NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 6 - Download Free PDF

NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here which is considered to be one of the most important study materials for the students studying in CBSE Class 10. Chapter 6 of NCERT Solutions for Class 10 Maths is structured under the first term of Class 10 Maths according to the updated CBSE Syllabus for 2021-22. It covers a vast topic including a number of rules and theorems. Students often tend to get confused about which theorem to use while solving one particular question. The solutions provided at BYJU’S are in such a way that every step while solving a problem, is explained clearly and in detail. The Solutions for NCERT Class 10 Maths are prepared by the subject experts to help students in their first term exam preparation in a better way. These solutions can be helpful not only for exam preparations but also in solving homework and assignments. The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks and, thus, the NCERT Solutions for Chapter 6 Triangles of Class 10 Maths 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 Class 10 first and second term examinations.

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

 
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Access Answers of NCERT Class 10 Maths Chapter 6 - Triangles

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 Solution: Ncert solutions class 10 chapter 6-1 Ncert solutions class 10 chapter 6-2 3. State whether the following quadrilaterals are similar or not: Ncert solutions class 10 chapter 6-3 Solution: From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.

Exercise 6.2 Page: 128

1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). Ncert solutions class 10 chapter 6-4 Solution: (i) Given, in △ ABC, DE∥BC ∴ AD/DB = AE/EC [Using Basic proportionality theorem] ⇒1.5/3 = 1/EC ⇒EC = 3/1.5 EC = 3×10/15 = 2 cm Hence, EC = 2 cm. (ii) Given, in △ ABC, DE∥BC ∴ AD/DB = AE/EC [Using Basic proportionality theorem] ⇒ AD/7.2 = 1.8 / 5.4 ⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10 ⇒ AD = 2.4 Hence, AD = 2.4 cm. 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;   Triangles Exercise 6.2 Answer 3 (i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm Therefore, by using Basic proportionality theorem, we get, PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5 So, we get, PE/EQ ≠ PF/FR Hence, EF is not parallel to QR. (ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm Therefore, by using Basic proportionality theorem, we get, PE/QE = 4/4.5 = 40/45 = 8/9 And, PF/RF = 8/9 So, we get here, PE/QE = PF/RF Hence, EF is parallel to QR. (iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm From the figure, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm And, FR = PR - PF = 2.56 - 0.36 = 2.20 cm So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i) And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii) So, we get here, PE/EQ = PF/FR Hence, EF is parallel to QR. 3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD Ncert solutions class 10 chapter 6-6 Solution: In the given figure, we can see, LM || CB, By using basic proportionality theorem, we get, AM/AB = AL/AC……………………..(i) Similarly, given, LN || CD and using basic proportionality theorem, ∴AN/AD = AL/AC……………………………(ii) From equation (i) and (ii), we get, AM/AB = AN/AD Hence, proved. 4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC Ncert solutions class 10 chapter 6-7 Solution: In ΔABC, given as, DE || AC Thus, by using Basic Proportionality Theorem, we get, ∴BD/DA = BE/EC ......................................................(i) In  ΔBAE, given as, DF || AE Thus, by using Basic Proportionality Theorem, we get, ∴BD/DA = BF/FE ......................................................(ii) From equation (i) and (ii), we get BE/EC = BF/FE Hence, proved. 5. In the figure, DE||OQ and DF||OR, show that EF||QR. Ncert solutions class 10 chapter 6-8 Solution: Given, In ΔPQO, DE || OQ So by using Basic Proportionality Theorem, PD/DO = PE/EQ……………… ..(i) Again given, in ΔPOR, DF || OR, So by using Basic Proportionality Theorem, PD/DO = PF/FR………………… (ii) From equation (i) and (ii), we get, PE/EQ = PF/FR Therefore, by converse of Basic Proportionality Theorem, EF || QR, in ΔPQR. 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. Ncert solutions class 10 chapter 6-9 Solution: Given here, In ΔOPQ, AB || PQ By using Basic Proportionality Theorem, OA/AP = OB/BQ…………….(i) Also given, In ΔOPR, AC || PR By using Basic Proportionality Theorem ∴ OA/AP = OC/CR……………(ii) From equation (i) and (ii), we get, OB/BQ = OC/CR Therefore, by converse of Basic Proportionality Theorem, In ΔOQR, 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). Ncert solutions class 10 chapter 6-10 Solution: Given, in ΔABC, D is the midpoint of AB such that AD=DB. A line parallel to BC intersects AC at E as shown in above figure such that DE || BC. We have to prove that E is the mid point of AC. Since, D is the mid-point of AB. ∴ AD=DB ⇒AD/DB = 1 …………………………. (i) In ΔABC, DE || BC, By using Basic Proportionality Theorem, Therefore, AD/DB = AE/EC From equation (i), we can write, ⇒ 1 = AE/EC ∴ AE = EC Hence, proved, E is the midpoint of AC. 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). Solution: Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that, AD=BD and AE=EC. Ncert solutions class 10 chapter 6-11 We have to prove that: DE || BC. Since, D is the midpoint of AB ∴ AD=DB ⇒AD/BD = 1...................................... (i)   Also given, E is the mid-point of AC. ∴ AE=EC ⇒ AE/EC = 1 From equation (i) and (ii), we get, AD/BD = AE/EC By converse of Basic Proportionality Theorem, DE || BC Hence, proved. 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. Solution: Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O. Ncert solutions class 10 chapter 6-12 We have to prove, AO/BO = CO/DO From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB In ΔADC, we have OE || DC Therefore, By using Basic Proportionality Theorem AE/ED = AO/CO ……………..(i) Now, In ΔABD, OE || AB Therefore, By using Basic Proportionality Theorem DE/EA = DO/BO…………….(ii) From equation (i) and (ii), we get, AO/CO = BO/DO ⇒AO/BO = CO/DO Hence, proved. 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. Solution: Given, Quadrilateral ABCD where AC and BD intersects each other at O such that, AO/BO = CO/DO. Ncert solutions class 10 chapter 6-13 We have to prove here, ABCD is a trapezium From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB In ΔDAB, EO || AB Therefore, By using Basic Proportionality Theorem DE/EA = DO/OB ……………………(i) Also, given, AO/BO = CO/DO ⇒ AO/CO = BO/DO ⇒ CO/AO = DO/BO ⇒DO/OB = CO/AO …………………………..(ii) From equation (i) and (ii), we get DE/EA = CO/AO Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB ⇒ AB || DC. Hence, quadrilateral ABCD is a trapezium with AB || CD.

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: Ncert solutions class 10 chapter 6-14 Solution: (i) Given, in ΔABC and ΔPQR, ∠A = ∠P = 60° ∠B = ∠Q = 80° ∠C = ∠R = 40° Therefore by AAA similarity criterion, ∴ ΔABC ~ ΔPQR (ii) Given, in  ΔABC and ΔPQR, AB/QR = 2/4 = 1/2, BC/RP = 2.5/5 = 1/2, CA/PA = 3/6 = 1/2 By SSS similarity criterion, ΔABC ~ ΔQRP (iii) Given, in ΔLMP and ΔDEF, LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6 MP/DE = 2/4 = 1/2 PL/DF = 3/6 = 1/2 LM/EF = 2.7/5 = 27/50 Here , MP/DE = PL/DF ≠ LM/EF Therefore, ΔLMP and ΔDEF are not similar. (iv) In ΔMNL and ΔQPR, it is given, MN/QP = ML/QR = 1/2 ∠M = ∠Q = 70° Therefore, by SAS similarity criterion ∴ ΔMNL ~ ΔQPR (v) In ΔABC and ΔDEF, given that, AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80° Here , AB/DF = 2.5/5 = 1/2 And, BC/EF = 3/6 = 1/2 ⇒ ∠B ≠ ∠F Hence, ΔABC and ΔDEF are not similar. (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. Ncert solutions class 10 chapter 6-15 Solution: As we can see from the figure, DOB is a straight line. Therefore, ∠DOC + ∠ COB = 180° ⇒ ∠DOC = 180° - 125° (Given, ∠ BOC = 125°) = 55° In ΔDOC, sum of the measures of the angles of a triangle is 180º Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180° ⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°) ⇒ ∠DCO = 55° It is given that, ΔODC ∝ ¼ ΔOBA, Therefore, ΔODC ~ ΔOBA. Hence, Corresponding angles are equal in similar triangles ∠OAB = ∠OCD ⇒ ∠ OAB = 55° ∠OAB = ∠OCD ⇒ ∠OAB = 55° 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 Solution: Ncert solutions class 10 chapter 6-16 In ΔDOC and ΔBOA, AB || CD, thus alternate interior angles will be equal, ∴∠CDO = ∠ABO Similarly, ∠DCO = ∠BAO Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal; ∴∠DOC = ∠BOA Hence, by AAA similarity criterion, ΔDOC ~ ΔBOA Thus, the corresponding sides are proportional. DO/BO = OC/OA ⇒OA/OC = OB/OD Hence, proved.   4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR. Ncert solutions class 10 chapter 6-17 Solution: In ΔPQR, ∠PQR = ∠PRQ ∴ PQ = PR ……………………...(i) Given, QR/QS = QT/PRUsing equation (i), we get QR/QS = QT/QP……………….(ii) In ΔPQS and ΔTQR, by equation (ii), QR/QS = QT/QP ∠Q = ∠Q ∴ ΔPQS ~ ΔTQR [By SAS similarity criterion] 5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS. Solution: Given, S and T are point on sides PR and QR of ΔPQR And ∠P = ∠RTS. Ncert solutions class 10 chapter 6-18 In ΔRPQ and ΔRTS, ∠RTS = ∠QPS (Given) ∠R = ∠R (Common angle) ∴ ΔRPQ ~ ΔRTS (AA similarity criterion) 6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC. Ncert solutions class 10 chapter 6-19 Solution: Given, ΔABE ≅ ΔACD. ∴ AB = AC [By CPCT] ……………………………....(i) And, AD = AE [By CPCT] …………………………...(ii) In ΔADE and ΔABC, dividing eq.(ii) by eq(i), AD/AB = AE/AC ∠A = ∠A [Common angle] ∴ ΔADE ~ ΔABC [SAS similarity criterion] 7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: Ncert solutions class 10 chapter 6-17 (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, Ncert solutions class 10 chapter 6-18 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: Ncert solutions class 10 chapter 6-19 (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, CA/PA = BC/MP 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. Ncert solutions class 10 chapter 6-20 (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) ⇒CD/GH = AC/FG (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. Ncert solutions class 10 chapter 6-21 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. Ncert solutions class 10 chapter 6-22 Solution: Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR i.e. AB/PQ = BC/QR = AD/PM We have to prove: ΔABC ~ ΔPQR As we know here, AB/PQ = BC/QR = AD/PM Ncert solutions class 10 chapter 6-23 ⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR) ⇒ ΔABD ~ ΔPQM [SSS similarity criterion] ∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal] ⇒ ∠ABC = ∠PQR In ΔABC and ΔPQR AB/PQ = BC/QR ………………………….(i) ∠ABC = ∠PQR …………………………...(ii) From equation (i) and (ii), we get, ΔABC ~ ΔPQR [SAS similarity criterion] 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD Solution: Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Ncert solutions class 10 chapter 6-24 In ΔADC and ΔBAC, ∠ADC = ∠BAC (Already given) ∠ACD = ∠BCA (Common angles) ∴ ΔADC ~ ΔBAC (AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. ∴ CA/CB = CD/CA ⇒ CA2 = CB.CD. Hence, proved. 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. Solution: Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that; AB/PQ = AC/PR = AD/PM We have to prove, ΔABC ~ ΔPQR Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN. Ncert solutions class 10 chapter 6-25 In ΔABD and ΔCDE, we have AD = DE  [By Construction.] BD = DC [Since, AP is the median] and, ∠ADB = ∠CDE [Vertically opposite angles] ∴ ΔABD ≅ ΔCDE [SAS criterion of congruence] ⇒ AB = CE [By CPCT] …………………………..(i) Also, in ΔPQM and ΔMNR, PM = MN [By Construction.] QM = MR [Since, PM is the median] and, ∠PMQ = ∠NMR [Vertically opposite angles] ∴ ΔPQM = ΔMNR [SAS criterion of congruence] ⇒ PQ = RN [CPCT] ……………………………...(ii) Now, AB/PQ = AC/PR = AD/PM From equation (i) and (ii), ⇒CE/RN = AC/PR = AD/PM ⇒ CE/RN = AC/PR = 2AD/2PM ⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN] ∴ ΔACE ~ ΔPRN [SSS similarity criterion] Therefore, ∠2 = ∠4 Similarly, ∠1 = ∠3 ∴ ∠1 + ∠2 = ∠3 + ∠4 ⇒ ∠A = ∠P ....................................................(iii) 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. Solution: Given, Length of the vertical pole = 6m Shadow of the pole = 4 m Let Height of tower = h m Length of shadow of the tower = 28 m Ncert solutions class 10 chapter 6-26 In ΔABC and ΔDEF, ∠C = ∠E (angular elevation of sum) ∠B = ∠F = 90° ∴ ΔABC ~ ΔDEF (AA similarity criterion) ∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional) ∴ 6/h = 4/28 ⇒h = (6×28)/4 ⇒ h = 6 × 7 ⇒ = 42 m Hence, the height of the tower is 42 m. 16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM. Solution: Given, ΔABC ~ ΔPQR Ncert solutions class 10 chapter 6-27 We know that the corresponding sides of similar triangles are in proportion. ∴AB/PQ = AC/PR = BC/QR……………………………(i) Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii) Since AD and PM are medians, they will divide their opposite sides. ∴ BD = BC/2 and QM = QR/2 ……………..………….(iii) From equations (i) and (iii), we get AB/PQ = BD/QM ……………………….(iv) In ΔABD and ΔPQM, From equation (ii), we have ∠B = ∠Q From equation (iv), we have, AB/PQ = BD/QM ∴ ΔABD ~ ΔPQM (SAS similarity criterion) ⇒AB/PQ = BD/QM = AD/PM Exercise 6.4 Page: 143 1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC. Solution: Given, ΔABC ~ ΔDEF, Area of ΔABC = 64 cm2 Area of ΔDEF = 121 cm2 EF = 15.4 cm Ncert solutions class 10 chapter 6-28 As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides, = AC2/DF2 = BC2/EF2 ∴ 64/121 = BC2/EF2 ⇒ (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. Ncert solutions class 10 chapter 6-29 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) = AB2/CD2 = (2CD)2/CD2 [∴ AB = 2CD] ∴ Area of (ΔAOB)/Area of (ΔCOD) = 4CD2/CD2 = 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. Ncert solutions class 10 chapter 6-30 Solution: Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O. We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO Let us draw two perpendiculars AP and DM on line BC. Ncert solutions class 10 chapter 6-31 We know that area of a triangle = 1/2 × Base × Height https://4.bp.blogspot.com/-9ywR15fTTyI/VUiJqLSptdI/AAAAAAAAFYk/1Y11QBtVU68/s1600/equation-2.PNG 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 Ncert solutions class 10 chapter 6-33 Now let us prove ΔABC ≅ ΔPQR. Since, ΔABC ~ ΔPQR ∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2 ⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR) ⇒ BC2/QR2 ⇒ 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. Ncert solutions class 10 chapter 6-34 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) = (1/4)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. Ncert solutions class 10 chapter 6-35 We have to prove: Area(ΔABC)/Area(ΔDEF) = AM2/DN2 Since, ΔABC ~ ΔDEF (Given) ∴ Area(ΔABC)/Area(ΔDEF) = (AB2/DE2) …………………………...(i) and, AB/DE = BC/EF = CA/FD ……………………………………...(ii) https://1.bp.blogspot.com/-ynQR15nRVwc/VUwp_lLVzpI/AAAAAAAAFZ0/hcsAT2o-iuE/s1600/equation-3.PNG 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) = AB2/DE2 = AM2/DN2 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. Solution: Ncert solutions class 10 chapter 6-37Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD. Area(ΔBQC) = ½ Area(ΔAPC) Since, ΔAPC and ΔBQC are both equilateral triangles, as per given, ∴ ΔAPC ~ ΔBQC [AAA similarity criterion] ∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2 Since, Diagonal = √2 side = √2 BC = AC Ncert solutions class 10 chapter 6-38 ⇒ area(ΔAPC) = 2 × area(ΔBQC) ⇒ area(ΔBQC) = 1/2area(ΔAPC) Hence, proved. 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.   Triangles Exercise 6.4 Answer 8 ∴ BD = DC = 1/2BC Let each side of triangle is 2a. As, ΔABC ~ ΔBDE ∴ Area(ΔABC)/Area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 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. Triangles Exercise 6.4 Answer 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) = AB2/DE ∴ Area(ΔABC)/Area(ΔDEF) = (4/9)= 16/81 = 16:81 Hence, the correct answer is (D).

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, 32 + 62 ≠ 82 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, 502 + 802 ≠ 1002 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, 122 + 52 = 132 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 PM2 = QM × MR. Solution: Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR https://3.bp.blogspot.com/-niqPczc0V0k/VUxc45H8iGI/AAAAAAAAFa0/1AGiUafEXZw/s1600/fig-21.PNG We have to prove, PM2 = QM × MR In ΔPQM, by Pythagoras theorem PQ2 = PM2 + QM2 Or, PM2 = PQ2 - QM2 ……………………………..(i) In ΔPMR, by Pythagoras theorem PR2 = PM2 + MR2 Or, PM2 = PR2 - MR2 ...............................................(ii) Adding equation, (i) and (ii), we get, 2PM2 = (PQ2 + PM2) - (QM2 + MR2) = QR2 - QM2 - MR2        [∴ QR2 = PQ2 + PR2] = (QM + MR)2 - QM2 - MR2 = 2QM × MR ∴ PM2 = QM × MR 3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB2 = BC × BD (ii) AC2 = BC × DC (iii) AD2 = BD × CD Ncert solutions class 10 chapter 6-42   Solution: (i) In ΔADB and ΔCAB, ∠DAB = ∠ACB (Each 90°) ∠ABD = ∠CBA (Common angles) ∴ ΔADB ~ ΔCAB [AA similarity criterion] ⇒ AB/CB = BD/AB ⇒ AB2 = 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 ⇒ AC2 =  DC × BC (iii) In ΔDCA and ΔDAB, ∠DCA = ∠DAB (Each 90°) ∠CDA = ∠ADB (common angles) ∴ ΔDCA ~ ΔDAB [AA similarity criterion] ⇒ DC/DA = DA/DA ⇒ AD2 = BD × CD 4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 . Solution: Given, ΔABC is an isosceles triangle right angled at C. Triangles Exercise 6.5 Answer 4 In ΔACB, ∠C = 90° AC = BC (By isosceles triangle property) AB2 = AC2 + BC2 [By Pythagoras theorem] = AC2 + AC2 [Since, AC = BC] AB2 = 2AC2 5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle. Solution: Given, ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2 Triangles Exercise 6.5 Answer 5 In ΔACB, AC = BC AB2 = 2AC2 AB2 = AC+ AC2 = AC2 + BC[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. Triangles Exercise 6.5 Answer 6 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, AB2 = AD+ BD2 (2a)2 = ADa ⇒ AD2 = 4a2 - a2 ⇒ AD2 = 3a2 ⇒ 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. Triangles Exercise 6.5 Answer 7 We have to prove, as per the question, AB+ BC+ CD2 + AD= AC+ BD2 Since, the diagonals of a rhombus bisect each other at right angles. Therefore, AO = CO and BO = DO In ΔAOB, ∠AOB = 90° AB2 = AO+ BO…………………….. (i) [By Pythagoras theorem] Similarly, AD2 = AO+ DO…………………….. (ii) DC2 = DO+ CO…………………….. (iii) BC2 = CO+ BO…………………….. (iv) Adding equations (i) + (ii) + (iii) + (iv), we get, AB+ AD+ DC+ BC2 = 2(AO+ BO+ DO+ CO2) = 4AO+ 4BO[Since, AO = CO and BO =DO] = (2AO)+ (2BO)2 = AC+ BD2 AB+ AD+ DC+ BC2 = AC+ BD2 Hence, proved. 8. In Fig. 6.54, O is a point in the interior of a triangle. Ncert solutions class 10 chapter 6-47 ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that: (i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 , (ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2. 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 Triangles Exercise 6.5 Answer 8 (i) By Pythagoras theorem in ΔAOF, we have OA2 = OF2 + AF2 Similarly, in ΔBOD OB2 = OD2 + BD2 Similarly, in ΔCOE OC2 = OE2 + EC2 Adding these equations, OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2 OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2. (ii) AF2 + BD2 + EC2 = (OA2 - OE2) + (OC2 - OD2) + (OB2 - OF2) ∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2. 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. Triangles Exercise 6.5 Answer 9 Let BA be the wall and AC be the ladder, Therefore, by Pythagoras theorem, AC2 = AB2 + BC2 102 = 82 + BC2 BC= 100 - 64 BC= 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. Triangles Exercise 6.5 Answer 10 Let AB be the pole and AC be the wire. By Pythagoras theorem, AC2 = AB2 + BC2 242 = 182 + BC2 BC= 576 - 324 BC= 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 Ncert solutions class 10 chapter 6-51  hours? Solution: Given, Speed of first aeroplane = 1000 km/hr Distance covered by first aeroplane flying due north in Ncert solutions class 10 chapter 6-52  hours (OA) = 1000 × 3/2 km = 1500 km Speed of second aeroplane = 1200 km/hr Distance covered by second aeroplane flying due west in Ncert solutions class 10 chapter 6-53  hours (OB) = 1200 × 3/2 km = 1800 km Triangles Exercise 6.5 Answer 11 In right angle ΔAOB, by Pythagoras Theorem, AB2 = AO2 + OB2 ⇒ AB2 = (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. Triangles Exercise 6.5 Answer 12 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, AP2 = PC2 + AC2 (12m)2 + (5m)2 = (AC)2 AC2 = (144+25) m2 = 169 m2 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 AE2 + BD2 = AB2 + DE2. Solution: Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Triangles Exercise 6.5 Answer 13 By Pythagoras theorem in ΔACE, we get AC2 + CE2 = AE2 .................................................(i) In ΔBCD, by Pythagoras theorem, we get BC2 + CD2 = BD2 ………………………………..(ii) From equations (i) and (ii), we get, AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii) In ΔCDE, by Pythagoras theorem, we get DE2 = CD2 + CE2 In ΔABC, by Pythagoras theorem, we get AB2 = AC2 + CB2 Putting the above two values in equation (iii), we get DE2 + AB2 = AE2 + BD2. 14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2. Ncert solutions class 10 chapter 6-57 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, AB2 = AD2 + BD2 ……………………....(i) AC2 = AD2 + DC2 ...................................(ii) Subtracting equation (ii) from equation (i), we get AB2 - AC2 = BD2 - DC2 = 9CD2 - CD2 [Since, BD = 3CD] = 8CD2 = 8(BC/4)[Since, BC = DB + CD = 3CD + CD = 4CD] Therefore, AB2 - AC2 = BC2/2 ⇒ 2(AB2 - AC2) = BC2 ⇒ 2AB2 - 2AC2 = BC2 ∴ 2AB2 = 2AC2 + BC2. 15.  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2. Solution: Given, ABC is an equilateral triangle. And D is a point on side BC such that BD = 1/3BC Triangles Exercise 6.5 Answer 15 Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC. ∴ BE = EC = BC/2 = a/2 And, AE = a√3/2 Given, BD = 1/3BC ∴ BD = a/3 DE = BE - BD = a/2 - a/3 = a/6 In ΔADE, by Pythagoras theorem, AD2 = AE2 + DE https://3.bp.blogspot.com/-GOGZB536EAw/VU7p43c1KbI/AAAAAAAAFe4/d1fu3QYmaRc/s1600/equation-5.PNG ⇒ 9 AD2 = 7 AB2 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. Solution: Given, an equilateral triangle say ABC, Triangles Exercise 6.5 Answer 16 Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC. ∴ BE = EC = BC/2 = a/2 In ΔABE, by Pythagoras Theorem, we get AB2 = AE2 + BE2 https://1.bp.blogspot.com/-9R825aQHYTo/VU7xTBhBVII/AAAAAAAAFfU/CQ6kaVAl580/s1600/equation-6.PNG 4AE2 = 3a2 ⇒ 4 × (Square of altitude) = 3 × (Square of one side) Hence, proved. 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. Triangles Exercise 6.5 Answer 17 We can observe that, AB2 = 108 AC2 = 144 And, BC2 = 36 AB2 + BC2 = AC2 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).

Exercise 6.6 Page: 152

1. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR Ncert solutions class 10 chapter 6-63 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) Ncert solutions class 10 chapter 6-64 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 that: (i) DM2 = DN . MC (ii) DN2 = DM . AN. Ncert solutions class 10 chapter 6-65 Solution:
  1. Let us join Point D and B.
Ncert solutions class 10 chapter 6-66 Given, BD ⊥AC, DM ⊥ BC and DN ⊥ AB Now from the figure we have, DN || CB, DM || AB and ∠B = 90 ° Therefore, DMBN is a rectangle. So, DN = MB and DM = NB The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC. ∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i) In ∆CDM, ∠1 + ∠2 + ∠DMC = 180° ⇒ ∠1 + ∠2 = 90° …………………………………….. (ii) In ∆DMB, ∠3 + ∠DMB + ∠4 = 180° ⇒ ∠3 + ∠4 = 90° …………………………………….. (iii) From equation (i) and (ii), we get ∠1 = ∠3 From equation (i) and (iii), we get ∠2 = ∠4 In ∆DCM and ∆BDM, ∠1 = ∠3 (Already Proved) ∠2 = ∠4 (Already Proved) ∴ ∆DCM ∼ ∆BDM (AA similarity criterion) BM/DM = DM/MC DN/DM = DM/MC (BM = DN) ⇒ DM2 = DN × MC Hence, proved. (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 ⇒ DN2 = AN × NB ⇒ DN2 = AN × DM (Since, NB = DM) Hence, proved. 3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2= AB2+ BC2+ 2 BC.BD. Ncert solutions class 10 chapter 6-67 Solution: By applying Pythagoras Theorem in ∆ADB, we get, AB2 = AD2 + DB2 ……………………… (i) Again, by applying Pythagoras Theorem in ∆ACD, we get, AC2 = AD2 + DC2 AC2 = AD2 + (DB + BC) 2 AC2 = AD2 + DB2 + BC2 + 2DB × BC From equation (i), we can write, AC2 = AB2 + BC2 + 2DB × BC Hence, proved. 4. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2= AB2+ BC2 – 2 BC.BD. Ncert solutions class 10 chapter 6-68 Solution: By applying Pythagoras Theorem in ∆ADB, we get, AB2 = AD2 + DB2 We can write it as; ⇒ AD2 = AB2 − DB2 ……………….. (i) By applying Pythagoras Theorem in ∆ADC, we get, AD2 + DC2 = AC2 From equation (i), AB2 − BD2 + DC2 = AC2 AB2 − BD2 + (BC − BD) 2 = AC2 AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD AC2 = AB2 + BC2 − 2BC × BD Hence, proved. 5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that : (i) AC2 = AD2 + BC.DM + 2 (BC/2) 2 (ii) AB2 = AD2 – BC.DM + 2 (BC/2) 2 (iii) AC2 + AB2 = 2 AD2 + ½ BC2 Ncert solutions class 10 chapter 6-69 Solution: (i) By applying Pythagoras Theorem in ∆AMD, we get, AM2 + MD2 = AD2 ………………. (i) Again, by applying Pythagoras Theorem in ∆AMC, we get, AM2 + MC2 = AC2 AM2 + (MD + DC) 2 = AC2 (AM2 + MD2 ) + DC2 + 2MD.DC = AC2 From equation(i), we get, AD2 + DC2 + 2MD.DC = AC2 Since, DC=BC/2, thus, we get, AD2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC2 AD2 + (BC/2) 2 + 2MD × BC = AC2 Hence, proved. (ii) By applying Pythagoras Theorem in ∆ABM, we get; AB2 = AM2 + MB2 = (AD2 − DM2) + MB2 = (AD2 − DM2) + (BD − MD) 2 = AD2 − DM2 + BD2 + MD2 − 2BD × MD = AD2 + BD2 − 2BD × MD = AD2 + (BC/2)2 - 2(BC/2) MD = AD2 + (BC/2)2 - BC MD Hence, proved. (iii) By applying Pythagoras Theorem in ∆ABM, we get, AM2 + MB2 = AB2 ………………….… (i) By applying Pythagoras Theorem in ∆AMC, we get, AM2 + MC2 = AC2 …………………..… (ii) Adding both the equations (i) and (ii), we get, 2AM2 + MB2 + MC2 = AB2 + AC2 2AM2 + (BD − DM) 2 + (MD + DC) 2 = AB2 + AC2 2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2 2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2 2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB2 + AC2 2AD2 + BC2/2 = AB2 + AC2 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. Ncert solutions class 10 chapter 6-70 By applying Pythagoras Theorem in ∆DEA, we get, DE2 + EA2 = DA2 ……………….… (i) By applying Pythagoras Theorem in ∆DEB, we get, DE2 + EB2 = DB2 DE2 + (EA + AB) 2 = DB2 (DE2 + EA2) + AB2 + 2EA × AB = DB2 DA2 + AB2 + 2EA × AB = DB2 ……………. (ii) By applying Pythagoras Theorem in ∆ADF, we get, AD2 = AF2 + FD2 Again, applying Pythagoras theorem in ∆AFC, we get, AC2 = AF2 + FC2 = AF2 + (DC − FD) 2 = AF2 + DC2 + FD2 − 2DC × FD = (AF2 + FD2) + DC2 − 2DC × FD AC2 AC2= AD2 + DC2 − 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, DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2 DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2 From equation (iv) and (vi), BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2 AB2 + BC2 + CD2 + DA2 = AC2 + BD2 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 Ncert solutions class 10 chapter 6-71 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/DP = PC/PB ∴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. Ncert solutions class 10 chapter 6-72 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 BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC. Ncert solutions class 10 chapter 6-73 Solution: In the given figure, let us extend BA to P such that; AP = AC. Now join PC. Ncert solutions class 10 chapter 6-74 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? Ncert solutions class 10 chapter 6-75 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. Ncert solutions class 10 chapter 6-76 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 Ncert solutions class 10 chapter 6-77 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, Triangles, is part of the Unit Geometry which constitutes 15 marks of the total marks of 80. On the basis of the updated CBSE Class 10 Syllabus for 2021-22, this chapter belongs to the Unit-Geometry under the first term and has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and problem-solving methods in this chapter is mandatory to score well in the first term 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 NCERT 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 figures, 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 Chapter 6 of the Class 10 Maths NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the both first and second examinations.

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.
Also Access 
NCERT Exemplar for Class 10 Maths Chapter 6 Triangle 
CBSE Notes for Class 10 Maths Chapter 6 Triangle
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Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 6
Why should we learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6?
The concepts covered in Chapter 6 of NCERT Solution Maths provide questions based on the exam pattern and model question paper. So it is necessary to learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6.
List out the important topics present in NCERT Solutions for Class 10 Maths Chapter 6?
The topics covered in the chapters are Introduction to the triangles, similar figures, similarity of triangles, criteria for similarity of triangles, areas of similar triangles and Pythagoras Theorem. These concepts are important from an exam perspective. It is strictly based on the latest syllabus of the CBSE for 2021-22, and also depends on the CBSE question paper design and marking scheme.
How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 6?
There are 6 exercises are there in NCERT Solutions for Class 10 Maths Chapter 6. The first exercise contains 3 questions, the second exercise contains 10 questions, the third exercise has 16 questions, the fourth exercise has 9 questions, the fifth exercise has 17 questions and the last or sixth exercise has ten questions based on the concepts of triangles.
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