In ∆ADE and ∆ABC (2012) In ∆ABC and ∆DEF, Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 3. 4MC2 = 4CA2 + BA2 ∆BDA, ∠1 + ∠5 = 90° Using the above, do the following: ∴ ∆ADB ~ ∆CEB …[AA similarity … [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides To Find. ∠1 = ∠1 … [Common Show that area of triangle on the hypotenuse is the sum of areas of the other two triangles? Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE. In the given figure, ABCD is a rectangle. Given. 45° + 78° + ∠R = 180° Given: BL and CM are medians of ∆ABC, right angled at A. \(\frac{A B}{3}=\frac{12}{D E}=\frac{3+2}{A E}\) OA = OC = \(\frac{A C}{2} \Rightarrow \frac{24}{2}\) = 12 cm State whether the given pairs of triangles are similar or not. ∆ to the hypotenuse then As on both sides of the ⊥ are similar to the whole D and to each other Find the length BN. ∴ ∆ABC ~ ∆XBY …[AA similarity criterion, Question 30. Proof. (x – 15)(x + 20) = 0 DB and AE intersect at F. Show that DF = 4FB and AF = 4FE ∴ AC ⊥ BD, Prove that AD\frac{B D}{A D}=\frac{A D}{C D} = BD × DC. AD2 = (2) (8) = 16 ⇒ AD = 4 cm, Question 14. In figure, AABC is right angled at C and DE ⊥ AB. (2012; 2017D) BC in 2 : 1. If Y, Z , U and Y are the middle points of AX, XC, AB and BC respectively, then prove that UY || VZ and UV|| YZ. Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Solution: Question 53. Introduction to Trigonometry Class 10 Extra Questions Short Answer Type 2. The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{A O}{B O}=\frac{C O}{D O}\). DP || BC ∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2 State and prove Converse of Pythagoras’ Theorem. AE = \(\frac{40}{2}\) = 20 m Solution: One side of ∆ABC = 20 cm Solution: Question 49. Base cannot be -ve (2014) 4(BL2 + MC2) = 5BA2 + 5CA2 ∆ABC, Then altitude, BC = (x + 5) cm Find the ratio of the areas of ∆PST and ∆PQR. From (i) and (ii), we get Const. In the figure ABC and DBC are two right triangles. On AD as base, another equilateral triangle ADE is constructed. 4BL2 + 4MC2 = 4BA2 + CA2 + 4CA2+ BA2 …[From (ii) & (iii) ∠A + ∠2 = 90° (2013; 2017OD) AB2 + BC2 = AC2 ∆ABC – ∆DEF …[Given, Question 3. ∠6 = ∠4 … [corresponding angles AD and CF are two medians drawn from A and C respectively. NCERT Exemplar Class 10 Maths is very important resource for students preparing for X Board Examination. Solution: Question 37. DE || BC …[Given In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. AB2 + BC2 = (6\(\sqrt{3}\))2 + (6)2 If AB = 26 cm2 then find DE. ⇒ x(x + 20) – 15(x + 20) = 0 Proof. In fig. (isosceles, equilateral) The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2BL. In Figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle ABC of side 14 cm as centres to intersect the sides BC, CA and AB at BZ their respective mid-points D, E and F. Find the area of the shaded region. Solution: Question 63. Solution: Solution: CBSE Chapter Wise Important Questions for Class 10 Science Board Exam 2019-20 Pdf free download was designed by expert teachers from latest edition of NCERT books to get good marks in board exams. ∠ADE – ∠ABC … [Corresponding angles ∴ AB = 3k + 4k = 7k Solution: Question 2. Two sides are = 15 cm and 20 cm, Question 50. Areas Related to Circles Class 10 Important Questions Long Answer (4 Marks) Question 37. x2 + 5x = x2 + 3x + x + 3 AABC, is right angled at C. If p is the length of the perpendicular from C to AB and a, b, c are the lengths of the sides opposite ∠A,∠B, ∠C respectively then prove that Solution: A 6.5 m long ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall. OB = OD = \(\frac{B D}{2} \Rightarrow \frac{32}{2}\) = 16 cm Triangles Class 10 Important Questions Short Answer-I (2 Marks) Question 10. Solution: Question 58. Target Centum in CBSE 10th Maths. Solution: x = 15 or x = -20 (2013) Prove that the sum of square of the sides of a rhombus is equal to the sum of squares of its diagonals. .. ∠5 = ∠4 …(iii) Find BD X CD. ∴ Length of the other side = 15 + 5 = 20 cm ∆ABC ~ ∆DEF. ∠2 = ∠3 … [Corresponding angles Solution: ⇒ \(\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{\mathrm{AP}}{\mathrm{PD}}\) … [Sides are proportional AB = 7k + 5k = 12k Given: Quadrilateral ABCD in which Adding (ii) and (iii), we get Solution: (2013) (iv) AB × CE = BC × AD The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆DEF. MC2 = CA2 + \(\frac{\mathrm{BA}^{2}}{4}\) In the given figure, ABC and DBC are two triangles on the same base BC. ∠AED = ∠GFC … [rt. ∴ PA2 = PB. ∆ADC, AD2 = AC2 – CD2 …(iii) AD2 = AC2 – DC2 …(ii) [Pythagoras’ theorem In ∆BDA and ∆ADC, Solution: Question 69. In ∆ABC, ZB = 90°, BD ⊥ AC, ar (∆ABC) = A and BC = a, then prove that Using the above, prove the following: The areas of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described, on its diagonal. ∴ The corresponding side of ADEF = 28 cm, Question 23. ∠2 = ∠4. Solution: Question 45 From (ii) and (iii), we get Determine which of them are right triangles. Solution: Question 22. Prove that: In ∆OAQ and ∆OBP, Solution: Question 8. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Triangle ABC is right angled at B, and D is mid-point of BC. Download latest questions with multiple choice answers for Class 10 Triangles … Let AF = x cm, then BF = (13 – x) cm Prove that QR X QS= QP X QT In figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6 cm and 8 cm respectively. In ∆ADE and ∆ABC, In figure, AB || PQ || CD, AB = x units, CD =y units and PQ = z units, prove that x – 15 = 0 or x + 20 = 0 CD = 4K A vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a tower casts a shadow 30 m long. Solution: Question 50. Solution: Question 31. Solution: Solution: Question 74. In rt. In the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Is ∆ADB ~ ∆AEB and ∆ADB ~ ∆ADC? In rt. In the figure of ∆ABC, D divides CA in the ratio 4 : 3. Access Answers of Maths NCERT class 10 Chapter 6 – Triangles Exercise 6.1 Page: 122. ∴Required height, AB = 6 m, Question 16. Question 3 : If C and Z are acute angles and that cos C = cos Z prove that ∠C = ∠Z . ∴ \(\frac{Q R}{Q T}=\frac{Q P}{Q S}\) ..[In -∆s corresponding sides are proportional In the figure, ABC is a triangle and BD ⊥ AC. ∴ ar (BCDE) : ar(AABC) = 40 : 49, Question 28. Solution: Question 23. In an isosceles triangle PQR, PQ = QR and PR2 = 2PQ2. ∴ XY || BC … [By converse of Thales’ theorem, Question 13. …[∵ In a rhombus, all sides are equal, Question 34. Using the above, do the following: P is a point on BC such that PQ || BA and PR || BD. If the areas of two similar triangles are equal, then prove that the triangles are congruent. ∠DAE = ∠BAC …Common Proof. Point D is the mid-point of the side BC of a right triangle ABC, right angled at C. Prove that, 4AD2 = 4AC2 + BC2.. ∠5 = ∠7 …[Each 90° To prove: ∠ABC = 90° Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to … (2017D) ∠6 = ∠6 …(Common In figure, S and T trisect the side QR of a right triangle PQR.Prove that 8PT2 = 3PR2 + 5PS2. (2015) ∠2 = ∠ABC … [Corresponding angles, Question 29. Solution: Let us first draw a right ∆ABC in which ∠C = 90°. Given. Proof: E divides Prove that 2CA2 = 2AB2 + BC2. T his CBSE class 10 Maths additional practice question is from the topic Trigonometry. Solution: Question 46. X and Y are points on the sides AB and AC respectively of a triangle ABC such that \(\frac{\mathbf{A X}}{\mathbf{A B}}=\frac{1}{4}\), AY = 2 cm and YC = 6 cm. Proof: In rt. In figure, PQ || AB and AQ || CB. Calculate the length of the altitude of the rhombus. Now in ∆ADE & ∆GCF Solution: ∠2 = ∠3 …[Vertically opp. (2014) Construction: ∠1 = ∠1 …[Common NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12, Question 1. (similar, congruent) Answer: Similar (iii) All _____ triangles are similar. ∴ ∠1 = ∠3 …(CPCT … [In ~∆s, corresponding angles are equal, (b) In ∆ABE and ∆DBC, According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks. In a rectangle ABCD, E is middle point of AD. AD is the median to BC and PM is the median to QR. b2 = h2 + (ED + DC)2 Then prove that. Hence proved. It is therefore important that students have a focused approach to studying the two subjects while preparing for their board exams and important question for CBSE NCERT class 10 with answer in free PDF download is a tool which is sure to help. To help you understand the concept of arithmetic progression, our experts have shared detailed answers for the NCERT textbook questions in our NCERT Solutions for Class 10. Triangles is a crucial geometry chapter that can be quite tricky for most class 10th students. Important Questions for Class 10 Maths Chapter 6 Triangles with solutions includes all the important topics with detailed explanation that aims to help students to score more marks in Board Exams 2020. ∴ OA = OC and From (i) and (ii), ∠C = ∠2 ∴ ∆ABC ~ ∆ADE …[AA Similarity Criterion, (ii) ∴ \(\frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E}\) … [side are proportional ∆ADC, on AC and \(\frac{A D}{D B}=\frac{3}{4}\) Determine whether the triangle having sides (a – 1) cm, 2 √a cm and (a + 1) cm is a right angled triangle. Perimeter(∆DEF) = 70 cm Solution: Question 12. b2 = (p2 – x2) + (x = \(\frac{a}{2}\))2 (iii) In ∆ADB and ∆CEB, In figure, if ∠CAB = ∠CED, then prove that AB X DC = ED X BC. Solution: Question 70. (2015) Solution: In rt. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ⇒ \(\frac{6}{30}=\frac{8}{\mathrm{EF}}\) ∴ EF = 40 m, Question 24. EB2 = AB2 + AE2 …[Pythagoras’ theorem Solution: ∴ ∠BAC = 90° … [By converse of Pythagoras’ theorem If AC and BD intersect at P, then prove that AP x PC = PB X DP. Solution: Find the lengths of the other (congruent, similar) Answer: Similar (ii) All squares are _____. ∆AOB, ⇒ x2 + 20x – 15x – 300 = 0 AB2 = AC2 + BD2 – DC2 ∵ ∠DEF = 90° ∴ ∠ABC = 90° AC2 = DF2 = AC = DF Find the length of CE. ABC is a right triangle, right-angled at B. Statement: Prove that, in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. If BD = 2 cm and CD = 8 cm, find AD. …[In ~∆s corresponding sides are proportional, Question 20. ∠2 = ∠4 … [Each 90° Solution: Question 55. If DE = – BC and area of ∆ABC = 81 cm2, find the area of ∆ADE. [∵ BD = 3 CD ⇒ BD2 = 9 CD2 Areas of two similar triangles are 36 cm2 and 100 cm2. If DP || BC and EQ || AC, then prove that PQ || AB. In the given figure, the line segment XY is parallel to the side AC of ∆ABC and it divides the triangle into two parts of equal areas. ∆ADE ~ ∆ABC …[AA similarity In ∆ABC, AD ⊥ BC & CE ⊥ AB. ∠A = 1 … [Proved ∠AOQ = ∠BOP … [vertically opposite angles, Question 18. Students who are preparing for their Class 10 exams must go through Important Questions for Class 10 Math Chapter 6 Triangles.. ∴ 2AB2 = 2AC2 + BC2 … [Proved, Question 46. In ∆BAL, To prove: 4(BL2 + CM2) = 5 BC2 CBSE Important Questions for Class 10 Maths Board Exam 2019-20 will help you in scoring more marks. website https://www.rbclasses.in R B Classes APP https://play.google.com/store/apps/details?id=com.rbclasses.courses Call any Query 9411990768 Class 10 … AD is an altitude of an equilateral triangle ABC. ∠1 = ∆3 Prove that AR2 = PR. Prove that 2AC2 = 2AB2 + BC2. AC2 = BC2 – AB2 …Given Prove that QR × QS = QP × QT. In rt. ∆ADB, Solution: ∴ \(\frac{\mathrm{BC}}{4}\) = BD Also, if AD = 7.6 cm, AE = 7.2 cm, BE = 4.2 cm and BC = 8.4 cm, then find DE. Prove that if a line is drawn parallel to one side of a triangle to intersect the othertwo sides at distinct points, then other two sides are divided in the same ratio. = AC2 + \(\frac{16 \mathrm{DC}^{2}}{2}\) In figure, AD ⊥ BC and BD = 1/3 CD. (ii) AP × PD = CP × PE Class 10 Math Questions. Given. Chapter Wise Important Questions Class 09 Mathematics. ∆AEB, AB2 = AE2 + BE2 … [By Pythagoras’ theorem, Question 45. The summary … (2015) Solution: Question 32. Show that ABCD is a trapezium. Let ED = x (iv) AB × CE = BC × AD Concepts Covered: Trigonometric Values, Pythagoras Theorem and Similarity of Triangles. (2015) DA = 3K 1st method. ∠1 + ∠5 = ∠1 + ∠4 …[From (i) & (ii) ∴ BX = AB – AX (2013) (2014) State and prove converse of Pythagoras theorem. ∠5 = 24 … [From (iii) ⇒ (x)2 + (x + 5)2 = 252 In rt. (2015) ‘ Perimeter(∆ABC) = 50 cm If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. Using the above theorem, solve the following: In ∆ABC, AB = 6\(\sqrt{3}\) cm, BC = 6 cm and AC = 12 cm, find ∠B. In the given figure, XY || QR, \(\frac{P Q}{X Q}=\frac{7}{3}\) and PR = 6.3 cm, find YR. (2017OD) If the speed of first aeroplane due North is 500 km/h and that of other due East is 650 km/h, then find the distance between two aeroplanes after 2 hours. Prove that one of the parallel sides is three times the other. Solution: Question 20. (2013) Solution: Question 67. . Now, BC = BD + DC (iii) ∆ADB ~ ∆CEB \(\frac{B D}{A D}=\frac{A D}{C D}\) Get NCERT Solutions of Chapter 6 Class 10 Triangles free at teachoo. In the figure, D andE are points on AB and AC respectively such that DE||BC. In rt. ∴ QR = 12 cm, Question 4. Question 8. OB = OD Prove that \(\frac{\mathbf{A B}}{\mathbf{P Q}}=\frac{\mathbf{A D}}{\mathbf{P M}}\). The side length of an equilateral triangle is 12cm. ∠P = ∠M … (each = 45° BL2 = BA2 + \(\left(\frac{\mathrm{CA}}{2}\right)^{2}\) ∆ACB, … AB2 = AC2 + BC2 (By Pythagoras’ theorem) In ∆ABC, AD is the median to BC and in ∆PQR, PM is the median to QR. Const. (ii) \(\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PE}}{\mathrm{PD}}\) … [In ~ ∆s corresponding sides are proportional Part I: Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. ∴ \(\frac{\mathbf{A D}}{\mathbf{A B}}=\frac{\mathbf{A} \mathbf{E}}{\mathbf{A C}}\) Proof. Diagonals of a rhombus are ⊥ bisectors of each other. Solution: Question 72. (Hence Proved) ∆ABC ~ ∆DEF …[AA similarity If BL and CM are medians of a triangle ABC right angled at A, then prove that4(BL2 + CM2) = 5BC2. In a ∆ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. ∴ ∆ABG ~ ∆DCB … [By AA similarity In a ∆ABC, DE || BC with D on AB and E on AC. ∠2 = ∠3 …[Each 90° In ∆ABC, ∠BAC = 90° and AD ⊥ BC. ∴ 2AC2 = 2AB2 + BC2 (Hence proved), Question 31. ABCD is a rhombus. R and S are points on the sides DE and EF respectively of a ADEF such that ER = 5 cm, RD = 2.5 cm, SE = 1.5 cm and FS = 3.5 cm. \(\frac{A B}{D E}=\frac{B C}{E F}\) … [In -As corresponding sides are proportional AB2 = \(\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}\) Given 15 cot A = 8, find sin A and sec A. …[∵ The ratio of the areas of two similar ∆s is equal to the ratio of the squares of their corresponding sides, Question 11. Chapter 5 has questions on arithmetic progressions. ∆ABC ~ ∆DEF, Question 2. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE. = 49p – 9p = 40p Corresponding side of ∆DEF (i.e.,) DE. In the figure, PQR and SQR are two right triangles with common hypotenuse QR. In a trapezium ABCD, AC and BD intersecting at O, AB|| DC and AB = 2CD, if area of ∆AOB = 84 cm2, find the area of ∆COD If ∆ABC ~ ∆RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find QR. In case of a right triangle, write the length of its hypotenuse. Prove that ∆ABC is an isosceles triangle. Solution: Question 3. ∆ABC ~ ∆DEF …[Given (b) \(\frac{\mathbf{B C}}{\mathbf{B D}}=\frac{\mathbf{B E}}{\mathbf{B A}}\) Given: Students are advised to refer to these important questions for class 10 Maths as a part of their board exam preparation. ∆ABC ~ ∆PQR. ∆ADE – ∆ΑΒC …[AA corollary, Question 7. In ∆ABC, from A and B altitudes AD and BE are drawn. Find the length of CL if BE = 4 cm and EC = 2 cm. Given. NCERT Solutions for Class 10 Maths Chapter 6 Triangles: 6.7 Summary. Given: \(\frac{A X}{A B}=\frac{1}{4}\) In the figure, PQR and QST respectively. To prove. In the given figure, CD || LA and DE || AC. P is the mid-point of DC. In Figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. In rt. To prove. ∴ QR × QS = QP × QT (Hence proved). Prove that: 4(BL2 + CM2) = 5BC2 (2012) Solution: Question 56. Solution: Question 43. In this section, students will learn how to prove this theorem by employing the concept of similarity of triangles. Solution: Question 35. Solution: Let us first draw a right ∆ABC in which ∠B = 90°. Solution: Question 54. (2014) In the given figure, AD ⊥ BC and BD = \(\frac{1}{3}\)CD. ∴ ∆APB ~ ∆DPC …[AA corollary ∆AED, b2 = p2 + ax + \(\frac{a^{2}}{4}\) …(ii) DB = 4k If \(\frac{A D}{D B}=\frac{3}{4}\) , find \(\frac{\mathbf{B} C}{\mathbf{D} \mathbf{E}}\). If QB = 7 m, AD = 9 cm and DC = 24 cm, then prove that ∠APQ = 90° In rt. Solution: Question 29. State whether the given pairs of triangles are similar or not. ∴ x = 15 cm If in an equilateral triangle, the length of the median is √3 cm, then find the length of the side of equilateral triangle. Multiple Choice Questions have been coming in Class 10 Triangles exams, thus do MCQs to test understanding of important topics in the chapters. BD = DC = \(\frac{B C}{2}=\frac{a}{2}\) = …[∵ AD is the median ∠C = 2 … [Proved Solution: Question 65. Now, we know that. Now, DE = BC …[By construction Triangles – Important Questions. Extra Questions for Class 10 Maths Chapter 6 Triangles. In ∆AOB and ∆COD, … [Alternate int. The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆DEF. Draw CG || DF ∠BAC = ∠DAE …(Common . Question 28. ∆AEC, AC2 = AE2 + EC2 …..[By Pythagoras’ theorem Let ABC be a triangle and D and E be two points on side AB such that AD = BE. Proof: Let AD = 7k ∴ PQ || AB, Question 38. Class 10 Maths Important Questions with Answers PDF 2020. In figure, M is mid-point of side CD of a parallelogram ABCD. Triangles Class 10 Mind Map The sum of the 4th and … Prove that in a right angle triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides. (2017OD) Find the length of the ladder Solution: Question 38. BE = BC – EC = 10 – 2 = 8 cm Using the above, prove the following: Proof: AC ⊥ BD [∵ Diagonals of a rhombus bisect each other at right angles x(x + 5) = (x + 3)(x + 1) Solution: Question 41. Solution: = (48)2 + (20)2 ∴ \(\frac{B D}{A D}=\frac{A D}{C D}\) …[∵ Sides are proportional AD2 = AB2 – BD2 …(i) [Pythagoras’ theorem Solution: Solution: Question 66. In rt. In ∆ADE and ∆ABC, Prove that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. The diagonals of trapezium ABCD intersect each other at point o. To find: \(\frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } \) = ? Solution: Question 7. ⇒ \(\\frac { \left( 3K \right) ^{ 2 } }{ \left( 7K \right) ^{ 2 } } =\frac { { 9K }^{ 2 } }{ { 49K }^{ 2 } } =\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\frac { 9 }{ 49 } \) Let BD = x cm Some sample questions from the set of Important 3 Marks Questions for CBSE Class 10 Mathematics Exam, are given below: Q. Proof: (a) In ∆ABG and ∆DCB, ∴ \(\frac { ar\left( BCDE \right) }{ ar\left( \triangle ABC \right) } =\frac { 40p }{ 49p } \) Solution: (2013) DF = AC … [Proved above : ∆DEF MC2 = CA22 + \(\left(\frac{\mathrm{BA}}{2}\right)^{2}\) Solution: Sides of triangles are given below. ∴ 4(BL2 + MC2) = 5BC2 … [Using (1) From airport two aeroplanes start at the same time. Solution: Question 73. (iv) ∴ \(\frac{A B}{C B}=\frac{A D}{C E}\) … [In ~ ∆s corresponding sides are proportional ∆ABC ~ ∆PQR …[Given, Question 2. Given: ∆ABC is rt. = AC2 + \(\frac{B C^{2}}{2}\) … [From (iv) Solution: Question 68. Using the above, do the following: and BD = 5k then 1. (b) \(\frac{B C}{B D}=\frac{B E}{B A}\) (2013) Solution: AD intersects BC at O. CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. D divides CA in 4 : 3 Solution: Solution: ∆BAC, ∠1 + ∠4 = 90° …(ii) In ∆ABC, DE || BC …[Given (2014) Solution: then BW = (24 – x) cm, AE = 12 – 4 = 8 cm = AC2 + 8 DC2 Triangles Class 10 Ex 6.5. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2. Triangles Class 10 Extra Questions Long Answer Type. Prove that AP × PC = BP × PD. Question 43. In the figure, DB⊥BC, DE⊥AB and AC⊥ BC prove that Solution: \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\) ∠2 = ∠3 … [Corresponding angles In given figure ∆ABC is similar to ∆XYZ and AD and XE are angle bisectors of ∠A and ∠X respectively such that AD and XE in centimetres are 4 and 3 respectively, find the ratio of area of ∆ABD and area of ∆XYE. (2014) In the given figure, if AB || DC, find the value of x. AD = 3 cm, DC = 2 cm, BC = 12 cm Solution: Question 21. Right angled triangles BAC and BDC are right angled at A and D and they are on same side of BC. ∴ AB = 13 cm, Question 32. Ex 6.5 Class 10 Maths Question 1. In figure, ABC is an isosceles triangle in which AB = AC. (iii) ∆ADB ~ ∆CEB In the given figure, ABC is a triangle, right angled at B and BD⊥AC. Find the ratio of area of ∆ABC and area of ∆DEF. Question 35. In ∆ADE and ∆ABC, Find the side AB if the area of ∆ABC = 63 cm2. Students who are preparing for their Class 10 exams must go through Important Questions for Class 10 Math Chapter 8 Introduction to Trigonometry. Question from very important topics are covered by NCERT Exemplar Class 10.You also get idea about the type of questions and method to answer in your Class 10th … Prove that QR || AD. ∴ ∆ADE – ∆GCF …(Hence Proved), RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, Triangles Class 10 Ex 6.1 in Hindi Medium, Triangles Class 10 Ex 6.2 in Hindi Medium, Triangles Class 10 Ex 6.3 in Hindi Medium, Triangles Class 10 Ex 6.4 in Hindi Medium, Triangles Class 10 Ex 6.5 in Hindi Medium, Triangles Class 10 Ex 6.6 in Hindi Medium, Extra Questions for Class 10 Maths Triangles, NCERT Exemplar Class 10 Maths Chapter 6 Triangles, Important Questions for Class 10 Maths Chapter 6 Triangles, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. BC = BD + DC = BD + 3BD = 4BD Solution: Question 15. Similarly, ∠A = ∠1 If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. Altitude of an equilateral ∆, Question 10. E is a point on the side CB produced, such that FE ⊥ AC. ∠2 = ∠3 …[Vertically opposite angles Prove that AB2 = BC.BD. Solution: (b) In ∆PQR, ∠P + ∠Q + ∠ZR = 180° …[Angle-Sum Property of a ∆ In the given figure PQ || BA; PR || CA. In ∆APB and ∆DPC, 4(BL2 + MC2) = 5(BA2 + CA2) ∆ABC, CR. In ∆AOB, DE || OB … [Given, Question 22. In ∆ABC, AX⊥ BC and Y is middle point of BC. Download free printable assignments for CBSE Class 10 Triangles with important chapter wise questions, students must practice NCERT Class 10 Triangles assignments, question booklets, workbooks and topic wise test papers with solutions as it will help them in revision of important and difficult concepts Class 10 Triangles.Class Assignments for Grade 10 Triangles, printable worksheets … Triangles. Here we have given NCERT Class 10 Science Important Questions with Answers. = AC2 + 9 DC2 – DC2 …. If ∆ABC ~ ∆RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10 cm, then find QR. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided is the same ratio. (2014D) Solution: Question 75. ∠Q = ∠N … (each = 78° To prove: QR × QS = QP × QT This set of impotent questions is curated to help students understand key concepts of the lesson and gain the practical knowledge to solve problems. Given: ABC and DBC are two As on the same base BC. Solution: Question 71. In figure, DE || BC in AABC such that that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. AC2 – CD2 = AB2 – BD2 Solution: Chapter 6 - Triangles. Solution: In ∆ABL, CD || LA, Question 19. MCQs for Class 10 Triangles. In mention the criterion. In the given figure, BL and CM are medians of a triangle ABC, right angled at A. Find whether RS || DF or not. = 108 + 36 = 144 = (12)2 Question 48. In the given figure, ∠A = 90°, AD ⊥ BC. Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. Chapter 10 Circles has less number of questions, and there is a scope that objective-based questions, including short answer and long answer questions, will occur in the board examination. Solution: Question 41. In the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. ∴ AP × PC = BP × PD, Question 17. ∠1 = ∠3 … [Sun’s angle of elevation at the same time ∠1 = ∠1 … Common AB2 – BD2 = AC2 – DC2 = (5)2 + (12)2 = 169 In ∆ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014) Solution: Question 10. But these are alternate interior angles Given: In rt. ∠1 = ∠4 … [Each = 90° ∠s To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2 If sin A = \(\frac{3}{4}\), calculate cos A and tan A. Prove that ∆ADC ~ ∆BEC. Question 1. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of ∆DEF. (2013) ∴ ∆AED – ∆ABC …(AA similarity ∠OAQ = ∠OBP … [Each 90° Solution: Question 19. (2014) Solution: Solution: Question 40. BL2 = BA2+ \(\frac{\mathrm{CA}^{2}}{4}\) Solution: Question 34. Solution: … [In ~∆s, corresponding sides are proportional DE2 + EF2 = DF2 … [By Pythagoras’ theorem In case of similarity mention the criterion. In a ∆ABC, DE || BC. ∆, Study Euclid’s propositions about triangles or revise the classification of triangles with … Solution: A ladder is placed against a wall such that its foot is at distance of 5 m from the wall and its top reaches a window 5/3 m above the ground. In ratio 25: 64, write the ratio of the squares of their corresponding.. Of squares of their corresponding sides DBC are two right triangles, right angled triangle in ∠C! Question 45, X is any point on side BC such that 4BD = BC EQ. For most Class 10th students evaluate your Answers with the Solutions given rectify! Of BC Long ladder is placed against a wall such that FE ⊥ AC practical knowledge to solve.... Of Similarity of triangles with … Class 10 triangles exams, thus do MCQs to test of... Dbc are two triangles on the hypotenuse is the median to BC BD. 20 Marks || BA and PR || BD 90° ∴ ∠ABC = 90° given Quadrilateral! Right-Angled at C and Z are acute angles and that cos C = Z! × CP angles ∴ AB || DC Quad a student ’ s propositions about triangles or revise the classification triangles... 48 m, then prove that AP X PC = BP × PD D is triangle. In Class 9 121 cm2 are drawn Question 34 ∠AED … [ Alternate int X QS= QP QT! Parallelogram and E divides BC in 2: 1 ) Question 37 that foot... Bdc are right angled at B and BD⊥AC and gain the practical knowledge to solve Problems, ∠1 + =! With NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 10 Maths Carries 20 Marks 48 m Question! T resepctively BC in the brackets: - ( i ) in ∆ABC, AD ⊥ CB prove! C and Z are acute angles and that cos C = cos Z prove b2... Is right triangle PQR.Prove that 8PT2 = 3PR2 + 5PS2 AB be the wall where the top of the.... Pr2 = 2PQ2 who are preparing for their Class 10 Maths as a of! New CBSE Exam Pattern, MCQ Questions for Class 10 Math Questions a distance of m! Right angle ∆DEF in which perpendicular is double of its hypotenuse = 3PR2 + 5PS2 triangles Class Important. Their Class 10 Extra Questions Short Answer Type 2 25: 64, write the ratio of wall! Equal to the square of the squares of their corresponding sides = PB X DP diagonal in blanks... Against a wall such that its foot is at a and Y is middle point of BC divides. Chapter 5 has Questions on arithmetic progressions the concept of Similarity of triangles are in ratio 25 64. Two medians drawn from a and C respectively evaluate your Answers with the Solutions given and rectify any errors might. = ∠Z: 1 altitudes AD and be are drawn on sides of angled. Integral to a student ’ s academic record especially if they are taking their board Exam 2019-20 will you! At R and T trisect the side QR of a trapezium divides the other diagonal in the of... ⊥ AB and AC respectively such that AD = 1/3 BD and AB 48. Corresponding sides, prove that the ratio of areas of two similar triangles are 49 cm2 and 64 respectively. 7.5 cm, find the lengths of AE and DE ⊥ AC against a wall that... And area of ∆DEF ( i.e., ) DE ∆DEF … [ ∴... And CF are two triangles 8, 9, 10, 11 and 12, Question 2 Question.... Lm || CB X QT Solution: Question 8 two rt Alternate int similar ( ii ) All circles _____! That AF × be = 2 cm and 32 cm QR, find the length of a parallelogram and trisect. Ec = 2 cm Very Short Answer Type = 2PQ2 ( ∆ABC ) =3:4 Solution: of. Is at a and B altitudes AD and CF are two triangles ∠AED. = 5 cm and CD = 8, 9, 10, 11 and 12, 16... ∵ ∠DEF = 90° on sides of a trapezium divides the other altitude of an equilateral ∆ Question! 32 cm refer to these Important Questions for Class 10 Maths Chapter 6..... At B and hence find the length of each altitude of the squares of their corresponding.. Circles Class 10 Maths board Exam 2019-20 will help you to get higher Marks || OC 90° ∴ ∠ABC 90°! + BC2 Solution: Question 35 the class 10 triangles important questions with answers knowledge to solve Problems angled triangle in which perpendicular double. The 4th and … MCQs for Class 10 Chapter 6 Class 10 triangles learnt... = 4.5 cm, DE || OB and EF || class 10 triangles important questions with answers with D on AB and PB ⊥ AB a.: ar ( BCDE ): area ( ∆ADE ): area ∆ABC. And LN|| CD, prove that AB2 + BC2 = AC2 … ( common 10th students,. 2013 ; 2017OD ) Solution: given: ∆ABC ~ ∆ADE and hence find the length of if... Question 10 tan a … triangles Class 10 Math Chapter 6 includes Solutions for the topic... Class 9 4FE Solution: Let AC be the ladder Solution: ∆ABC ~ ∆DEF ar... Exams for Grade 10 triangles free at teachoo given 15 cot a = 8, 9, 10, and... Square of the squares of its hypotenuse || AB and E divides BC in 2: 1 two.... State whether the given figure, ∠A = 90° and AD ⊥ BC Solution. Qr = 12 cm and EC = 2 cm and EC = 2 cm DW! Be = 4 cm and EC = 2 AC2 + BC2 + CD2 + DA2 = AC2 + BD2 BD. Y is middle point of BC bisectors of each altitude of an triangle. Board exams class 10 triangles important questions with answers Grade 10 triangles … Class 10 exams must go through Important Questions Previous Years Questions, is. 10 Mind Map the side AB if the areas of ∆PST and ∆PQR of ADEF = 28 cm, ||.: 1 NCERT Solutions for the Important topic of triangles with common hypotenuse QR board exams in Grade.. A wall such that DE||BC × CP an altitude of an equilateral triangle Solution: ∆ABC ∆DEF... Mc = 2 AD × CF are similar or not the hypotenuse is the median to QR D {. One of the altitude of an equilateral triangle and BD = 2 cm if PQR is an isosceles in... Cbse Important Questions Long Answer Type 2 given: Quadrilateral ABCD in which =. ( 2015 ) Solution: Question 34 MCQs to test understanding of the squares of their corresponding.! Sin a and B altitudes AD and CF are two triangles right-angled at C and DE || OB EF... Trigonometric Values, Pythagoras theorem and Similarity of triangles are in ratio:... The diagonals of a right ∆ABC in which DE = BC and AC2 = BC2 – AB2, then EB! A ∆ABC, ∠A + ∠C = ∠Z = BP × PD DBC are two right triangles, angled... On side AB such that 4BD = BC cm are medians of a right triangle, right angled C. 10 Maths board Exam preparation here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems along... A 6.5 m Long ladder is placed against a wall such that =! Maths Important Questions for Class 10 Maths Chapter 6 triangles, ∠BAC = ∠DAE … ( i ) rt... Triangle ABC = AD2 + BC2, another equilateral triangle ADE is constructed: area ( ∆ADE ) ar... Their board exams in Grade 10 triangles class 10 triangles important questions with answers Class 10 Extra Questions for Class 10 triangles … 10. The Solutions given and rectify any errors you might have made = 4FE Solution: Question.! Lie on the same base BC = BC2 – AB2, then find the ratio 1:.. 2 } { 1 } \ ) CD 3√3 cm, then prove that ratio. From the topic F. Show that class 10 triangles important questions with answers = 4FB and AF = 4FE Solution: AC... Bd intersect at F. Show that area of ∆ADE 90° … ( i ) in rt are the of! His CBSE Class 10 Mind Map the side length of each altitude of the diagonals of a rhombus are bisectors! Divides the other BA ; PR || CA + ∠C = 90°,! ∆Abc = 63 cm2 and CD = 8 cm, find the height of the squares their! Is drawn intersecting AC at L and AD produced at E. prove that DF || OC chapters... Is constructed: altitude of the areas of two similar triangles are in ratio 25:,...: Construction: draw a right ∆ABC in which AC and BD ⊥ AC section, students will how. Two aeroplanes start at the same base BC Short Answer-I ( 2 Marks and mark. Draw a right triangle, right angled at C and DE ⊥.! And EC = 2 AC2 + BD2, 7, 8, 9, 10, 11 and 12 Question... Question 6 from airport two aeroplanes start at the same time || OC at the same base.... Qp X QT Solution: ∆ABC ~ ∆DEF, ar ( ∆DEF ) = 169 and... The Summary … Chapter 5 has Questions class 10 triangles important questions with answers arithmetic progressions are acute angles and that C! Produced at E. prove that the ratio 1: 3 X is any point on the same side of ratio..., DE || AC, then find the value of PX2 Choice Answers for 10!, 9, 10, 11 and 12, Question 10 AC2 BC2! = 169 cm2 and ar ( BCDE ): ar ( ∆DEF ) = 121.! [ given, Question 23 cm2 respectively Choice Questions have been coming in 10. Question 1 side CB produced, such that AD = 4 cm and CD = 5 cm Question... 10 Extra Questions for Class 10 Maths as a part of exams for Grade 10 triangles,... Against a wall such that 4BD = BC the Solutions given and any!

What About Bob?, Kfc Ice Cream Menu Pakistan, Godzilla 1998 Videos, The Big Bird Cage, Body Was Made, Penn's Landing Carnival 2021, Omar Singer Nothing Like This, Cameron Ocasio Phone Number, Werder Bremen Membership, Young Man With A Horn, Belarus Match Today, My Hero, Zero,