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NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.6 are designed to help students understand and apply the properties of similar triangles and Pythagoras' Theorem in real-life situations. These solutions offer a step-by-step approach to solving problems that require applying the similarity criteria (AA, SAS, SSS) and the Pythagorean Theorem to find unknown sides and solve geometric problems.

Exercise 6.6 is centered around applications of similarity theorems and involves proof-based questions as well as solving real-world problems. This exercise helps students to:

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

Triangles

Medium

State whether the following quadrilaterals are similar or not:

Triangles

Medium

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

Triangles

Medium

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

Triangles

Medium

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

Triangles

Medium

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

Triangles

Medium

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?

Triangles

Medium

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]

Triangles

Medium

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

[CBSE - 2020]

Triangles

Easy

\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]

Triangles

Medium

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.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

Understand the relationship between sides and angles of similar triangles
Use Pythagoras’ Theorem to solve practical problems
Apply these concepts to geometrical proofs and calculations

These solutions provide clear, detailed explanations that will make it easier for students to approach the topic with confidence, ensuring solid exam preparation.

Q1. In Fig. 6.50, PS is the bisector of ∠QPR of ΔPQR. Prove that:

$\frac{QS}{PQ} = \frac{SR}{PR}$

$PQ QS = PR SR$

Reasoning:

As we know, 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 (Basic Proportionality Theorem).

Solution:

Draw a line parallel to PS through R, which intersects QP produced at T.
PS || RT.
In ΔQPR, we have the following:

∠QPS = ∠SPR (alternate interior angles)
∠QPR = ∠PTR (corresponding angles)
∠QPS = ∠PTR (by AA similarity criterion)

From the above angles, we can conclude that triangles are similar:

ΔQPR ~ ΔQMP
ΔPQR ~ ΔPRM
Using the similarity, we have:

$\frac{QS}{PQ} = \frac{SR}{PR}$

$PQ QS = PR SR$

Q2. In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC, and DN ⊥ AB. Prove that:

(i)

$\frac{DM^2}{MC} = \frac{DN^2}{AN}$

$MC D M ^{2} = A N D N ^{2}$
(ii)

$\frac{DM^2}{MC} = \frac{DN^2}{AN}$

$MC D M ^{2} = A N D N ^{2}$

Solution:

By applying the AA similarity criterion and the Basic Proportionality Theorem (BPT), we can solve this.
(i) From the rectangle DMBN, we have that

$DM = BN$

$D M = BN$ and

$DN = BM$

$D N = BM$ .
(ii) From similarity of triangles and applying the Pythagoras theorem, we obtain the required ratio for the areas and corresponding sides.

Q3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that:

$AC^2 + AB^2 = BC^2 + BD^2$

$A C^{2} + A B^{2} = B C^{2} + B D^{2}$

Solution:

In ΔADC, since ∠ADC = 90°, we apply the Pythagorean theorem.
For ΔABC, the squares of the sides satisfy the Pythagorean relationship.
Using properties of similar triangles and the angles involved, we can derive the required relation.

Q4. In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that:

$AC^2 + AB^2 = BC^2 - 2BC \cdot BD$

$A C^{2} + A B^{2} = B C^{2} - 2 BC \cdot B D$

Solution:

Again, we use the Pythagorean theorem for triangles involving right angles.
The side relations are established from similarity and proportionality between the triangles formed by the perpendiculars and sides.

Q5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i)

$\frac{BC^2}{AC^2 + AD^2} = BC \cdot DM$

$A C ^{2} + A D ^{2} B C ^{2} = BC \cdot D M$
(ii)

$\frac{BC^2}{AB^2 + AD^2} = BC \cdot DM$

$A B ^{2} + A D ^{2} B C ^{2} = BC \cdot D M$
(iii)

$AC^2 + AB^2 = BC^2 + AD^2$

$A C^{2} + A B^{2} = B C^{2} + A D^{2}$

Solution:

(i) In ΔAMC, since ∠AMC = 90°, using the Pythagorean theorem:

$AC^2 + AM^2 = CM^2$

$A C^{2} + A M^{2} = C M^{2}$

(ii) Similarly, applying Pythagoras' theorem in ΔAMB and using side relations, we can conclude the second equation.
(iii) The sum of squares on both sides of the triangle is equal, leading to the third equation.

Q6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Solution:

In parallelogram ABCD, where AB = CD and AD = BC, we draw AE ⊥ DF.
Using the Pythagorean theorem and properties of right triangles formed by the diagonals and sides, we get:

$AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2$

$A C^{2} + B D^{2} = A B^{2} + B C^{2} + C D^{2} + D A^{2}$

Thus, we prove the required relation.

Q7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that:

(i) ΔAPC ∼ ΔDPB
(ii)

$AP \cdot PB = CP \cdot DP$

$A P \cdot PB = CP \cdot D P$

Solution:

Using properties of similar triangles and angles in the same segment, we show that:
(i) Triangles ΔAPC and ΔDPB are similar by the AA similarity criterion.
(ii) From the properties of similar triangles, we use the proportionality rule to get:

$AP \cdot PB = CP \cdot DP$

$A P \cdot PB = CP \cdot D P$

Q8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) ΔPAC ∼ ΔPDB
(ii)

$PA \cdot PB = PC \cdot PD$

$P A \cdot PB = PC \cdot P D$

Solution:

(i) Using the exterior angle of a cyclic quadrilateral and the AA similarity criterion, we show that:

$\angle PAC = \angle PDB$

$∠ P A C = ∠ P D B$

(ii) Applying the proportionality of similar triangles, we derive the required relation:

$PA \cdot PB = PC \cdot PD$

$P A \cdot PB = PC \cdot P D$

Q9. In Fig. 6.63, D is a point on side BC of ΔABC such that $BD = BA$

$B D = B A$ and $CD = CA$

$C D = C A$ . Prove that AD is the bisector of ∠BAC.

Solution:

We extend BA to E such that AE = AC, and join CE.
In ΔAEC, AE = AC, so ∠ACE = ∠AEC.
We apply the converse of the Basic Proportionality Theorem (BPT) to prove that AD is the angle bisector.

Q10. 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. How much string does she have out?

Solution:

Using the Pythagorean theorem, we find that the string length is:

$AB = \sqrt{1.8^2 + 2.4^2} = 3 \, \text{m}$

$A B = 1.8^{2} + 2.4^{2} = 3 m$

After pulling in the string at the rate of 5 cm per second for 12 seconds, the new horizontal distance of the fly is:

$AE = 3.0 - 0.6 = 2.4 \, \text{m}$

$A E = 3.0 - 0.6 = 2.4 m$

We can apply the Pythagorean theorem to calculate the new position of the fly after 12 seconds.

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.6

Q1. What is the focus of Exercise 6.6?
Answer:
Exercise 6.6 focuses on applying the properties of similar triangles, Pythagoras' Theorem, and the relationship between sides and angles in different geometric problems. It involves solving problems related to real-life applications and proof-based questions.

Q2. How do I prove triangles are similar in this exercise?
Answer:
To prove triangles are similar, you can use the following criteria:

AA (Angle-Angle): If two triangles have their corresponding angles equal, they are similar.
SSS (Side-Side-Side): If the corresponding sides of two triangles are proportional, they are similar.
SAS (Side-Angle-Side): If two sides of one triangle are proportional to two sides of another triangle, and the included angle is equal, then the triangles are similar.

Q3. What is the importance of Pythagoras' Theorem in this exercise?
Answer:
Pythagoras' Theorem is used to determine the relationship between the sides of a right-angled triangle. The theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. This helps in solving problems involving right-angled triangles.

Q4. How do I apply similarity in solving problems?
Answer:

Identify the triangles that are similar in the problem.
Use the appropriate similarity criteria (AA, SAS, or SSS).
Set up proportionality equations based on the similar triangles.
Solve for the unknown values of sides or angles.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions offer clear, detailed explanations and step-by-step methods for applying similarity criteria and the Pythagorean Theorem. By practicing these solutions, students will improve their problem-solving abilities and be well-prepared for geometry questions in Class 10 exams.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

Q1. In Fig. 6.50, PS is the bisector of ∠QPR of ΔPQR. Prove that:

Reasoning:

Solution:

Q2. In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC, and DN ⊥ AB. Prove that:

Solution:

Q3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that:

Solution:

Q4. In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that:

Solution:

Q5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

Solution:

Q6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Solution:

Q7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that:

Solution:

Q8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

Solution:

Q9. In Fig. 6.63, D is a point on side BC of ΔABC such that $BD = BA$

$B D = B A$ and $CD = CA$

$C D = C A$ . Prove that AD is the bisector of ∠BAC.

Solution:

Q10. 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. How much string does she have out?

Solution:

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.6

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NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6 Triangles

Q1. In Fig. 6.50, PS is the bisector of ∠QPR of ΔPQR. Prove that:

Reasoning:

Solution:

Q2. In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC, and DN ⊥ AB. Prove that:

Solution:

Q3. In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that:

Solution:

Q4. In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that:

Solution:

Q5. In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

Solution:

Q6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Solution:

Q7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that:

Solution:

Q8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

Solution:

Q9. In Fig. 6.63, D is a point on side BC of ΔABC such that BD=BABD = BA

BD=BA and CD=CACD = CA

CD=CA. Prove that AD is the bisector of ∠BAC.

Solution:

Q10. 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. How much string does she have out?

Solution:

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.6

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Q9. In Fig. 6.63, D is a point on side BC of ΔABC such that $BD = BA$

$B D = B A$ and $CD = CA$

$C D = C A$ . Prove that AD is the bisector of ∠BAC.