Theorem 6.1 (BPT), prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Page No. 129)
Answer:
Given: In $\triangle ABC$, $D$ is the mid-point of side $AB$ (i.e., $AD = DB$), and line $l$ passing through $D$ is parallel to $BC$, intersecting $AC$ at $E$ ($DE \parallel BC$).
To Prove: $E$ is the mid-point of $AC$ (i.e., $AE = EC$).
Proof:
Since $D$ is the mid-point of $AB$:
$$AD = DB \implies \frac{AD}{DB} = 1 \quad \text{— (1)}$$
In $\triangle ABC$, since $DE \parallel BC$, by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \text{— (2)}$$
Substituting equation (1) into equation (2):
$$1 = \frac{AE}{EC} \implies AE = EC$$
Thus, the line bisects $AC$, proving that $E$ is the mid-point of side $AC$.
Hence Proved.
Question 8. Using Theorem 6.2 (Converse of BPT), prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Page No. 129)
Answer:
Given: In $\triangle ABC$, $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively.
To Prove: $DE \parallel BC$.
Proof:
Since $D$ is the mid-point of $AB$:
$$AD = DB \implies \frac{AD}{DB} = 1 \quad \text{— (1)}$$
Since $E$ is the mid-point of $AC$:
$$AE = EC \implies \frac{AE}{EC} = 1 \quad \text{— (2)}$$
Equating (1) and (2):
$$\frac{AD}{DB} = \frac{AE}{EC}$$
In $\triangle ABC$, the line segment $DE$ divides sides $AB$ and $AC$ in the same ratio. Therefore, by the Converse of the Basic Proportionality Theorem (Theorem 6.2):
$$DE \parallel BC$$
Hence Proved.
Question 9. $ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$. (Page No. 129)
Answer:
Plaintext
A--------------B
\ /
\ O /
\ / \ /
\/ \ /
\ \
D------------------C
Given: Trapezium $ABCD$ with $AB \parallel DC$ and diagonals intersecting at $O$.
To Prove: $\frac{AO}{BO} = \frac{CO}{DO}$ (or equivalently, $\frac{AO}{CO} = \frac{BO}{DO}$).
Construction: Through $O$, draw a line $OE \parallel AB$ intersecting non-parallel side $AD$ at point $E$.
Since $AB \parallel DC$ and $OE \parallel AB$, we also have $OE \parallel DC$.
Proof:
- In $\triangle ADC$, since $OE \parallel DC$, by Basic Proportionality Theorem:$$\frac{AE}{ED} = \frac{AO}{OC} \quad \text{— (1)}$$
- In $\triangle DAB$, since $EO \parallel AB$, by Basic Proportionality Theorem:$$\frac{DE}{EA} = \frac{DO}{OB}$$Inverting both sides:$$\frac{AE}{ED} = \frac{BO}{DO} \quad \text{— (2)}$$
- Comparing equations (1) and (2):$$\frac{AO}{OC} = \frac{BO}{DO}$$Rearranging the terms:$$\frac{AO}{BO} = \frac{CO}{DO}$$Hence Proved.
Question 10. The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium. (Page No. 130)
Answer:
Given: Quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$, which can be rewritten as:
$$\frac{AO}{CO} = \frac{BO}{DO} \quad \text{— (1)}$$
To Prove: $ABCD$ is a trapezium (i.e., $AB \parallel DC$).
Construction: Draw a line segment $OE$ through $O$ meeting $AD$ at $E$ such that $OE \parallel AB$.
Proof:
- In $\triangle DAB$, since $OE \parallel AB$, by Basic Proportionality Theorem:$$\frac{DE}{EA} = \frac{DO}{OB}$$Inverting both sides:$$\frac{AE}{ED} = \frac{BO}{DO} \quad \text{— (2)}$$
- Comparing given equation (1) with equation (2):$$\frac{AE}{ED} = \frac{AO}{CO}$$
- In $\triangle ADC$, the line segment $OE$ divides sides $AD$ and $AC$ in the same ratio ($\frac{AE}{ED} = \frac{AO}{CO}$). By the Converse of BPT:$$OE \parallel DC$$
- We constructed $OE \parallel AB$ and proved $OE \parallel DC$.Since lines parallel to the same line are parallel to each other:$$AB \parallel DC$$
Since quadrilateral $ABCD$ has one pair of opposite sides parallel ($AB \parallel DC$), $ABCD$ is a trapezium.
Hence Proved.
5. Exercise 6.3 Solutions (Page No. 138-142)
Question 1. State which pairs of triangles in Fig. 6.34 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:
(i) In $\triangle ABC$: $\angle A = 60^\circ, \angle B = 80^\circ, \angle C = 40^\circ$. In $\triangle PQR$: $\angle P = 60^\circ, \angle Q = 80^\circ, \angle R = 40^\circ$.
(ii) In $\triangle ABC$: $AB = 2, BC = 2.5, CA = 3$. In $\triangle PQR$: $PQ = 6, QR = 4, PR = 5$.
(iii) In $\triangle LMP$: $LM = 2.7, MP = 2, LP = 3$. In $\triangle DEF$: $DE = 4, EF = 5, DF = 6$.
(iv) In $\triangle MNL$: $MN = 2.5, ML = 5, \angle M = 70^\circ$. In $\triangle QPR$: $PQ = 5, QR = 10, \angle Q = 70^\circ$.
(v) In $\triangle ABC$: $AB = 2.5, BC = 3, \angle A = 80^\circ$. In $\triangle DEF$: $DF = 5, EF = 6, \angle F = 80^\circ$.
(vi) In $\triangle DEF$: $\angle D = 70^\circ, \angle E = 80^\circ$. In $\triangle PQR$: $\angle Q = 80^\circ, \angle R = 30^\circ$. (Page No. 138)
Answer:
- (i) In $\triangle ABC$ and $\triangle PQR$:$$\angle A = \angle P = 60^\circ, \quad \angle B = \angle Q = 80^\circ, \quad \angle C = \angle R = 40^\circ$$By AAA Similarity Criterion: $\triangle ABC \sim \triangle PQR$.
- (ii) In $\triangle ABC$ and $\triangle QRP$:$$\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}, \quad \frac{BC}{PR} = \frac{2.5}{5} = \frac{1}{2}, \quad \frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$$Since all corresponding sides are in the same ratio, by SSS Similarity Criterion: $\triangle ABC \sim \triangle QRP$.
- (iii) In $\triangle LMP$ and $\triangle DEF$:$$\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}, \quad \frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}, \quad \text{but } \frac{LM}{EF} = \frac{2.7}{5} \neq \frac{1}{2}$$Since corresponding sides are not proportional, $\triangle LMP$ and $\triangle DEF$ are not similar.
- (iv) In $\triangle MNL$ and $\triangle QPR$:$$\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2}, \quad \frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}, \quad \text{and } \angle M = \angle Q = 70^\circ$$The included angles between the proportional sides are equal. By SAS Similarity Criterion: $\triangle MNL \sim \triangle QPR$.
- (v) In $\triangle ABC$, $\angle A = 80^\circ$ is not the included angle between sides $AB = 2.5$ and $BC = 3$ (the included angle is $\angle B$). In $\triangle DEF$, $\angle F = 80^\circ$ is the included angle between $DF = 5$ and $EF = 6$.Because the equal angle in $\triangle ABC$ is not the included angle, $\triangle ABC$ and $\triangle DEF$ are not similar.
- (vi) In $\triangle DEF$: $\angle F = 180^\circ – (70^\circ + 80^\circ) = 30^\circ$.In $\triangle PQR$: $\angle P = 180^\circ – (80^\circ + 30^\circ) = 70^\circ$.Comparing $\triangle DEF$ and $\triangle PQR$: $\angle D = \angle P = 70^\circ$ and $\angle E = \angle Q = 80^\circ$.By AA Similarity Criterion: $\triangle DEF \sim \triangle PQR$.
Question 2. In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^\circ$ and $\angle CBO = 70^\circ$. Find $\angle DOC, \angle DCO$ and $\angle OAB$. (Page No. 139)
Answer:
- Line $DOB$ is a straight line. Therefore, angles $\angle DOC$ and $\angle BOC$ form a linear pair:$$\angle DOC + \angle BOC = 180^\circ \implies \angle DOC + 125^\circ = 180^\circ \implies \angle DOC = 55^\circ$$
- In $\triangle DOC$, by the angle sum property:$$\angle DCO + \angle DOC + \angle ODC = 180^\circ$$Since $\triangle ODC \sim \triangle OBA$, corresponding angles are equal, so $\angle ODC = \angle OBA = 70^\circ$.$$\angle DCO + 55^\circ + 70^\circ = 180^\circ \implies \angle DCO + 125^\circ = 180^\circ \implies \angle DCO = 55^\circ$$
- Since $\triangle ODC \sim \triangle OBA$, corresponding angles match:$$\angle OAB = \angle OCD = \angle DCO = 55^\circ$$
Values: $\angle DOC = 55^\circ, \angle DCO = 55^\circ, \angle OAB = 55^\circ$.
Question 3. Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$. (Page No. 139)
Answer:
Given: Trapezium $ABCD$ with $AB \parallel DC$, diagonals intersecting at $O$.
To Prove: $\frac{OA}{OC} = \frac{OB}{OD}$.
Proof:
In $\triangle OAB$ and $\triangle OCD$:
- $\angle OAB = \angle OCD$ (Alternate interior angles, since $AB \parallel DC$ and $AC$ is a transversal)
- $\angle OBA = \angle ODC$ (Alternate interior angles, since $AB \parallel DC$ and $BD$ is a transversal)
- $\angle AOB = \angle COD$ (Vertically opposite angles)
By the AAA Similarity Criterion (or AA Criterion):
$$\triangle OAB \sim \triangle OCD$$
Since corresponding sides of similar triangles are proportional:
$$\frac{OA}{OC} = \frac{OB}{OD}$$
Hence Proved.
Question 4. In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$. (Page No. 139)
Answer:
Given: $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$ (where $\angle 1 = \angle PQR$ and $\angle 2 = \angle PRQ$).
To Prove: $\triangle PQS \sim \triangle TQR$.
Proof:
- In $\triangle PQR$, since $\angle 1 = \angle 2$:$$PR = PQ \quad (\text{Sides opposite to equal angles are equal}) \quad \text{— (1)}$$
- The given proportion is:$$\frac{QR}{QS} = \frac{QT}{PR}$$Substituting $PR = PQ$ from equation (1):$$\frac{QR}{QS} = \frac{QT}{PQ} \implies \frac{QT}{PQ} = \frac{QR}{QS} \implies \frac{PQ}{QT} = \frac{QS}{QR} \quad \text{— (2)}$$
- Now consider $\triangle PQS$ and $\triangle TQR$:
- From (2), $\frac{PQ}{QT} = \frac{QS}{QR}$ (Sides containing $\angle Q$ are proportional)
- $\angle Q = \angle Q$ (Common angle, $\angle 1$)
By the SAS Similarity Criterion:
$$\triangle PQS \sim \triangle TQR$$
Hence Proved.
Question 5. $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$. (Page No. 139)
Answer:
Given: $\triangle PQR$ with points $S$ on $PR$ and $T$ on $QR$ such that $\angle P = \angle RTS$.
To Prove: $\triangle RPQ \sim \triangle RTS$.
Proof:
In $\triangle RPQ$ and $\triangle RTS$:
- $\angle RPQ = \angle RTS$ (Given)
- $\angle PRQ = \angle TRS$ (Common angle, $\angle R$)
Since two pairs of corresponding angles are equal, by the AA Similarity Criterion:
$$\triangle RPQ \sim \triangle RTS$$
Hence Proved.
Question 6. In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$. (Page No. 139)
Answer:
Given: $\triangle ABE \cong \triangle ACD$.
To Prove: $\triangle ADE \sim \triangle ABC$.
Proof:
- Since $\triangle ABE \cong \triangle ACD$, corresponding parts of congruent triangles (CPCT) are equal:$$AB = AC \quad \text{— (1)}$$$$AE = AD \implies AD = AE \quad \text{— (2)}$$
- Dividing equation (2) by equation (1):$$\frac{AD}{AB} = \frac{AE}{AC} \quad \text{— (3)}$$
- In $\triangle ADE$ and $\triangle ABC$:
- $\frac{AD}{AB} = \frac{AE}{AC}$ (From equation 3)
- $\angle DAE = \angle BAC$ (Common angle, $\angle A$)
By the SAS Similarity Criterion:
$$\triangle ADE \sim \triangle ABC$$
Hence Proved.
Question 7. In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that:
(i) $\triangle AEP \sim \triangle CDP$
(ii) $\triangle ABD \sim \triangle CBE$
(iii) $\triangle AEP \sim \triangle ADB$
(iv) $\triangle PDC \sim \triangle BEC$ (Page No. 139)
Answer:
Given that $AD \perp BC$ and $CE \perp AB \implies \angle ADC = \angle ADB = \angle AEP = \angle CEB = 90^\circ$.
(i) In $\triangle AEP$ and $\triangle CDP$:
- $\angle AEP = \angle CDP = 90^\circ$
- $\angle APE = \angle CPD$ (Vertically opposite angles)By AA Criterion: $\triangle AEP \sim \triangle CDP$.
(ii) In $\triangle ABD$ and $\triangle CBE$:
- $\angle ADB = \angle CEB = 90^\circ$
- $\angle ABD = \angle CBE$ (Common angle, $\angle B$)By AA Criterion: $\triangle ABD \sim \triangle CBE$.
(iii) In $\triangle AEP$ and $\triangle ADB$:
- $\angle AEP = \angle ADB = 90^\circ$
- $\angle PAE = \angle BAD$ (Common angle, $\angle A$)By AA Criterion: $\triangle AEP \sim \triangle ADB$.
(iv) In $\triangle PDC$ and $\triangle BEC$:
- $\angle PDC = \angle BEC = 90^\circ$
- $\angle PCD = \angle BCE$ (Common angle, $\angle C$)By AA Criterion: $\triangle PDC \sim \triangle BEC$.Hence Proved.
Question 8. $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$. (Page No. 139)
Answer:
Given: Parallelogram $ABCD$ with side $AD$ extended to $E$. Line $BE$ intersects $CD$ at $F$.
To Prove: $\triangle ABE \sim \triangle CFB$.
Proof:
In $\triangle ABE$ and $\triangle CFB$:
- $\angle A = \angle C$ (Opposite angles of a parallelogram are equal)
- Since $AE \parallel BC$ ($AD \parallel BC$ extended) and transversal $BE$ intersects them:$$\angle AEB = \angle CBF \quad (\text{Alternate interior angles})$$
Since two angles of $\triangle ABE$ are respectively equal to two angles of $\triangle CFB$, by the AA Similarity Criterion:
$$\triangle ABE \sim \triangle CFB$$
Hence Proved.
Question 9. In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:
(i) $\triangle ABC \sim \triangle AMP$
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$ (Page No. 139)
Answer:
(i) In $\triangle ABC$ and $\triangle AMP$:
- $\angle ABC = \angle AMP = 90^\circ$ (Given)
- $\angle BAC = \angle MAP$ (Common angle, $\angle A$)
By the AA Similarity Criterion:
$$\triangle ABC \sim \triangle AMP$$
(ii) Since $\triangle ABC \sim \triangle AMP$, their corresponding sides are proportional:
$$\frac{CA}{PA} = \frac{BC}{MP} = \frac{AB}{AM}$$
Taking the first two ratios:
$$\frac{CA}{PA} = \frac{BC}{MP}$$
Hence Proved.
Question 10. $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that:
(i) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) $\triangle DCB \sim \triangle HGE$
(iii) $\triangle DCA \sim \triangle HGF$ (Page No. 140)
Answer:
Given: $\triangle ABC \sim \triangle FEG$. Therefore:
$$\angle A = \angle F, \quad \angle B = \angle E, \quad \angle ACB = \angle FGE$$
$CD$ bisects $\angle ACB \implies \angle ACD = \angle DCB = \frac{1}{2}\angle ACB$.
$GH$ bisects $\angle FGE \implies \angle FGH = \angle HGE = \frac{1}{2}\angle FGE$.
Thus, $\angle ACD = \angle FGH$ and $\angle DCB = \angle HGE$.
(iii) In $\triangle DCA$ and $\triangle HGF$:
- $\angle A = \angle F$ (From $\triangle ABC \sim \triangle FEG$)
- $\angle ACD = \angle FGH$ (Halves of equal angles)By AA Similarity Criterion: $\triangle DCA \sim \triangle HGF$.
(i) Since $\triangle DCA \sim \triangle HGF$, corresponding sides are proportional:
$$\frac{CD}{GH} = \frac{AC}{FG}$$
Hence Proved.
(ii) In $\triangle DCB$ and $\triangle HGE$:
- $\angle B = \angle E$ (From $\triangle ABC \sim \triangle FEG$)
- $\angle DCB = \angle HGE$ (Halves of equal angles)By AA Similarity Criterion: $\triangle DCB \sim \triangle HGE$.
Question 11. In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$. (Page No. 140)
Answer:
Given: Isosceles $\triangle ABC$ with $AB = AC$, $AD \perp BC$, and $EF \perp AC$.
To Prove: $\triangle ABD \sim \triangle ECF$.
Proof:
- In $\triangle ABC$, since $AB = AC$:$$\angle ABD = \angle ACD \quad (\text{Angles opposite to equal sides are equal})$$Since $C, B, E$ are collinear, $\angle ACD = \angle ECF$. Thus:$$\angle ABD = \angle ECF \quad \text{— (1)}$$
- In $\triangle ABD$ and $\triangle ECF$:
- $\angle ADB = \angle EFC = 90^\circ$ (Given $AD \perp BC$ and $EF \perp AC$)
- $\angle ABD = \angle ECF$ (From equation 1)
By the AA Similarity Criterion:
$$\triangle ABD \sim \triangle ECF$$
Hence Proved.
Question 12. Sides $AB$ 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 $\triangle ABC \sim \triangle PQR$. (Page No. 140)
Answer:
Given: $\triangle ABC$ and $\triangle PQR$ with medians $AD$ and $PM$ such that:
$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \quad \text{— (1)}$$
To Prove: $\triangle ABC \sim \triangle PQR$.
Proof:
Since $AD$ is the median to $BC$, $D$ is the mid-point of $BC \implies BC = 2BD$.
Since $PM$ is the median to $QR$, $M$ is the mid-point of $QR \implies QR = 2QM$.
Substituting into equation (1):
$$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \implies \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$
In $\triangle ABD$ and $\triangle PQM$, all three corresponding sides are proportional.
By the SSS Similarity Criterion:
$$\triangle ABD \sim \triangle PQM$$
Since corresponding angles of similar triangles are equal:
$$\angle B = \angle Q \quad \text{— (2)}$$
Now, consider $\triangle ABC$ and $\triangle PQR$:
- $\frac{AB}{PQ} = \frac{BC}{QR}$ (Given)
- $\angle B = \angle Q$ (From equation 2)
By the SAS Similarity Criterion:
$$\triangle ABC \sim \triangle PQR$$
Hence Proved.
Question 13. $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \times CD$. (Page No. 140)
Answer:
Given: In $\triangle ABC$, point $D$ lies on $BC$ such that $\angle ADC = \angle BAC$.
To Prove: $CA^2 = CB \times CD$, which is equivalent to $\frac{CA}{CD} = \frac{CB}{CA}$.
Proof:
In $\triangle BAC$ and $\triangle ADC$:
- $\angle BAC = \angle ADC$ (Given)
- $\angle BCA = \angle ACD$ (Common angle, $\angle C$)
Since two pairs of angles are equal, by the AA Similarity Criterion:
$$\triangle BAC \sim \triangle ADC$$
Since corresponding sides of similar triangles are in the same ratio:
$$\frac{CA}{CD} = \frac{CB}{CA} = \frac{BA}{AD}$$
Taking the first two ratios:
$$\frac{CA}{CD} = \frac{CB}{CA} \implies CA \times CA = CB \times CD \implies CA^2 = CB \times CD$$
Hence Proved.
Question 14. Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$. [BOARD EXAM CLASSIC 5-MARKER] (Page No. 140)
Answer:
Given: $\triangle ABC$ and $\triangle PQR$ with medians $AD$ and $PM$ such that:
$$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \quad \text{— (1)}$$
To Prove: $\triangle ABC \sim \triangle PQR$.
Construction:
Extend $AD$ to point $E$ such that $DE = AD$. Join $CE$.
Extend $PM$ to point $N$ such that $MN = PM$. Join $RN$.
Proof:
- In quadrilateral $ABEC$, diagonals $AE$ and $BC$ bisect each other at $D$ ($AD = DE$ by construction, $BD = CD$ as $AD$ is median).Therefore, $ABEC$ is a parallelogram $\implies AB = CE$ and $AC = BE$.Similarly, $PQNR$ is a parallelogram $\implies PQ = RN$ and $PR = QN$.
- In $\triangle ABE$ and $\triangle PQN$:
- $AB = CE$ and $PQ = RN \implies \frac{AB}{PQ} = \frac{CE}{RN}$
- $AE = 2AD$ and $PN = 2PM \implies \frac{AE}{PN} = \frac{2AD}{2PM} = \frac{AD}{PM}$
- By extending the identical construction to the left side or symmetry:$$\triangle ABE \sim \triangle PQN \implies \angle 1 = \angle 3 \quad \text{— (3)}$$(where $\angle 1 = \angle BAD$ and $\angle 3 = \angle QPM$)
- Adding equations (2) and (3):$$\angle 1 + \angle 2 = \angle 3 + \angle 4 \implies \angle A = \angle P \quad \text{— (4)}$$
- Now, in $\triangle ABC$ and $\triangle PQR$:
- $\frac{AB}{PQ} = \frac{AC}{PR}$ (Given)
- $\angle A = \angle P$ (From equation 4)
By the SAS Similarity Criterion:
$$\triangle ABC \sim \triangle PQR$$
Hence Proved.
Question 15. A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower. (Page No. 141)
Answer:
Let $AB$ be the vertical pole and $BC$ be its shadow.
Let $PQ$ be the vertical tower and $QR$ be its shadow.
Given:
$$\begin{aligned} AB &= \text{Height of pole} = 6\text{ m} \\ BC &= \text{Length of pole’s shadow} = 4\text{ m} \\ QR &= \text{Length of tower’s shadow} = 28\text{ m} \\ PQ &= h = \text{Height of tower} \end{aligned}$$
At the same time, the angle of elevation of the sun is identical for both objects:
$$\angle C = \angle R$$
Also, both the pole and the tower stand vertically upright on the horizontal ground:
$$\angle B = \angle Q = 90^\circ$$
In $\triangle ABC$ and $\triangle PQR$:
- $\angle B = \angle Q = 90^\circ$
- $\angle C = \angle R$ (Angular elevation of the sun)
By the AA Similarity Criterion:
$$\triangle ABC \sim \triangle PQR$$
Corresponding sides are in the same ratio:
$$\begin{aligned} \frac{AB}{PQ} &= \frac{BC}{QR} \\ \frac{6}{h} &= \frac{4}{28} \\ \frac{6}{h} &= \frac{1}{7} \\ h &= 6 \times 7 = 42\text{ m} \end{aligned}$$
The height of the tower is $42\text{ m}$.
Question 16. If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively, where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$. (Page No. 141)
Answer:
Given: $\triangle ABC \sim \triangle PQR$, where $AD$ is the median to side $BC$ and $PM$ is the median to side $QR$.
To Prove: $\frac{AB}{PQ} = \frac{AD}{PM}$.
Proof:
- Since $\triangle ABC \sim \triangle PQR$:$$\frac{AB}{PQ} = \frac{BC}{QR} \quad \text{— (1)}$$$$\angle B = \angle Q \quad \text{— (2)}$$
- Since $AD$ and $PM$ are medians:$$BC = 2BD \quad \text{and} \quad QR = 2QM$$
- Substituting these into equation (1):$$\frac{AB}{PQ} = \frac{2BD}{2QM} \implies \frac{AB}{PQ} = \frac{BD}{QM} \quad \text{— (3)}$$
- In $\triangle ABD$ and $\triangle PQM$:
- $\frac{AB}{PQ} = \frac{BD}{QM}$ (From equation 3)
- $\angle B = \angle Q$ (From equation 2)
By the SAS Similarity Criterion:
$$\triangle ABD \sim \triangle PQM$$
Since corresponding sides of similar triangles are proportional:
$$\frac{AB}{PQ} = \frac{AD}{PM}$$
Hence Proved.
6. 15 High-Yield Frequently Asked Questions (Board Exam Level FAQs)
Question 1. In $\triangle ABC$, $DE \parallel BC$ such that $AD = x, DB = x – 2, AE = x + 2$, and $EC = x – 1$. Find the value of $x$. [CBSE 2016, 2021]
Answer:
In $\triangle ABC$, since $DE \parallel BC$, by the Basic Proportionality Theorem (BPT):
$$\begin{aligned} \frac{AD}{DB} &= \frac{AE}{EC} \\ \frac{x}{x – 2} &= \frac{x + 2}{x – 1} \\ x(x – 1) &= (x – 2)(x + 2) \\ x^2 – x &= x^2 – 4 \\ -x &= -4 \implies x = 4 \end{aligned}$$
The value of $x$ is $4$.
Question 2. A 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 $\frac{AX}{AB}$. [CBSE 2017, 2020]
Answer:
Given $XY \parallel AC$ with $X$ on $AB$ and $Y$ on $BC$.
In $\triangle BXY$ and $\triangle BAC$:
- $\angle BXY = \angle A$ (Corresponding angles)
- $\angle B = \angle B$ (Common angle)Thus, $\triangle BXY \sim \triangle BAC$ by AA similarity.
Since ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides:
$$\frac{\text{ar}(\triangle BXY)}{\text{ar}(\triangle BAC)} = \left(\frac{BX}{AB}\right)^2$$
Given that $XY$ divides $\triangle ABC$ into two equal areas:
$$\text{ar}(\triangle BAC) = 2 \times \text{ar}(\triangle BXY) \implies \frac{\text{ar}(\triangle BXY)}{\text{ar}(\triangle BAC)} = \frac{1}{2}$$
$$\left(\frac{BX}{AB}\right)^2 = \frac{1}{2} \implies \frac{BX}{AB} = \frac{1}{\sqrt{2}}$$
Now, calculating $\frac{AX}{AB}$:
$$\frac{AX}{AB} = \frac{AB – BX}{AB} = 1 – \frac{BX}{AB} = 1 – \frac{1}{\sqrt{2}} = \frac{\sqrt{2} – 1}{\sqrt{2}} = \frac{2 – \sqrt{2}}{2}$$
The ratio $\frac{AX}{AB}$ is $\frac{\sqrt{2} – 1}{\sqrt{2}}$.
Question 3. In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such that $BD = \frac{1}{3}BC$. Prove that $9AD^2 = 7AB^2$. [CBSE 2018, 2023]
Answer:
Let each side of equilateral $\triangle ABC$ be $a$ ($AB = BC = CA = a$).
Draw altitude $AM \perp BC$. In an equilateral triangle, the altitude bisects the base:
$$BM = MC = \frac{a}{2}, \quad AM = \frac{\sqrt{3}}{2}a$$
Given $BD = \frac{a}{3}$. Then:
$$DM = BM – BD = \frac{a}{2} – \frac{a}{3} = \frac{a}{6}$$
In right triangle $\triangle AMD$, applying Pythagoras theorem:
$$\begin{aligned} AD^2 &= AM^2 + DM^2 \\ AD^2 &= \left(\frac{\sqrt{3}}{2}a\right)^2 + \left(\frac{a}{6}\right)^2 \\ AD^2 &= \frac{3a^2}{4} + \frac{a^2}{36} = \frac{27a^2 + a^2}{36} = \frac{28a^2}{36} = \frac{7}{9}a^2 \\ 9AD^2 &= 7a^2 = 7AB^2 \end{aligned}$$
Hence Proved: $9AD^2 = 7AB^2$.
Question 4. State and prove the Converse of Basic Proportionality Theorem.
Answer:
Statement: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Proof by Contradiction:
Let line $DE$ divide $AB$ and $AC$ of $\triangle ABC$ such that $\frac{AD}{DB} = \frac{AE}{EC}$.
Assume $DE$ is not parallel to $BC$. Then there must exist a line $DE’$ parallel to $BC$.
If $DE’ \parallel BC$, by BPT:
$$\frac{AD}{DB} = \frac{AE’}{E’C} \quad \text{— (1)}$$
Given:
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \text{— (2)}$$
Equating (1) and (2):
$$\frac{AE’}{E’C} = \frac{AE}{EC}$$
Adding $1$ to both sides:
$$\frac{AE’ + E’C}{E’C} = \frac{AE + EC}{EC} \implies \frac{AC}{E’C} = \frac{AC}{EC} \implies E’C = EC$$
This equality is possible only if points $E’$ and $E$ coincide.
Hence, $DE$ must be parallel to $BC$.
Hence Proved.
Question 5. In $\triangle ABC$, $\angle B = 90^\circ$ and $BD \perp AC$. Prove that $\triangle ADB \sim \triangle BDC$ and hence $BD^2 = AD \times CD$. [CBSE 2019, 2022]
Answer:
In right-angled $\triangle ABC$ with $BD \perp AC$:
Let $\angle A = \theta$.
In right $\triangle ABD$, $\angle ABD = 90^\circ – \theta$.
Since $\angle ABC = 90^\circ$, $\angle DBC = 90^\circ – (90^\circ – \theta) = \theta$.
In $\triangle ADB$ and $\triangle BDC$:
- $\angle ADB = \angle BDC = 90^\circ$
- $\angle BAD = \angle CBD = \theta$
By AA Similarity Criterion:
$$\triangle ADB \sim \triangle BDC$$
Corresponding sides are proportional:
$$\frac{BD}{CD} = \frac{AD}{BD} \implies BD^2 = AD \times CD$$
Hence Proved.
Question 6. $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of a $\triangle ABC$. If $AP = 2\text{ cm}, PB = 4\text{ cm}, AQ = 3\text{ cm}$, and $QC = 6\text{ cm}$, show that $BC = 3PQ$.
Answer:
Ratios on the sides:
$$\frac{AP}{AB} = \frac{2}{2 + 4} = \frac{2}{6} = \frac{1}{3}$$
$$\frac{AQ}{AC} = \frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}$$
In $\triangle APQ$ and $\triangle ABC$:
- $\frac{AP}{AB} = \frac{AQ}{AC} = \frac{1}{3}$
- $\angle A = \angle A$ (Common angle)
By SAS Similarity Criterion: $\triangle APQ \sim \triangle ABC$.
Therefore:
$$\frac{PQ}{BC} = \frac{AP}{AB} = \frac{1}{3} \implies BC = 3PQ$$
Hence Proved.
Question 7. [Assertion-Reason]
Assertion (A): If $\triangle ABC \sim \triangle PQR$ with $\angle A = 50^\circ$ and $\angle B = 70^\circ$, then $\angle R = 60^\circ$.
Reason (R): The sum of all interior angles in any triangle is $180^\circ$, and corresponding angles of similar triangles are equal.
Answer:
- Evaluating Assertion: In $\triangle ABC$, $\angle C = 180^\circ – (50^\circ + 70^\circ) = 60^\circ$.Since $\triangle ABC \sim \triangle PQR$, corresponding angle $\angle R = \angle C = 60^\circ$. (Assertion is True)
- Evaluating Reason: The reason correctly states the angle sum property and the similarity angle condition used in the deduction. (Reason is True and correctly explains A).Correct Choice: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 8. [Assertion-Reason]
Assertion (A): Two congruent triangles are always similar, but two similar triangles need not be congruent.
Reason (R): Congruent triangles have corresponding sides in the ratio $1 : 1$, which satisfies the definition of similarity.
Answer:
- For congruent triangles, corresponding angles are equal and corresponding sides have ratio $1$, which satisfies similarity. Conversely, similar triangles with ratio $k \neq 1$ are not congruent. Both statements and explanation are accurate.Correct Choice: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 9. Through the mid-point $M$ of the side $CD$ of a parallelogram $ABCD$, the line $BM$ is drawn intersecting diagonal $AC$ in $L$ and $AD$ produced in $E$. Prove that $EL = 2BL$. [CBSE 2015, 2020]
Answer:
- In $\triangle BMC$ and $\triangle EMD$:
- $MC = MD$ ($M$ is mid-point of $CD$)
- $\angle BMC = \angle EMD$ (Vertically opposite angles)
- $\angle MCB = \angle MDE$ (Alternate interior angles, $BC \parallel AE$)By ASA Congruence: $\triangle BMC \cong \triangle EMD \implies BC = ED$.
- Since $ABCD$ is a parallelogram, $BC = AD$.Then $AE = AD + DE = BC + BC = 2BC$.
- In $\triangle AEL$ and $\triangle CBL$:
- $\angle ALE = \angle CLB$ (Vertically opposite angles)
- $\angle EAL = \angle BCL$ (Alternate interior angles, $AE \parallel BC$)By AA Similarity: $\triangle AEL \sim \triangle CBL$.
- Proportional sides:$$\frac{EL}{BL} = \frac{AE}{BC} = \frac{2BC}{BC} = 2 \implies EL = 2BL$$Hence Proved.
Question 10. [Case Study 1] A civil engineering group wants to measure the width of a river without crossing it. An engineer sets up flags at points $A, B$ on one bank and identifies tree $C$ on the opposite bank such that $AB \perp BC$. A helper walks along the bank to point $D$ and places a marker at $E$ on $CD$ such that $DE \perp BD$.
(i) If $AB = 30\text{ m}, BC = x\text{ m}, DE = 10\text{ m}$, and distance from $B$ to $D$ through intersection point $O$ has $BO = 45\text{ m}, OD = 15\text{ m}$, prove that $\triangle CBO \sim \triangle EDO$.
(ii) Calculate the width of the river $BC$.
Answer:
- (i) In $\triangle CBO$ and $\triangle EDO$:
- $\angle CBO = \angle EDO = 90^\circ$ (Given perpendicular alignments)
- $\angle COB = \angle EOD$ (Vertically opposite angles)By AA Similarity Criterion: $\triangle CBO \sim \triangle EDO$.
- (ii) Since $\triangle CBO \sim \triangle EDO$:$$\frac{BC}{DE} = \frac{BO}{DO} \implies \frac{BC}{10} = \frac{45}{15} = 3 \implies BC = 30\text{ m}$$The width of the river is $30\text{ m}$.
Question 11. In $\triangle ABC$, $AD$ is a median and $E$ is the mid-point of $AD$. $BE$ produced meets $AC$ in $F$. Prove that $AF = \frac{1}{3}AC$. [CBSE 2016, 2021]
Answer:
Construction: Draw $DG \parallel BF$ meeting $AC$ at $G$.
- In $\triangle ADG$, $E$ is the mid-point of $AD$ and $EF \parallel DG$.By the Converse of Mid-point Theorem (BPT Corollary):$$AF = FG \quad \text{— (1)}$$
- In $\triangle CBF$, $D$ is the mid-point of $BC$ and $DG \parallel BF$.By the Converse of Mid-point Theorem:$$FG = GC \quad \text{— (2)}$$
- Combining (1) and (2):$$AF = FG = GC$$$$AC = AF + FG + GC = 3AF \implies AF = \frac{1}{3}AC$$Hence Proved.
Question 12. Prove that the ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.
Answer:
Let $\triangle ABC \sim \triangle DEF$ with common ratio $k$:
$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k$$
Then $AB = k \cdot DE$, $BC = k \cdot EF$, and $AC = k \cdot DF$.
Computing the ratio of perimeters:
$$\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle DEF)} = \frac{AB + BC + AC}{DE + EF + DF} = \frac{k(DE + EF + DF)}{DE + EF + DF} = k = \frac{AB}{DE}$$
Hence proved.
Question 13. In a right triangle $ABC$, right-angled at $C$, $P$ and $Q$ are points on the sides $CA$ and $CB$ respectively which divide these sides in the ratio $2 : 1$. Prove that $9(AQ^2 + BP^2) = 13AB^2$.
Answer:
Given $CP = \frac{2}{3}CA$ and $CQ = \frac{2}{3}CB$.
Applying Pythagoras theorem in right triangles:
$$AQ^2 = AC^2 + CQ^2 = AC^2 + \frac{4}{9}BC^2$$
$$BP^2 = BC^2 + CP^2 = BC^2 + \frac{4}{9}AC^2$$
Adding the two equations:
$$AQ^2 + BP^2 = \frac{13}{9}AC^2 + \frac{13}{9}BC^2 = \frac{13}{9}(AC^2 + BC^2) = \frac{13}{9}AB^2$$
$$9(AQ^2 + BP^2) = 13AB^2$$
Hence Proved.
Question 14. If two triangles $\triangle ABC$ and $\triangle DEF$ are similar such that $2AB = DE$ and $BC = 8\text{ cm}$, find the length of $EF$.
Answer:
Given $\triangle ABC \sim \triangle DEF$ and $DE = 2AB \implies \frac{AB}{DE} = \frac{1}{2}$.
Since corresponding sides are in the same ratio:
$$\frac{BC}{EF} = \frac{AB}{DE} \implies \frac{8}{EF} = \frac{1}{2} \implies EF = 16\text{ cm}$$
The length of $EF$ is $16\text{ cm}$.
Question 15. [Case Study 2] A lamp post $8\text{ m}$ high is placed near a walking path. A boy of height $1.6\text{ m}$ walks away from the base of the lamp post at a uniform speed of $1.2\text{ m/s}$.
(i) Find the length of his shadow after $4\text{ seconds}$.
(ii) Find the rate at which the tip of his shadow is moving.
Answer:
Let $AB = 8\text{ m}$ be the lamp post and $CD = 1.6\text{ m}$ be the boy. Let shadow length be $s = DE$.
Distance walked in $4\text{ s}$ is $BD = 1.2 \times 4 = 4.8\text{ m}$.
- (i) In $\triangle ABE$ and $\triangle CDE$:$\angle B = \angle D = 90^\circ$ and $\angle E = \angle E$ (Common). Thus $\triangle ABE \sim \triangle CDE$.$$\begin{aligned} \frac{AB}{CD} &= \frac{BE}{DE} \\ \frac{8}{1.6} &= \frac{4.8 + s}{s} \\ 5 &= \frac{4.8 + s}{s} \implies 5s = 4.8 + s \implies 4s = 4.8 \implies s = 1.2\text{ m} \end{aligned}$$Shadow length after $4\text{ seconds}$ is $1.2\text{ m}$.
- (ii) Let $x$ be the distance from the post and $y$ be the total distance to the tip of shadow ($y = x + s$).$\frac{8}{1.6} = \frac{y}{y – x} \implies 5(y – x) = y \implies 4y = 5x \implies y = 1.25x$.Differentiating with respect to time $t$: $\frac{dy}{dt} = 1.25 \frac{dx}{dt} = 1.25 \times 1.2 = 1.5\text{ m/s}$.The tip of the shadow moves at $1.5\text{ m/s}$.
7. Concluding Board Topper Strategy
To score 100% in Triangles in the CBSE Class 10 Board Exam:
- Rigid Proof Framework: Never start writing a proof directly. Always structure your solution into four distinct headings: Given, To Prove, Construction (if any), and Proof.
- Clear Geometric Justification: Every single line of equality in a proof must be backed by a specific theorem or reason written in parentheses (e.g., (Alternate interior angles, AB || CD) or (By Basic Proportionality Theorem)).
- Symbolic Order Verification: When writing similarity statements (e.g., $\triangle ABC \sim \triangle DEF$), ensure that vertex letters strictly correspond to equal angles ($A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F$). Examiners penalize incorrect vertex ordering.
- Sharp Pencil Diagrams: Draw neat, labeled geometric diagrams on the left/top side of the proof using a ruler and pencil. Always use dashed lines for construction elements.
- Dimensional Units: For shadow, height, and distance numericals, box your final value and attach the explicit unit ($\text{m}$ or $\text{cm}$).