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). Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,∴ ΔAPC ~ ΔBQC [AAA similarity criterion]∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2, ⇒ area(ΔAPC) = 2 × area(ΔBQC)⇒ area(ΔBQC) = 1/2area(ΔAPC). For each of the following cases, state whether EF || QR :
Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m. Let us say now, the fly is at point D after 12 seconds. In case of a right triangle, write the length of its hypotenuse. 15. Give two different examples of pair of. 11. 4. (ii) Trapezium and square. Prove that. State whether the following quadrilaterals are similar or not: 1. 1. It is given that DE || BC. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. 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. Therefore, the distance of the foot of the ladder from the base of the wall is 6 m. 10. If ΔABC ~ ΔFEG, Show that: (i) CD/GH = AC/FG(ii) ΔDCB ~ ΔHGE(iii) ΔDCA ~ ΔHGF. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. \( \frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64} \)
Given, in ΔABC, D is the midpoint of AB such that AD=DB. PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
Given: ABC and DBC are triangles on the same base BC. (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. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF. Chapter 6 of NCERT Solutions for Class 10 Maths is the triangles and it covers a vast topic including a number of rules and theorems. (ii) by eq(i), ∠A = ∠A [Common angle]∴ ΔADE ~ ΔABC [SAS similarity criterion]. 12. Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that; In right angle triangle, ADB and ADC, by Pythagoras theorem, Subtracting equation (ii) from equation (i), we get, = 9CD2 – CD2 [Since, BD = 3CD] = 8CD2. 6.60, AD is a median of a triangle ABC and AM ⊥ BC.
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. 9. 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? (ii)
(i), And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii), 3. ABC is an isosceles triangle with AC = BC. In Fig. 6.64)? In Fig. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig. 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. Show that ΔABC ~ ΔPQR. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD. State which pairs of triangles in Figure, are similar. 7. 12. 7. Tick the correct answer and justify: In ΔABC, AB = 6√, 1. Given, in figure, PS is the bisector of ∠QPR of ∆PQR. Solution: Given, ΔABC ~ ΔDEF,Area of ΔABC = 64 cm2Area of ΔDEF = 121 cm2, 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. (ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2. Given that ΔABC is an isosceles triangle right angled at C. Given that ΔABC is an isosceles triangle having AC = BC and AB. ∴ 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. 6. AN. Download the PDF for NCERT Class 10 Maths Chapter 6 using the link prevailing on our page and practice them as and when you … 8. ABC is an equilateral triangle of side 2a. Hence, by Pythagoras theorem ΔABC is right angle triangle. We have to prove that E is the mid point of AC. ∴ Area(ΔABC)/Area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 4/1 = 4:1, 9. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. 11. Show that ΔABC ~ ΔPQR. ∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (congruent, similar), (ii) All squares are __________. 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 for Class 10 Maths Chapter 6 Triangles is also available exercisewise that is given below. 2. Areas of these triangles are in the ratio(A) 2 : 3(B) 4 : 9(C) 81 : 16(D) 16 : 81. If the areas of two similar triangles are equal, prove that they are congruent. Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Hence, having a clear understanding of the concepts, theorems and the problem-solving methods in this chapter is mandatory to score well in the Board examination of class 10 maths. Show that CA2 = CB.CD. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. The Main Strength of the Vedantu’s NCERT Solutions for Class 10 Maths Chapter 6 Triangles, Lies in the Following Points: Solutions are delivered keeping in mind the age group of … Fill in the blanks using the correct word given in brackets :
13. 9. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. ΔABC and ΔBDE are two equilateral triangle. 8. 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. Encourages the children to come up with diverse solutions to problems. Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR, ⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. You can view them online or download PDF file for future use. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.