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## ncert solutions for class 10 maths chapter 6

Distance covered by first aeroplane flying due north in  hours (OA) = 100 × 3/2 km = 1500 km, Distance covered by second aeroplane flying due west in  hours (OB) = 1200 × 3/2 km = 1800 km. (similar, congruent), (iii) All __________ triangles are similar. By applying Pythagoras Theorem in ∆ADB, we get. 8. In the fig 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC. (ii) By applying Pythagoras Theorem in ∆ABM, we get; (iii) By applying Pythagoras Theorem in ∆ABM, we get. \) 1. (ii) by eq(i), ∠A = ∠A [Common angle]∴ ΔADE ~ ΔABC [SAS similarity criterion]. Also please like, and share it with your friends! 9. Fill in the blanks using correct word given in the brackets:-(i) All circles are __________. Therefore they are not similar. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. 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. NCERT Book. Prove that 9AD2 = 7AB2. AB || CD, thus alternate interior angles will be equal, Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;∴∠DOC = ∠BOA, Hence, by AAA similarity criterion,ΔDOC ~ ΔBOA. In the given figure, we can see, LM || CB. 6.64)? Therefore, ΔLMP and ΔDEF are not similar. In the fig 6.18, if LM || CB and LN || CD, prove that AM. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 6 Triangles. $$\frac{P F}{F R}=\frac{8}{9}$$ 5. 7. In a triangle if the square of one side is equal to the sum of the square of the other two sides, then the angle opposite to the first side is a right angle. NCERT Solutions for Class 10 Maths Chapter 6 Triangles Class 10 are two of the important lessons for your final board exams and our faculties have put in a lot of effort to provide you with revised solutions and the important facts related to the chapter. 16. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter. You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. Given, Two poles of heights 6 m and 11 m stand on a plane ground. Since ΔABC ~ ΔDEF∴ ∠B = ∠EAB/DE = BM/EN [Already Proved in equation (i)]∴ ΔABC ~ ΔDEF [SAS similarity criterion]⇒ AB/DE = AM/DN …………………………………………………..(iii)∴ ΔABM ~ ΔDENAs the areas of two similar triangles are proportional to the squares of the corresponding sides.∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2Hence, proved. It explains Basic Proportionality Theorem and different theorems are discussed performing various activities. You can view them online or download PDF file for future use. To have a much better understanding of the exercises and Chapter refer to NCERT solutions for class 10 Maths Chapter 6. In ΔOQR, BC || QR. In this chapter, we shall study about those figures which have the same shape but not necessarily the same size. ABC is an equilateral triangle of side 2a. For each of the following cases, state whether EF || QR. Find the ratio of the area of ΔDEF and ΔABC. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. On the basis of the CBSE Class 10 syllabus, this chapter belongs to the unit that has the second-highest weightage. 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). (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 cmSolution: (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)2Therefore, the above equation satisfies, Pythagoras theorem. If the areas of two similar triangles are equal, prove that they are congruent. Show that BC || QR. Therefore, by converse of Basic Proportionality Theorem. Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR, Or, PM2 = PR2 – MR2 ………………………………………..(ii), = QR2 – QM2 – MR2        [∴ QR2 = PQ2 + PR2], 3. Two poles of heights 6 m and 11 m stand on a plane ground. E and F are points on the sides PQ and PR respectively of a ΔPQR. In Fig. Show that AO, 10. These solutions can be helpful not only for the exam preparation but also in solving the homework and assignments. Let AD = x cm D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. 10. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). (isosceles, equilateral) Show that CA2 = CB.CD. • Pythagoras Theorem: We studied about Pythagoras theorem in earlier class which states, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given, ABC is an equilateral triangle of side 2a. To get fastest exam alerts and government job alerts in India, join our Telegram channel. We have to prove: Area(ΔABC)/Area(ΔDEF) = AM2/DN2. Hence, it is right angled triangle.Length of Hypotenuse = 25 cm. Prove that AD is the bisector of ∠ BAC. Given, in figure, PS is the bisector of ∠QPR of ∆PQR. PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm It is given that DE || BC Given, ABCD is a square whose one diagonal is AC. NCERT Solutions for Class 10 Maths Chapter 6 Triangles is also available exercisewise that is given below. There are total 6 exercises in the Chapter 5 Class 10. $$\begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{1.5}{3}=\frac{1}{x}} \\ {x=\frac{3 \times 1}{1.5}} \\ {x=2} \\ {\therefore \mathrm{EC}=2 \mathrm{cm}}\end{array}$$ Let EC = x cm Similarly, given, LN || CD and using basic proportionality theorem, 4. This solution contains questions, answers, images, explanations of the complete Chapter 6 titled Triangles of Maths taught in Class 10. ∴ AP/DM = AO/DO⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO. Studyrankers is a free educational platform for cbse k-12 students. Download the PDF for NCERT Class 10 Maths Chapter 6 using the link prevailing on our page and practice them as and when you … 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°. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. ΔABC and ΔBDE are two equilateral triangle. Find each of its altitudes. After you have studied lesson, you must be looking for answers of its questions. A ladder 10 m long reaches a window 8 m above the ground. 3. ∴ AD/DB = AE/EC [Using Basic proportionality theorem], (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, 2. And D is a point on side BC such that BD = 1/3BC. Prove that 2AB2 = 2AC2 + BC2. In Fig. (iii) Equilateral (iv) (a) Equal $$\frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}$$ The given two figures are not similar because their corresponding angles are not equal.