In fig 6.53 abd is a triangle
Web∠ ABD = ∠ CBA (Common angle) Therefore, ΔADB ΔCAB (AA similarity) AB 2 = CB * BD (ii) Let ∠ CAB = x. In ΔCBA, ∠ CBA = 180 o – 90 o – x. ∠ CBA = 90 o – x. Similarly, in ΔCAD. ∠ … WebIn Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB 2 = BC . BD (ii) AC 2 = BC . DC (iii) AD 2 = BD . CD . maths; triangle; class-10; 1 Answer +1 vote . answered by Alisha selected by admin . Best answer. Answer Welcome to Edusaint Q&A, where you can ask questions and receive answers from other members of the ...
In fig 6.53 abd is a triangle
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Weba= 3 b= 6 c =6. 1. The triangle perimeter is the sum of the lengths of its three sides. p= a+b+c = 3+6 +6 = 15. 2. Semiperimeter of the triangle. The semiperimeter of the triangle … WebQ3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
WebState which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : ... In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that. AB 2 = BC . BD; AC 2 = BC . DC; AD 2 = BD . CD; Solution. I. In ∆ABD ... Weba= 6 b= 6 c =6. 1. The triangle perimeter is the sum of the lengths of its three sides. p= a+b+c = 6+6 +6 = 18. 2. Semiperimeter of the triangle. The semiperimeter of the triangle …
WebSides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (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 cm WebIn Fig. 6.53,ABDis a triangle right angled at Aand AC⊥BD. Show that (i) AB2=BC⋅BD(ii) AC2=BC⋅DC(iii) AD2=BD⋅CD Video Explanation Answer Answered By toppr How satisfied are you with the answer? This will help us to improve better answr Get Instant Solutions, 24x7 No Signup required download app More Questions by difficulty EASYMEDIUMHARD
WebIn Fig. 6.53,ABD is a triangle right angled at A Show that (i) AB2=BC⋅BD(ii) AC2=BC. DC(11¨)AD2=BD⋅CDB<2 Fig. 6.53 Open in App Solution Verified by Toppr Was this answer …
Web(3) In Fig. 6.53, ABD is a triangle right angled at A and AC⊥BD . Show that (i) AB 2=BC.BD (ii) AC 2=BC.DC (iii) AD 2=BD.CD Solution Verified by Toppr Was this answer helpful? 0 0 Similar questions A shown in the following figure, the area of square ABCD is 16cm 2 and the area of square CIPO is 9cm 2. prom the bookWebSep 15, 2024 · ∠B is common between the two triangles. So, BCA∼ BAD ...AA test of similarity ....(I) Hence, AB. BC = AD. AC = BD. AB...C.S.S.T. And, ∠BAC=∠BDA ....C.A.S.T … prom theme thronesWebIn Fig. 6.53, \\mathrm{ABD} is a triangle right angled at \\mathrm{A} and \\mathrm{AC} \\perp \\mathrm{BD}. Show that (i) \\mathrm{AB}^{2}=\\mathrm{BC} \\cdot \\mathr... prom the filmWebJan 24, 2024 · In Fig, 6.53. ABD is a triangle right angled at A and AC 1. BD. (i) AB2=BC∘, BD ABCis an isosceles triangle right angled at C. Prove that AB2=2AC2. 5. ABCis an isosceles … labels of lungsWebIn Fig. 6.53, A B D is a triangle right angled at A and A C ⊥ B D. Show that (i) A B 2 = B C ⋅ B D (ii) A C 2 = B C ⋅ D C (iii) A D 2 = B D ⋅ C D Answer View Answer Discussion You must be … labels of moonbaseprom themeWebIn Fig 6.53 ABD Is A Triangle Right Angled At A And AC Perpendicular To BD Show That (I) AB²=BC.BD AK MtCourse by Anand Kushwaha Hello Students, ... labels of maps