**NCERT Class 10 Mathematics Chapter 6 Triangles** **Solutions,** **NCERT Solutions For Class 10 Maths, CBSE Solutions For Class 10 Mathematics** to each chapter is provided in the list so that you can easily browse throughout different chapter **NCERT Class 10 Mathematics Chapter 6 Triangles****Notes** and select needs one.

**NCERT Class 10 Mathematics Chapter 6 Triangles**

**NCERT Class 10 Mathematics Chapter 6 Triangles**Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. **NCERT****Class 10 Mathematics Textual Question Answer. **These solutions are part of NCERT All Subject Solutions. Here we have given

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**Solutions****NCERT Class 10 Mathematics Chapter 6 Triangles****Triangles**

**Triangles****Chapter – 6**

Exercise 6.1 |

**1. Fill in the blanks using correct word given in the brackets:**

**(i) All circles are __________. (congruent, similar)**

Ans: Similar.

**(ii) All squares are __________. (similar, congruent)**

Ans: similar.

**(iii) All __________ triangles are similar. (isosceles, equilateral)**

Ans: Equilateral.

**(iv) Two polygons of the same number of sides are similar, if:**

**(a) their corresponding angles are __________ and **

Ans: Equal.

**(b) Their corresponding sides are __________. (equal, proportional)**

Ans: Proportional.

**2. Give two different examples of pair of:**

**(i) Similar figures.**

Ans: Two equilateral triangles with sides 1 cm and 2 cm

**(ii) Non-similar figures**

Ans:

**3. State whether the following quadrilaterals are similar or not: **

Ans:

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

Exercise 6.2. |

**1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (i).**

Ans:

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

**2. E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR.**

**(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm**

Ans:

**(ii) PE = 4 cm QR = 4.5cm PF = 8 cm and RF = 9cm**

Ans:

**(iii) PQ = 1.28cm PR = 2.56cm PE = 0.18cm and PF = 0.36cm**

Ans:

**3. In the fig DE || AC and DF || AE. Prove that AM/MB = AN/AD**

Ans:

In the given figure, LM || CB By using basic proportionality theorem, we obtain

**4.**

Ans:

**5.In the following figure, DE || OQ and DF || OR, show that EF || QR. **

Ans:

**6. 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.**

Ans:

**7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).**

Ans:

**8. Using Converse of basic proportionality theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).**

Ans:

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. i.e., AP

= PB and AQ = QC It can be observed that

**9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO**

Ans:

**10.**

Ans: Let us consider the following figure for the given question.

Exercise 6.3 |

**1. State which pairs of triangles in the following figure 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)**

Ans: (i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ∆ABC ∼ ∆PQR [By AAA similarity criterion]

Ans: Do yourself.

Ans: The given triangles are not similar as the corresponding sides are not proportional.

Ans: The given triangles are not similar as the corresponding sides are not proportional.

**(v) **

Ans: The given triangles are not similar as the corresponding sides are not proportional.

**(vi)**

In ∆DEF,

∠D +∠E +∠F = 180º (Sum of the measures of the angles of a triangle is 180º.)

70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ∆PQR,

∠P +∠Q +∠R = 180º

(Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ∆DEF and ∆PQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ∆DEF ∼ ∆PQR [By AAA similarity criterion]

**2. In the following figure, ∆ODC ∼ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB**

Ans: Do yourself.

**3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangle.show that AO/OC = OB/OD**

Ans:

In ∆DOC and ∆BOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠ DOC = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ∆DOC ∼ ∆BOA [AAA similarity criterion]

**4.**

Ans:

**5. S and T are point on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ∼ ∆RTS.**

Ans:

In ∆RPQ and ∆RST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ∆RPQ ∼ ∆RTS (By AA similarity criterion)

** 6. In the following figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ∼ ∆ABC.**

Ans:

It is given that ∆ABE ≅ ∆ACD.

∴ AB = AC [By CPCT] …………………(1)

And, AD = AE [By CPCT] …………….(2)

In ∆ADE and ∆ABC,

**7. In the following figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:**

**(i) ∆AEP – ∆CDP**

Ans:

In ∆AEP and ∆CDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

∆AEP – ∆CDP

**(ii) ∆ABD – ∆CBE **

Ans:

In ∆ABD and ∆CBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

∆ABD – ∆CBE

**(iii) ∆AEP – ∆ADB**

Ans:

In ∆PDC and ∆BEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

∆PDC – ∆BEC

**(iv) ∆PDC – ∆BEC**

Ans:

In ∆PDC and ∆BEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

∆PDC – ∆BEC

**8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ∼ ∆CFB**

Ans:

In ∆ABE and ∆CFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ∆ABE – ∆CFB (By AA similarity criterion)

**9. In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:**

Ans: In ∆ABC and ∆AMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ∆ABC – ∆AMP (By AA similarity criterion)

**10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC – ∆FEG, Show that:**

Ans:

**11. 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 ∆ABD – ∆ECF**

Ans: It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ∆ABD and ∆ECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Proved above)

∴ ∆ABD ∼ ∆ECF (By using AA similarity criterion)

**12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see the given figure). Show that ∆ABC – ∆PQR.**

Ans:

**13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that:**

Ans:

In ∆ADC and ∆BAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ∆ADC ∼ ∆BAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in Proportion.

**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**.

Ans: Do yourself.

**15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. **

Ans:

Let AB and CD be a tower and a pole respectively.

Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, ∠DCF = ∠BAE

And, ∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and pole are vertical to the ground)

∴ ∆ABE ∼ ∆CDF (AAA similarity criterion)

Therefore, the height of the tower will be 42 metres.

**16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM**

Ans: