NCERT solutions for Class 10 Maths Chapter 6 Triangles is provided here which is considered to be one of the most important study material for the students studying in CBSE Class 10. NCERT Solutions for Class 10 Maths Chapter 6 is provided here along with the downloadable PDF. Chapter 6 of class 10 is the triangles and it covers a vast topic including a number of rules and theorems. Students often tend to get confused which theorem to use while solving one particular question.

The NCERT Solutions for Class 10 Maths are prepared by the subject experts to help students in their board exam preparation in a better way. The solutions provided at BYJU’S are in such a way that every step while solving a problem, is explained clearly and in detail. These solutions can be helpful not only for the exam preparation but also in solving the homework and assignments. The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks and, thus, the solutions of chapter 6 – Triangles is one of the best resources to prepare and equip oneself to solve any type of questions in the exam, from the chapter. It is highly recommended for the students to practice these NCERT Solutions on a regular basis to excel in the Class 10 Board examination.

### Download PDF of NCERT Solutions for Class 10 Maths 6- Triangles

### Access Answers of Maths NCERT class 10 Chapter 6 – Triangles

### Class 10 Maths Chapter 6 Exercise 6.1 Page: 122

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

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

Answer: Similar

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

Answer: Similar

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

Answer: 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)**

Answer: (a) Equal

(b) Proportional

**2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures**

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

Solution: From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.

### Class 10 Maths Chapter 6 Exercise 6.2 Page: 128

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

**2. E and F are points on the sides PQ and PR respectively of a ΔPQR. 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**

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

**(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm**

**Solution: Given, **in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;

3. In the figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD

**4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC**

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

**6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.**

**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).**

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

**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.**

**10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.**

### Class 10 Maths Chapter 6 Exercise 6.3 Page: 138

**1. State which pairs of triangles in 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:**

**Solution:**

(i) Given, in ΔABC and ΔPQR,

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore by AAA similarity criterion,

∴ ΔABC ~ ΔPQR

### Class 10 Maths Chapter 6 Exercise 6.3 Page: 139

(vi) In ΔDEF, by sum of angles of triangles, we know that,

∠D + ∠E + ∠F = 180°

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

⇒ ∠F = 30°

Similarly, In ΔPQR,

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

⇒ ∠P = 70°

Now, comparing both the triangles, ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Therefore, by AAA similarity criterion,

Hence, ΔDEF ~ ΔPQR

2. ** **In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

**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 triangles, show that AO/OC = OB/OD**

**4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.**

**5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.**

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

**(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC**

**Solution: **Given, altitudes AD and CE of ΔABC intersect each other at the point P.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,

ΔPDC ~ ΔBEC

**8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.**

**Solution: **Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In ΔABE and ΔCFB,

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

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

∴ ΔABE ~ ΔCFB (AA similarity criterion)

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

**(i) ΔABC ~ ΔAMP**

**(ii) CA/PA = BC/MP**

**Solution: **Given, ABC and AMP are two right triangles, right angled at B and M respectively.

(i) In ΔABC and ΔAMP, we have,

∠CAB = ∠MAP (common angles)

∠ABC = ∠AMP = 90° (each 90°)

∴ ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence,

**10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:**

**(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF**

**Solution: **Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.

(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (AA similarity criterion)

**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 ΔABD ~ ΔECF.**

**Solution: **Given, ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Already proved)

∴ ΔABD ~ ΔECF (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 ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.**

**13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA ^{2} = CB.CD**

**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 ΔABC ~ ΔPQR.**

Now, In ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

∠A = ∠P

∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

**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.**

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

### Class 10 Maths Chapter 6 Exercise 6.4 Page: 143

**1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm ^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.**

**Solution: **Given, ΔABC ~ ΔDEF,

Area of ΔABC = 64 cm^{2}

Area of ΔDEF = 121 cm^{2}

EF = 15.4 cm

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,

= AC^{2}/DF^{2} = BC^{2}/EF^{2}

∴ 64/121 = BC^{2}/EF^{2}

⇒ (8/11)^{2} = (BC/15.4)^{2}

⇒ 8/11 = BC/15.4

⇒ BC = 8×15.4/11

⇒ BC = 8 × 1.4

⇒ BC = 11.2 cm

**2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.**

**Solution: Given, **ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

Area of (ΔAOB)/Area of (ΔCOD) = AB^{2}/CD^{2}

= (2CD)^{2}/CD^{2} [∴ AB = CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD^{2}/CD = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

**3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.**

**Solution: **Given, ABC and DBC are two triangles on the same base BC. ADintersects BC at O.

We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 × Base × Height

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ~ ΔDMO (AA similarity criterion)

∴ AP/DM = AO/DO

⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.

**4. If the areas of two similar triangles are equal, prove that they are congruent.**

**Solution: Say **ΔABC and ΔPQR are two similar triangles and equal in area

Now let us prove ΔABC ≅ ΔPQR.

Since, ΔABC ~ ΔPQR

∴ Area of (ΔABC)/Area of (ΔPQR) = BC^{2}/QR^{2}

⇒ BC^{2}/QR^{2} =1 [Since, Area(ΔABC) = (ΔPQR)

⇒ BC^{2}/QR^{2}

⇒ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

**5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.**

**Solution: **Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC.

In ΔABC,

F is the mid point of AB (Already given)

E is the mid-point of AC (Already given)

So, by the mid-point theorem, we have,

FE || BC and FE = 1/2BC

⇒ FE || BC and FE || BD [BD = 1/2BC]

Since, opposite sides of parallelogram are equal and parallel

∴ BDEF is parallelogram.

Similarly in ΔFBD and ΔDEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common sides)

BD = FE (Opposite sides of parallelogram BDEF)

∴ ΔFBD ≅ ΔDEF

Similarly, we can prove that

ΔAFE ≅ ΔDEF

ΔEDC ≅ ΔDEF

As we know, if triangles are congruent, then they are equal in area.

So,

Area(ΔFBD) = Area(ΔDEF) ……………………………**(i)**

Area(ΔAFE) = Area(ΔDEF) ……………………………….**(ii)**

and,

Area(ΔEDC) = Area(ΔDEF) ………………………….**(iii)**

Now,

Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF) + Area(ΔAFE) + Area(ΔEDC) ………**(iv)**

Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF)

From equation **(i)**, **(ii)** and **(iii)**,

⇒ Area(ΔDEF) = ¼ Area(ΔABC)

⇒ Area(ΔDEF)/Area(ΔABC) = 1/4

Hence, Area(ΔDEF):Area(ΔABC) = 1:4

**6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.**

**Solution: **Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.

We have to prove: Area(ΔABC)/Area(ΔDEF) = AM^{2}/DN^{2}

Since, ΔABC ~ ΔDEF (Given)

∴ Area(ΔABC)/Area(ΔDEF) = (AB^{2}/DE^{2}) ……………………………**(i)**

and, AB/DE = BC/EF = CA/FD ………………………………………**(ii)**

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

∴ ∠B = ∠E

AB/DE = BM/EN [Already Proved in equation **(i)**]

∴ ΔABC ~ ΔDEF [SAS similarity criterion]

⇒ AB/DE = AM/DN …………………………………………………..**(iii)**

∴ ΔABM ~ ΔDEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides.

∴ area(ΔABC)/area(ΔDEF) = AB^{2}/DE^{2} = AM^{2}/DN^{2}

Hence, proved.

^{}**7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.**

**Tick the correct answer and justify:**

**8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4**

**Solution: Given,**ΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.

∴ BD = DC = 1/2BC

Let each side of triangle is 2*a*.

As, ΔABC ~ ΔBDE

∴ Area(ΔABC)/Area(ΔBDE) = AB^{2}/BD^{2} = (2*a*)^{2}/(*a*)^{2} = 4*a*^{2}/*a*^{2} = 4/1 = 4:1

Hence, the correct answer is (C).

**9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81**

**Solution: **Given, Sides of two similar triangles are in the ratio 4 : 9.

Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = AB^{2}/DE^{2 }

∴ Area(ΔABC)/Area(ΔDEF) = (4/9)^{2 }= 16/81 = 16:81

Hence, the correct answer is (D).

### Class 10 Maths Chapter 6 Exercise 6.5 Page: 150

**1. Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.****(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm**

**Solution:**

(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)^{2} + (24)^{2} = (25)^{2}

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64

Or, 3^{2} + 6^{2} ≠ 8^{2}

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfies Pythagoras theorem.

(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50^{2} + 80^{2} ≠ 100^{2}

As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle does not satisfies Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given, sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Thus, 144 +25 = 169

Or, 12^{2} + 5^{2} = 13^{2}

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Hence, length of the hypotenuse of this triangle is 13 cm.

**2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM ^{2} = QM × MR.**

**Solution: **Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR

.

We have to prove, PM^{2} = QM × MR

In ΔPQM, by Pythagoras theorem

PQ^{2} = PM^{2} + QM^{2}

Or, PM^{2} = PQ^{2} – QM^{2} ……………………………..**(i)**

In ΔPMR, by Pythagoras theorem

PR^{2} = PM^{2} + MR^{2}

Or, PM^{2} = PR^{2} – MR^{2} ………………………………………..**(ii)**

Adding equation, **(i)** and **(ii)**, we get,

2PM^{2} = (PQ^{2} + PM^{2}) – (QM^{2} + MR^{2})

= QR^{2} – QM^{2} – MR^{2 } [∴ QR^{2} = PQ^{2} + PR^{2}]

= (QM + MR)^{2} – QM^{2} – MR^{2}

= 2QM × MR

∴ PM^{2} = QM × MR

**3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB ^{2} = BC × BD
(ii) AC^{2} = BC × DC
(iii) AD^{2} = BD × CD**

**Solution:**

(i) In ΔADB and ΔCAB,

∠DAB = ∠ACB (Each 90°)

∠ABD = ∠CBA (Common angles)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB^{2} = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° – 90° – x

∠CBA = 90° – x

Similarly, in ΔCAD

∠CAD = 90° – ∠CBA

= 90° –* *x

∠CDA = 180° – 90° – (90° – x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each 90°)

∴ ΔCBA ~ ΔCAD [AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC^{2} = DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB (Each 90°)

∠CDA = ∠ADB (common angles)

∴ ΔDCA ~ ΔDAB [AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD^{2} = BD × CD

**4. ABC is an isosceles triangle right angled at C. Prove that AB ^{2} = 2AC^{2} .**

**Solution:** Given, ΔABC is an isosceles triangle right angled at C.

In ΔACB, ∠C = 90°

AC = BC (By isosceles triangle property)

AB^{2} = AC^{2} + BC^{2} [By Pythagoras theorem]

= AC^{2} + AC^{2} [Since, AC = BC]

AB^{2} = 2AC^{2}

**5. ABC is an isosceles triangle with AC = BC. If AB ^{2} = 2AC^{2}, prove that ABC is a right triangle.**

**Solution:** Given, ΔABC is an isosceles triangle having AC = BC and AB^{2} = 2AC^{2}

In ΔACB,

AC = BC

AB^{2} = 2AC^{2}

AB^{2} = AC^{2 }+ AC^{2}

= AC^{2} + BC^{2 }[Since, AC = BC]

Hence, by Pythagoras theorem ΔABC is right angle triangle.

**6. ABC is an equilateral triangle of side 2a. Find each of its altitudes**.

**Solution: **Given, ABC is an equilateral triangle of side 2a.

Draw, AD ⊥ BC

In ΔADB and ΔADC,

AB = AC

AD = AD

∠ADB = ∠ADC [Both are 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled ΔADB,

AB^{2} = AD^{2 }+ BD^{2}

(2*a*)^{2} = AD^{2 }+ *a*^{2 }

⇒ AD^{2 }= 4*a*^{2} – *a*^{2}

⇒ AD^{2 }= 3*a*^{2}

⇒ AD^{ }= √3a

**7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.**

**Solution: Given, **ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,

AB^{2 }+ BC^{2 }+ CD^{2} + AD^{2 }= AC^{2 }+ BD^{2}

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In ΔAOB,

∠AOB = 90°

AB^{2} = AO^{2 }+ BO^{2 }…………………….. **(i)** [By Pythagoras theorem]

Similarly,

AD^{2} = AO^{2 }+ DO^{2 }…………………….. **(ii)**

DC^{2} = DO^{2 }+ CO^{2 }…………………….. **(iii)**

BC^{2} = CO^{2 }+ BO^{2 }…………………….. **(iv)**

Adding equations **(i) + (ii) + (iii) + (iv)**, we get,

AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = 2(AO^{2 }+ BO^{2 }+ DO^{2 }+ CO^{2 })

= 4AO^{2 }+ 4BO^{2 }[Since, AO = CO and BO =DO]

= (2AO)^{2 }+ (2BO)^{2} = AC^{2 }+ BD^{2}

AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = AC^{2 }+ BD^{2}

Hence, proved.

Page No: 151

**8. In Fig. 6.54, O is a point in the interior of a triangle.**

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:

(i) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2} ,

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

**Solution: **Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA^{2} = OF^{2} + AF^{2}

Similarly, in ΔBOD

OB^{2} = OD^{2} + BD^{2}

Similarly, in ΔCOE

OC^{2} = OE^{2} + EC^{2}

Adding these equations,

OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2 }+ EC^{2}

OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}.

(ii) AF^{2} + BD^{2} + EC^{2} = (OA^{2} – OE^{2}) + (OC^{2} – OD^{2}) + (OB^{2} – OF^{2})

∴ AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

**9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.**

**Solution: Given, a ladder 10 m long reaches a window 8 m above the ground.**

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC^{2} =^{ }AB^{2} + BC^{2}

10^{2} = 8^{2} + BC^{2}

BC^{2 }= 100 – 64

BC^{2 }= 36

BC^{ }= 6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

**10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?**

**Solution: Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.**

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

AC^{2} =^{ }AB^{2} + BC^{2}

24^{2} = 18^{2} + BC^{2}

BC^{2 }= 576 – 324

BC^{2 }= 252

BC^{ }= 6√7m

Therefore, the distance from the base is 6√7m.

**11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?**

**Solution: Given,**

Speed of first aeroplane = 1000 km/hr

Distance covered by first aeroplane flying due north in ** ** hours (OA) = 100 × 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane flying due west in** ** hours (OB) = 1200 × 3/2 km = 1800 km

In right angle ΔAOB, by Pythagoras Theorem,

AB^{2} =^{ }AO^{2} + OB^{2}

⇒ AB^{2} =^{ }(1500)2 + (1800)2

⇒ AB = √2250000 + 3240000

= √5490000

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

**12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.**

**Solution: **Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP^{2} =^{ }PC^{2} + AC^{2}

(12m)^{2} + (5m)^{2} = (AC)^{2}

AC^{2} = (144+25)m^{2} = 169 m^{2}

AC = 13m

Therefore, the distance between their tops is 13 m.

**13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE ^{2} + BD^{2} = AB^{2} + DE^{2}.**

**Solution: **Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC^{2} +^{ }CE^{2} = AE^{2} ………………………………………….**(i)**

In ΔBCD, by Pythagoras theorem, we get

BC^{2} +^{ }CD^{2} = BD^{2} ………………………………..**(ii)**

Fromequations **(i)** and **(ii)**, we get,

AC^{2} +^{ }CE^{2} + BC^{2} +^{ }CD^{2} = AE^{2} + BD^{2} …………..**(iii)**

In ΔCDE, by Pythagoras theorem, we get

DE^{2} =^{ }CD^{2} + CE^{2}

In ΔABC, by Pythagoras theorem, we get

AB^{2} =^{ }AC^{2} + CB^{2}

Putting the above two values in equation **(iii)**, we get

DE^{2} + AB^{2} = AE^{2} + BD^{2}.

### Class 10 Maths Chapter 6 Exercise 6.5 Page: 151

**14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB ^{2} = 2AC^{2} + BC^{2}.**

**Solution: **Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB^{2} =^{ }AD^{2} + BD^{2} ……………………….**(i)**

AC^{2} =^{ }AD^{2} + DC^{2} ……………………………..**(ii)**

Subtracting equation **(ii)** from equation **(i)**, we get

AB^{2} – AC^{2} = BD^{2} – DC^{2}

= 9CD^{2} – CD^{2} [Since, BD = 3CD]

= 9CD^{2} = 8(BC/4)^{2 }[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB^{2} – AC^{2} = BC^{2}/2

⇒ 2(AB^{2} – AC^{2}) = BC^{2}

⇒ 2AB^{2} – 2AC^{2} = BC^{2}

∴ 2AB^{2} = 2AC^{2} + BC^{2}.

**15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD ^{2} = 7AB^{2}.**

**16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.**

**17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°**

**(B) 60°
(C) 90° **

**(D) 45°**

**Solution: Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.**

We can observe that,

AB^{2} = 108

AC^{2} = 144

And, BC^{2} = 36

AB^{2} + BC^{2} = AC^{2}

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).

### Class 10 Maths Chapter 6 Exercise 6.6 Page: 152

**1. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that ⋅**

Solution: Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given, PS is the angle bisector of ∠QPR. Therefore,

∠QPS = ∠SPR………………………………..(i)

As per the constructed figure,

∠SPR=∠PRT(Since, PS||TR)……………(ii)

∠QPS = ∠QRT(Since, PS||TR) …………..(iii)

From the above equations, we get,

∠PRT=∠QTR

Therefore,

PT=PR

In △QTR, by basic proportionality theorem,

QS/SR = QP/PT

Since, PT=TR

Therefore,

QS/SR = PQ/PR

Hence, proved.

**2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove **

(ii) In right triangle DBN,

∠5 + ∠7 = 90° ……………….. (iv)

In right triangle DAN,

∠6 + ∠8 = 90° ………………… (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)

From equation (iv) and (vi), we get,

∠6 = ∠7

From equation (v) and (vi), we get,

∠8 = ∠5

In ∆DNA and ∆BND,

∠6 = ∠7 (Already proved)

∠8 = ∠5 (Already proved)

∴ ∆DNA ∼ ∆BND (AA similarity criterion)

AN/DN = DN/NB

⇒ DN^{2} = AN × NB

⇒ DN^{2} = AN × DM (Since, NB = DM)

Hence, proved.

**3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that**

**AC ^{2}= AB^{2}+ BC^{2}+ 2 BC.BD.**

Solution: By applying Pythagoras Theorem in ∆ADB, we get,

AB^{2} = AD^{2} + DB^{2} ……………………… (i)

Again, by applying Pythagoras Theorem in ∆ACD, we get,

AC^{2} = AD^{2} + DC^{2}

AC^{2} = AD^{2} + (DB + BC) ^{2}

AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB × BC

From equation (i), we can write,

AC^{2} = AB^{2} + BC^{2} + 2DB × BC

Hence, proved.

**4. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that**

**AC ^{2}= AB^{2}+ BC^{2} – 2 BC.BD.**

Solution: By applying Pythagoras Theorem in ∆ADB, we get,

AB^{2} = AD^{2} + DB^{2}

We can write it as;

⇒ AD^{2} = AB^{2} − DB^{2} ……………….. (i)

By applying Pythagoras Theorem in ∆ADC, we get,

AD^{2} + DC^{2} = AC^{2}

From equation (i),

AB^{2} − BD^{2} + DC^{2} = AC^{2}

AB^{2} − BD^{2} + (BC − BD)^{ 2} = AC^{2}

AC^{2} = AB^{2} − BD^{2} + BC^{2} + BD^{2} −2BC × BD

AC^{2}= AB^{2} + BC^{2} − 2BC × BD

Hence, proved.

5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC^{2} = AD^{2} + BC.DM + 2 (BC/2)^{ 2}

(ii) AB^{2} = AD^{2} – BC.DM + 2 (BC/2)^{ 2}

(iii) AC^{2} + AB^{2} = 2 AD^{2} + ½ BC^{2}

Solution:

(i) By applying Pythagoras Theorem in ∆AMD, we get,

AM^{2} + MD^{2} = AD^{2} ………………. (i)

Again, by applying Pythagoras Theorem in ∆AMC, we get,

AM^{2} + MC^{2} = AC^{2}

AM^{2} + (MD + DC)^{ 2} = AC^{2}

(AM^{2} + MD^{2} ) + DC^{2} + 2MD.DC = AC^{2}

From equation(i), we get,

AD^{2} + DC^{2} + 2MD.DC = AC^{2}

Since, DC=BC/2, thus, we get,

AD^{2 }+ (BC/2)^{ 2 }+ 2MD.(BC/2)^{ 2} = AC^{2}

AD^{2 }+ (BC/2)^{ 2 }+ 2MD × BC = AC^{2}

Hence, proved.

(ii) By applying Pythagoras Theorem in ∆ABM, we get;

AB^{2} = AM^{2} + MB^{2}

=> (AD^{2} − DM^{2} ) + MB^{2}

=> (AD^{2} − DM^{2} ) + (BD − MD)^{ 2}

=> AD^{2} − DM^{2} + BD^{2} + MD^{2} − 2BD × MD

=> AD^{2} + BD^{2} − 2BD × MD

=> AD^{2 } + (BC/2)^{ 2 }– 2(BC/2) MD

=> AD^{2 } + (BC/2)^{ 2 }– BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in ∆ABM, we get,

AM^{2} + MB^{2} = AB^{2} ………………….… (i)

By applying Pythagoras Theorem in ∆AMC, we get,

AM^{2} + MC^{2} = AC^{2} …………………..… (ii)

Adding both the equations (i) and (ii), we get,

2AM^{2} + MB^{2} + MC^{2} = AB^{2} + AC^{2}

2AM^{2} + (BD − DM)^{ 2} + (MD + DC)^{ 2} = AB^{2} + AC^{2}

2AM^{2}+BD^{2} + DM^{2} − 2BD.DM + MD^{2} + DC^{2} + 2MD.DC = AB^{2} + AC^{2}

2AM^{2} + 2MD^{2} + BD^{2} + DC^{2} + 2MD (− BD + DC) = AB^{2} + AC^{2}

2(AM^{2}+ MD^{2}) + (BC/2)^{ 2} + (BC/2)^{ 2} + 2MD(-BC/2 + BC/2)^{ 2} = AB^{2} + AC^{2}

2AD^{2 }+ BC^{2}/2 = AB^{2} + AC^{2}

**6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.**

Solution:

Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we get,

DE^{2} + EA^{2} = DA^{2} ……………….… (i)

By applying Pythagoras Theorem in ∆DEB, we get,

DE^{2} + EB^{2} = DB^{2}

DE^{2} + (EA + AB)^{ 2} = DB^{2}

(DE^{2} + EA^{2} ) + AB^{2} + 2EA × AB = DB^{2}

DA^{2} + AB^{2} + 2EA × AB = DB^{2} ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we get,

AD^{2} = AF^{2} + FD^{2}

Again, applying Pythagoras theorem in ∆AFC, we get,

AC^{2} = AF^{2} + FC^{2} = AF^{2} + (DC − FD)^{ 2}

= AF^{2} + DC^{2} + FD^{2} − 2DC × FD

= (AF^{2} + FD^{2} ) + DC^{2} − 2DC × FD AC^{2}

AC^{2}= AD^{2} + DC^{2} − 2DC × FD ………………… (iii)

Since ABCD is a parallelogram,

AB = CD ………………….…(iv)

And BC = AD ………………. (v)

In ∆DEA and ∆ADF,

∠DEA = ∠AFD (Each 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common Angles)

∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)

⇒ EA = DF ……………… (vi)

Adding equations (i) and (iii), we get,

DA^{2} + AB^{2} + 2EA × AB + AD^{2} + DC^{2} − 2DC × FD = DB^{2} + AC^{2}

DA^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA × AB − 2DC × FD = DB^{2} + AC^{2}

From equation (iv) and (vi),

BC^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA × AB − 2AB × EA = DB^{2} + AC^{2}

AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

**7. In Figure, two chords AB and CD intersect each other at the point P. Prove that : **

**(i) ∆APC ~ ∆ DPB **

**(ii) AP . PB = CP . DP**

Solution: Firstly let us join CB, in the given figure.

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

Therefore,

∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that ∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

AP/DP = PC/PB = CA/BD

∴AP. PB = PC. DP

Hence, proved.

**8. 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) ∆ PAC ~ ∆ PDB **

**(ii) PA . PB = PC . PD.**

**Solution:**

(i) In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.

∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

(ii) We have already proved above,

∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

Therefore,

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

∴ AP. PB = PC. DP

**9. In Figure, D is a point on side BC of ∆ ABC such that ⋅ Prove that AD is the bisector of ∠ BAC.**

**Solutions:** In the given figure, let us extend BA to P such that;

AP = AC.

Now join PC.

Given, BD/CD = AB/AC

= BD/CD = AP/AC

By using the converse of basic proportionality theorem, we get,

AD || PC

∠BAD = ∠APC (Corresponding angles) ……………….. (i)

And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)

By the new figure, we have;

AP = AC

⇒ ∠APC = ∠ACP ……………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

∠BAD = ∠APC

Therefore, AD is the bisector of the angle BAC.

Hence, proved.

**10. 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 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?**

**Solution:**

Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the

horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.

To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;

AC2 = AB2 + BC2

AB2 = (1.8 m) 2 + (2.4 m) 2

AB2 = (3.24 + 5.76) m2

AB2 = 9.00 m2

⇒ AB = √9 m = 3m

Thus, the length of the string out is 3 m.

As its given, she pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let us say now, the fly is at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m

In ∆ADB, by Pythagoras Theorem,

AB2 + BD2 = AD2

(1.8 m) 2 + BD2 = (2.4 m) 2

BD2 = (5.76 − 3.24) m2 = 2.52 m2

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m = 2.787 m

= 2.79 m

## NCERT Solutions for Class 10 Maths Chapter 6- Triangles

NCERT Solutions Class 10 Maths Chapter 6, Triangle, is part of the Unit Geometry which constitutes 15 marks of the total marks of 80. On the basis of the CBSE Class 10 syllabus, this chapter belongs to the unit that has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and the problem-solving methods in this chapter is mandatory to score well in the Board examination of class 10 maths.

Main topics covered in this chapter include:

6.1 Introduction

From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.

6.2 Similar Figures

In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing relevant activity. Similar figures are two figures having the same shape but not necessarily the same size.

6.3 Similarity of Triangles

The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.

6.4 Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.

6.5 Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.

6.6 Pythagoras Theorem

You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities and make use of it while solving certain problems.

6.7 Summary

The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.

**List of Exercises in class 10 Maths Chapter 6:**

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles including the detailed explanation of similar figure, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn the chapter 6 of NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure learn the NCERT Solutions for Class 10 effectively to score high in the board examination.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

- Helps to ensure that the students use the concepts in solving the problems.
- Encourages the children to come up with diverse solutions to problems.
- Hints are given for those questions which are difficult to solve.
- Helps the students in checking if the solutions they gave for the questions are correct or not.