RS Aggarwal Solutions Class 10 Chapter 4 Triangles


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RS Aggarwal Solutions Class 10 Chapter 4 Triangles





Exercise 4A





Question 1





Solution:





From given triangle, points D and E are on the sides AB and AC respectively such that DE || BC.





(i)AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm.





By Thale’s Theorem:





AD/DB = AE/EC





Here DB = AB – AD = 10 – 3.6 = 6.4





=> EC = 4.6/3.6 x 6.4





or EC = 8





And, AC = AE + EC





AC = 4.5 + 8 = 12.5





(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.





By Thale’s Theorem:





AD/DB = AE/EC





Add 1 on both sides





AD/DB + 1 = AE/EC + 1





(AD + DB)/DB = (AE + EC)/EC





AB/DB = AC/EC





or DB = (ABxEC)/ AC





= (13.3 x 5.1)/11.90





= 5.7





=> BD = 5.7





And, AD = AB – DB





AD = 13.3 – 5.7





AD = 7.6 cm





(iii) AD/DB = 4/7 or AD = 4 cm, DB = 7 cm, and AC = 6.6





By Thale’s Theorem:





AD/DB = AE/EC





Add 1 on both sides





AD/DB + 1 = AE/EC + 1





(AD + DB)/DB = (AE + EC)/EC





(4+7)/7 = AC/EC = 6.6/EC





EC = (6.6 x 7)/11





= 4.2





And, AE = AC – EC





AE = 6.6 – 4.2





AE = 2.4 cm





(iv)





AD/AB = 8/15 or AD = 8 cm, AB = 15 cm, and EC = 3.5 cm





By Thale’s Theorem:





AD/AB = AE/AC





8/15 = AE/(AE+EC) = AE/(AE+3.5)





8(AE + 3.5) = 15AE





7 AE = 28





or AE = 4 cm





Question 2





Solution:





From figure, D and E are the points on the sides AB and AC respectively and DE || BC





then AD/DB = AE/EC





(i) AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.





x/(x-2) = (x+2)/(x-1)





x(x-1) = (x+2)(x-2)





Solving above equation, we get





x = 4 cm





(ii) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.





AD/DB = AE/EC





4/(x-4) = 8/(3x-19)





4(3x-19) = 8(x-4)





Solving, we get x = 11 cm





(iii)





AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) and EC = 3x cm.





AD/DB = AE/EC





(7x-4)/(3x + 4) = (5x – 2)/3x





(7x – 4) (3x) = (5x – 2) (3x + 4)





21x² – 12x – 15x² – 20x + 6x = -8





6x² – 26x + 8 = 0





(x – 4) (3x – 1) = 0





Either x – 4 = 0 or (3x – 1) = 0





=> x = 4 or 1/3 (not possible)





So, x = 4





Question 3:





Solution:





From figure, D and E are the points on the sides AB and AC of ∆ABC





(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm





AD/DB = 5.7/9.8 = 3/5





and AE/EC = 4.8/8 = 3/5





=> AD/DB = AE/EC





=> DE || BC





(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm and AE = 4.2 cm.





AD = AB – BD = 11.7 – 6.5 = 5.2 cm and





EC = AC – AE = 11.2 – 4.2 = 7 cm





AD/DB = 5.2/7 = 4/5





AE/EC = 4.2/7 = 3/5





=> AD/DB ≠ AE/EC





=> DE is not parallel to BC





(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm and EC = 4 cm.





DB = AB – AD = 10.8 – 6.3 = 4.5 cm





AE =AC – EC = 9.6 – 4 = 5.6 cm





AD /DB = 6.3/4.5 = 7/5





AE/EC = 5.6/4 = 7/5





=> AD/DB = AE/EC





=> DE || BC





(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm.





DB = AB – AD = 12 – 7.2 = 4.8 cm and





EC = AC – AE = 10 – 6,4 =3.6 cm





AD /DB = 7.2/4.8 = 3/2 and





AE/EC = 6.4/3.6 = 16/9





=> AD/DB ≠ AE/EC





=> DE is not parallel to BC





Question 4:





Solution:





(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.





AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC,





BD/DC = AB/AC





Substituting given values, we get





5.6/DC = 6.4/8





DC = 7 cm





(ii) If AB = 10 cm, AC = 14 cm and BC – 6 cm, find BD and DC.





By angle-bisector theorem





BD/DC = AB/AC = 10/14





Let BD = x cm and DC = (6-x) (As BC = 6 cm given)





x/(6-x) = 10/14





14x = 10(6 – x)





14x = 60 – 10x





14x + 10x = 60





or x = 2.5





Or BD = 2.5





Then DC = 6 – 2.5 = 3.5 cm





(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.





BD/DC = AB/AC





Here DC = BC – BD = 6 – 3.2 = 2.8





=> DC = 2.8





3.2/2.8 = 5.6/AC





=> AC = 4.9 cm





(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.





BD/DC = AB/AC





BD/3 = 5.6/4





=> BD = 4.2





Now, BC = BD + DC = 4.2 + 3 = 7.2





BC is 7.2 cm.





Question 5:





Solution:





M is a point on the side BC of a parallelogram ABCD





(i)Consdier ∆DMC and ∆NMB,





∠DCM = ∠NBMalternate angles
∠DMC = ∠NMBvertically opposite angles
∠CDM = ∠MNBalternate angles




By By AAA-similarity:





∆DMC ∼ ∆NMB





From similarity of the triangle:





DM/MN = DC/BN





(ii)





From (i), DM/MN = DC/BN





Add 1 on both sides





DM/MN + 1 = DC/BN + 1





(DM+MN)/MN = (DC+BN)/BN





Since AB = CD





(DM+MN)/MN = (AB+BN)/BN





DN/DM = AN/DC





Hence proved.





Question 6:





Solution:





 RS Aggarwal Solutions Class 10 Maths Chapter 7 Exercise 7A-question-6




Here, AB || DC





M and N are the mid points of sides AD and BC respectively.





MN is joined.





To prove : MN || AB or DC.





Produce AD and BC to meet at P.





Now, In ∆PAB





DC ||AB





PD/DA = PC/CB





PD/2PM = PC/2CN





M and N are midpoints of AD and BC respectively.





PD/PM = PC/CN





MN||DC But DC ||AB





Therefore, MN || DC ||AB





Question 7:





Solution:





From given statement:





In Δ ADC





EO || AB || DC





By thales theorem: AE/ED = AO/OC …(1)





In Δ DAB,





EO || AB





So, By thales theorem: DE/EA = DO/OB …(2)





From (1) and (2)





AO/OC = DO/OB





(5x – 7) / (2x + 1) = (7x-5) / (7x+1)





(5x – 7)(7x + 1) = (7x – 5)(2x + 1)





35x^2 + 5x – 49x – 7 = 14x^2 – 10x + 7x – 5





35x^2 – 14x^2 – 44x + 3x – 7 + 5 = 0





21x^2 – 42x + x – 2 = 0





21(x – 2) + (x – 2) = 0





(21x + 1)(x – 2) = 0





Either (21x + 1) = 0 or (x – 2) = 0





x = -1/21 (does not satisfy) or x = 2





=> x = 2.





Question 8:





Solution:





In ∆ABC, M and N are points on the sides AB and AC respectively such that BM = CN and if ∠B = ∠C.





We know that, sides opposite to equal angles are equal.





AB = AC





BM = CN ( given)





AB – BM = AC – CN





=> AM = An





From ∆ABC





AM/MB = AN/NC





Therefore, MN ||BN





Exercise 4B





Question 1:





Solution:





Two triangles are similarity of their corresponding angles are equal and corresponding sides are proportional.





(i) In ∆ABC and ∆PQR





∠A = ∠Q = 50°





∠B = ∠P = 60° and





∠C = ∠R = 70°





∆ABC ~ ∆QPR (By AAA)





(ii) In ∆ABC and ∆DEF





In ∆ABC and ∆DEF





AB = 3 cm, BC = 4.5





DF = 6 cm, DE = 9 cm





∆ABC is not similar to ∆DEF





(iii) In ∆ABC and ∆PQR





In ∆ABC and ∆PQR





AC = 8 cm BC = 6 cm





Included ∠C = 80°





PQ = 4.5 cm, QR = 6 cm





and included ∠Q = 80°





AC/QR = 8/6 = 4/3





and BC/PQ = 6/4.5 = 4/3





=> AC/QR = BC/PQ





and ∠C = ∠Q = 80°





∆ABC ~ ∆PQR (By SAS)





(iv)In ∆DEF and ∆PQR





DE = 2.5, DF = 3 and EF = 2





PQ = 4, PR = 6 and QR = 5





DE/QR = 2.5/5 = 1/2





DF/PR = 3/6 = 1/2





and EF/PQ = 2/4 = 1/2





∆DEF ~ ∆PQR (By SSS)





(v) In ∆ABC and ∆MNR





∠A = 80°, ∠C = 70°





So, ∠B = 180° – (80° + 70°) = 30°





∠M = 80°, ∠N = 30°, and ∠R = 180° – (80° + 30°) = 70°





Now, in ∆ABC





∠A = ∠M – 80°, ∠B = ∠N = 30°





and ∠C = ∠R = 70°





∆ABC ~ ∆MNR (By AAA or AA)





Question 2:





Solution:





Here ∆ODC ~ ∆OBA, so





∠D = ∠B = 70°





∠C = ∠A





∠COD = ∠AOB





(i) But ∠DOC + ∠BOC = 180° (Linear pair)





∠DOC + 115°= 180°





∠DOC = 180° – 115° = 65°





(ii) ∠DOC + ∠CDO + ∠DCO = 180° (Angles of a triangle)





65° + 70° + ∠DCO = 180°





135° + ∠DCO = 180°





∠DCO = 180° – 135° = 45°





(iii) ∠AOB = ∠DOC = 65° (vertically opposite angles)





∠OAB = ∠DCO = 45° (Since ∆ODC ~ ∆OBA)





(iv) ∠OBA = ∠CDO = 70° (Since ∆ODC ~ ∆OBA)





Question 3:





Solution:





Since ∆OAB ~ ∆OCD





AB = 8 cm, BO = 6.4 cm OC = 3.5 cm, CD = 5 cm





Let OD = y and OA = x





rs aggarwal class 10 maths exercise 7b question 6




Question 4:
Solution:





From given figure,





∠ADE = ∠B





To prove:





∆ADE ~ ∆ABC and





find DE





Given: AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm





Now, In ∆ADE and ∆ABC





∠ADE = ∠B (given)





∠A = ∠A (common)





∆ADE ~ ∆ABC (By AA)





Again,





AD/AB = DE/BC





rs aggarwal class 10 maths exercise 7b question 8




Question 5:
Solution:





Form given statement: ∆ABC ~ ∆PQR,





PQ = 12 cm





Perimeter of ∆ABC = AB + BC + CA = 32 cm





Perimeter of ∆PQR = PQ + QR + RP = 24 cm





Now,





rs aggarwal class 10 maths exercise 7b question 9




Question 6:
Solution:





Form given statement: ∆ABC ~ ∆DEF





BC = 9.1 cm, EF = 6.5 cm and Perimeter of ∆DEF = 25 cm





rs aggarwal class 10 maths exercise 7b question 10




Perimeter of ∆ABC is 35 cm





Question 7:
Solution:





∠CAB = 90° and AD ⊥ BC





If AC = 75 cm, AB = 1 m or 100 cm, BC = 1.25 m or 125 cm





∠BDA = ∠BAC = 90°





∠DBA = ∠CBA [common angles]





By AA





∆BDA ~ ∆BAC





And,





rs aggarwal class 10 maths exercise 7b question 12




Question 8:
Solution:





From given:





∠ABC = 90°, BD ⊥ AC.





AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm





In ∆ABC and ∆BDC,





∠ABC = ∠BDC (each 90°)





∠BCA = ∠BCD (common angles)





∆ABC ~ ∆BDC (AA axiom)





So, corresponding sides are proportional





AB/BD = BC/CD





=> 5.7/3.8 = BC/5.4





=> BC = 8.1





Exercise 4C





Question 1:





Solution:





Area of ∆ABC = 64 cm² and





area of ∆DEF = 121 cm²





EF = 15.4 cm





rs aggarwal class 10 maths exercise 7c-1




Question 2:
Solution:





The areas of two similar triangles ABC and PQR are in the ratio 9 : 16.





BC = 4.5 cm





rs aggarwal class 10 maths exercise 7c-2




Question 3:
Solution:





∆ABC ~ ∆PQR





ar (∆ABC) = 4ar (∆PQR)





rs aggarwal class 10 maths exercise 7c-3




Question 4:
Solution:





Areas of two similar triangles are 169 cm² and 121 cm² (given)





Longest side of largest triangle = 26 cm





Let longest side of smallest triangle is x cm





rs aggarwal class 10 maths exercise 7c-4




Longest side of smallest triangle is 22 cm





Question 5:
Solution:





Area of ∆ABC = 100 cm²





area of ∆DEF = 49 cm²





Altitude of ∆ABC is 5 cm





rs aggarwal class 10 maths exercise 7c-5




AL ⊥ BC and DM ⊥ EF





Let DM = x cm





rs aggarwal class 10 maths exercise 7c-6




Or x = 3.5





Altitude of smaller triangle is 3.5 cm





Question 6:
Solution:





Corresponding altitudes of two similar triangles are 6 cm and 9 cm (given)





We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.





Ratio in the areas of two similar triangles = (6)² : (9)² = 36 : 81 = 4 : 9





Question 7:
Solution:





Areas of two similar triangles are 81 cm² and 49 cm²





Altitude of the first triangle = 6.3 cm





Let altitude of second triangle = x cm





Area of ∆ABC = 81 cm² and area of ∆DEF = 49cm²





Altitude AL = 6 – 3 cm





Let altitude DM = x cm





rs aggarwal class 10 maths exercise 7c-7




Altitude of second triangle is 4.9 cm





Question 8:
Solution:





Areas of two similar triangles are 100 cm² and 64 cm²





Median DM of ∆DEF = 5.6 cm





Let median AL of ∆ABC = x





rs aggarwal class 10 maths exercise 7c-8




Corresponding median of the other triangle is 7 cm.





Question 9:
Solution:





In ∆ABC, PQ is a line which meets AB in P and AC in Q.





Given: AP = 1 cm, PB = 3 cm, AQ = 1.5 cm QC = 4.5 cm





Now, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3





=> AP/PB = AQ/QC





From figure: AB = AP + PB = 1+3 = 4 cm





AC = AQ + QC = 1.5 + 4.5 = 6 cm





In ∆APQ and ∆ABC,





AP/AB = AQ/AC





angle A (common)





∆APQ and ∆ABC are similar triangles.





Now,





rs aggarwal class 10 maths exercise 7c-10




Which implies,





area of ∆APQ = 1/16 of the area of ∆ABC





Hence Proved.





Question 10:
Solution:





DE || BC





DE = 3 cm, BC = 6 cm





area (∆ADE) = 15 cm²





Now,





In ∆ABC





DE ||BC. Therefore triangles, ∆ADE and ∆ABC are similar.





rs aggarwal class 10 maths exercise 7c-12




Area of ∆ABC is 60 cm2.





Exercise 4D





Question 1:
Solution:





A given triangle to be right-angled, if it satisfies Pythagorean Theorem. That is, the sum of the squares of the two smaller sides must be equal to the square of the largest side.





(i) 9 cm, 16 cm, 18 cm





Longest side = 18





Now (18)² = 324





and (9)² + (16)² = 81 + 256 = 337





324 ≠ 337





It is not a right triangle.





(ii) 1 cm, 24 cm, 25 cm





Longest side = 25 cm





(25)² = 625





and (7)² x (24)² = 49 + 576 = 625





625 = 625





It is a right triangle.





(iii) 1.4 cm, 4.8 cm, 5 cm





Longest side = 5 cm





(5)² = 25





and (1.4)² + (4.8)² = 1.96 + 23.04 = 25.00 = 25





25 = 25





It is a right triangle.





(iv) 1.6 cm, 3.8 cm, 4 cm





Longest side = 4 cm





(4 )² = 16





and (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00 = 17





16 ≠ 17





It is not a right triangle.





(v) (a- 1) cm, 2√a cm, (a + 1) cm





Longest side = (a + 1) cm





(a + 1)² = a² + 2a + 1





and (a – 1)² + (2 √a )² = a² – 2a + 1 + 4a = a² + 2a + 1





a² + 2a + 1 = a² + 2a + 1





It is a right triangle.





Question 2:
Solution:





A man goes 80 m from O to east side and reaches A, then he goes 150 m due north from A and reaches B.





Draw a figure based on given instructions:





rs aggarwal class 10 maths exercise 7d-1




From right ∆OAB,





By Pythagoras Theorem:





OB2 = OA2+ AB2





= (80)² + (150)²





= 6400 + 22500





= 28900





or OB = √28900 = 170





Man is 170 m away from the starting point.





Question 3:





Solution:





A man goes 10 m due south from O and reaches A and then 24 m due west from A and reaches B.





Draw a figure based on given instructions:





rs aggarwal class 10 maths exercise 7d-2




From right ∆OAB,





By Pythagoras Theorem:





OB2 = OA2+ AB2





= (10)² + (24)²





= 676





or OB = 26





Man is 26 m away from the starting point.





Question 4:
Solution:





Height of the window = 12 m





Length of a ladder = 13 m





In the figures,





rs aggarwal class 10 maths exercise 7d-3




Let AB is ladder, A is window of building AC





By Pythagoras Theorem:





AB2 = AC2 + BC2





(13)² = (12)² + x²





169 = 144 + x²





x² = 169 – 144 = 25





or x = 5





Distance between foot of ladder and building = 5 m.





Question 5:
Solution:





Height of window AC = 20 m





Let length of ladder AB = x m





Distance between the foot of the ladder and the building (BC) = 15 m





In the figure:





rs aggarwal class 10 maths exercise 7d-4




By Pythagoras Theorem:





AB2 = AC2 + BC2





x² = 20² + 15²





= 400 + 225





= 625





or x = 25





Length of ladder is 25 m





Question 6:
Solution:





Height of first pole AB = 9 m and





Height of second pole CD = 14 m





Let distance between their tops CA = x m





rs aggarwal class 10 maths exercise 7d-5




From A, draw AE || BD meeting CD at E.





Then EA = DB = 12 m CE = CD – ED = CD – AB = 14-9 = 5 m





In right ∆AEC,





AC² = AE² + CE²





= 122 + 52





= 144 + 25





= 169





or AC = 13





Distance between pole’s tops is 13 m





Question 7:





Solution:





Length of wire = AC = 24 m





Height of the pole = AB = 18 m





Let Distance between the base of the pole and other end of the wire = BC = x m





rs aggarwal class 10 maths exercise 7d-6




In right ∆ABC,





By Pythagoras Theorem:





AC2 = AB2 + BC2





(24)² = (18)² + x²





576 = 324 + x²





x² = 576 – 324 = 252





or x = 6v7





BC is 6v7m





Question 8:
Solution:





In ∆PQR, O is a point in it such that





OP = 6 cm, OR = 8 cm and ∠POR = 90°





PQ = 24 cm, QR = 26 cm





Now,





In ∆POR, ∠O = 90°





PR² = PO² + OR²





= (6)² + (8)²





= 36 + 64





= 100





PR = 10





Greatest side QR is 26 cm





QR² = (26)² = 676





and PQ² + PR² = (24)² + (10)²





= 576 + 100





= 676





Which implies, 676 = 676





QR² = PQ² + PR²





∆PQR is a right angled triangle and right angle at P.





Question 9:
Solution:





In isosceles ∆ABC, AB = AC = 13 cm





Consider AL is altitude from A to BC and AL = 5 cm





Now, in right ∆ALB





AB2 = AL2 + BL2





(13)² = (5)² + BL²





169 = 25 + BL²





BL² = 169 – 25 = 144





or BL = 12





Since L is midpoint of BC, then





BC = 2 x BC = 2 x 12 = 24





BC is 24 cm





Question 10:





Solution:





In an isosceles ∆ABC in which AB = AC = 2a units, BC = a units





AD is the altitude. Therefore, D is the midpoint of BC





=> BD = a/2





We have two right triangles: ΔADB and ΔADC





By Pythagoras theorem,





AB2 = BD2 + AD2





(2a)2 = (a/2)2 + AD2





rs aggarwal class 10 maths exercise 7d-8




Exercise 4E





Question 1:





Solution:





Two properties for similarity of two triangles are:





(i) Angle-Angle-Angle (AAA) property.





(ii) Angle-Side-Angle (ASA) property.





Question 2:
Solution:





In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally.





Question 3:
Solution:





If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side.





Question 4:
Solution:





The line joining the midpoints of two sides of a triangle, is parallel to the third side.





Question 5:
Solution:





In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar.





Question 6:
Solution:





In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.





Question 7:
Solution:





In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar.





Question 8:
Solution:





In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.





Question 9:
Solution:





In a right angled triangle, the square on the hypotenuse is equal to the sum of squares on the other two sides.





Question 10:
Solution:





In a triangle, if the square on the longest side is equal to the sum of the squares on the other two sides, then the angle opposite to the hypotenuse is a right angle.





Question 11:
Solution:





The ratio of their areas will be 1 : 4.





Question 12:
Solution:





In two triangles ∆ABC and ∆PQR,





AB = 3 cm, AC = 6 cm, ∠A = 70°





PR = 9 cm, ∠P = 70° and PQ= 4.5 cm





Now,





∠A = ∠P = 70° (Same)





AC/PR = 6/9 = 2/3 and





AB/PQ = 3/4.5 = 2/3





=> AC/PR = AB/PQ





Both ∆ABC and ∆PQR are similar.





Question 13
Solution:





∆ABC ~ ∆DEF (given)





2AB = DE, BC = 6 cm (given)





∠E = ∠B and ∠D = ∠A and ∠F = ∠C





2AB = DE





=> AB/DE = 1/2





Therefore,





AB/DE = BC/EF





1/2 = 6/EF





or EF = 12 cm





Question 14:
Solution:





From figure:





DE || BC





AD = x cm, DB = (3x + 4) cm





AE = (x + 3) cm and EC = (3x + 19) cm





In ∆ABC





AD/DB = AE/EC





x/(3x+4) = (x+3)/(3x+19)





3x2 + 19x – 3x2– 9x – 4x = 12





x = 2





Question 15:
Solution:





Let AB is the ladder and A is window.





Then, AB = 10 m and AC = 8 m





Let BC = x





In right ∆ABC,





By Pythagoras Theorem:





AB2 = AC2 + BC2





(10)² = 8² + x²





100 = 64 + x²





x² = 100 – 64 = 36





or x = 6





Therefore, Distance between foot of ladder and base of the wall is 6 m.





Complete RS Aggarwal Solutions Class 10





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