In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. (ii) As, ΔABC ~ ΔAMP (AA similarity criterion). 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. 13. In Fig. In the figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD. There are total 6 exercises in the Chapter 5 Class 10. NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here which is considered to be one of the most important study material for the students studying in CBSE Class 10. (ii) Trapezium and square. 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. By applying Pythagoras Theorem in ∆ADC, we get. Again, applying Pythagoras theorem in ∆AFC, we get, DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2, DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2, BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2. This theorem is also known as Baudhayan Theorem. To download our free pdf of Chapter 6 Triangles Maths NCERT Solutions for Class 10 to help you to score more marks in … 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 180ºTherefore, ∠DCO + ∠ CDO + ∠ DOC = 180°⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)⇒ ∠DCO = 55°, Hence, Corresponding angles are equal in similar triangles, 3. (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(Recall that you have proved it in Class I. In the given figure, let us extend BA to P such that; By using the converse of basic proportionality theorem, we get, ∠BAD = ∠APC (Corresponding angles) ……………….. (i), And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii). 2. \( \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=1.5 \)
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 10 so that you can refer them as and when required. 6. 5. 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? How far apart will be the two planes after. 5. Hence, By Pythagoras theorem ΔABC is right angle triangle. ABC is an equilateral triangle of side 2a. Chapter Triangles does not require you to have vast knowledge about triangles. Sorry!, This page is not available for now to bookmark. 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°. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.. For a better understanding of this chapter, you should … Show that ΔABC ~ ΔPQR. 1. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. NCERT Solutions for class 10 Maths chapter 6 are given below. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. It will help the students to enhance their knowledge about the chapter triangles and the mathematician involved. (ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm, (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given), Here, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm, And, FR = PR - PF = 2.56 - 0.36 = 2.20 cm, In ΔPQR, EF || QR. State whether the following quadrilaterals are similar or not: 1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. 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. 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. Prove that. Again, by applying Pythagoras Theorem in ∆AMC, we get. Hence, the distance between two aeroplanes will be 300√61 km. 6.61, two chords AB and CD intersect each other at the point P. Prove that : 8. 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. 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. To have a much better understanding of the exercises and Chapter refer to NCERT solutions for class 10 Maths Chapter 6. Show that PM, 3. Give two different examples of pair of. 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. Show that ΔRPQ ~ ΔRTS. (equal, proportional), 2. (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. Main topics covered in this chapter include: From your earlier classes, you are familiar with triangles and many of their properties. Two poles of heights 6 m and 11 m stand on a plane ground. Exercise 6.2 Class 10 Maths NCERT Solutions were … In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. After you have studied lesson, you must be looking for answers of its questions. \( \begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{x}{7.2}=\frac{1.8}{5.4}} \\ {x=\frac{1.8 \times 7.2}{5.4}} \\ {x=2.4} \\ {\therefore \mathrm{AD}=2.4 \mathrm{cm}}\end{array} \), Ans : (i)
Say ΔABC and ΔPQR are two similar triangles and equal in area, Since, ΔABC ~ ΔPQR∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)⇒ BC2/QR2⇒ BC = QRSimilarly, we can prove thatAB = PQ and AC = PRThus, ΔABC ≅ ΔPQR [SSS criterion of congruence]. MC (ii) DN2 = DM . Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii). Hence, quadrilateral ABCD is a trapezium with AB || CD. 9. Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.∴ BE = EC = BC/2 = a/2In ΔABE, by Pythagoras Theorem, we getAB2 = AE2 + BE2, 4AE2 = 3a2⇒ 4 × (Square of altitude) = 3 × (Square of one side), 17. (i). In Fig. If the distance between the feet of the poles is 12 m, find the distance between their tops. NCERT Solutions Class 10 Maths Chapter 6 Triangles. Prove that: (i) DM2 = DN . If AB2 = 2AC2, prove that ABC is a right triangle. By applying Pythagoras Theorem in ∆ADF, we get. (congruent, similar)
Two figures having the same shape (and not necessarily the same size) are called similar figures. ABC is an equilateral triangle of side 2a. Sides of two similar triangles are in the ratio 4 : 9. 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. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. 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. Given, ΔABC is an isosceles triangle right angled at C. 5. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals. 16. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Determine which of them are right triangles? Chapter 1 – Real Numbers; Chapter 2 – Polynomials; Chapter 3 – Pair of Linear Eq in two Variables; Chapter 4 – Quadratic Equations; Chapter 5 – Arithmetic Progressions; Chapter 6 – Triangles; Chapter 7 – Coordinate Geometry; Chapter 8 – Introduction to Trigonometry These NCERT Solutions for Chapter 6 Triangles Class 10 Maths will help you in scoring more marks in the examinations. In the figure, DE||AC and DF||AE. By applying Pythagoras Theorem in ∆DEB, we get. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.