WebOct 23, 2024 · A Then (а) BD.CD = BC² (б) AB.AC = BC² (c) BD.CD = AD² (d) AB.AC = AD² Answer 11. D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6 Answer 12. WebBC = 2(5) + 2 = 10 + 2 = 12. AC = 3x = 3(5) = 15. DF = 6(5) = 30. EF = 3(5) + 9 = 15 + 9 = 24. Therefore, the length of the sides of the triangle are AB = 9 cm, BC = 12 cm, AC = 15 cm, DE = 18 cm, EF = 24 cm and DF = 30 cm. Try This: Let ABC and DEF be two triangles in which AB = DE, BC = FD and CA = EF. The two triangles are congruent under ...
In the given figure, DE ∥ CB. Determine AC and AE. - Toppr
WebJan 8, 2024 · Find a quadratic polynomial whose zeroes are 5+ under root 2 and 5- under root 2. Convert it into rational number in form P/Q4.05 If x is a number whose simplest form is p/q , where p and q are integers and q not= 0,then x is a non- terminating repeating decimal only when q is no … WebIn given ΔABC points D and E are mid points of AB and AC and also BC = 6 cm then find DE. asked Mar 23, 2024 in Triangles by AdvaitMogarkar (43.7k points) triangles; class-10; … random chats for kids
In Fig.DE II BC such that AE = (1/4) AC. If AB = 6 cm, find AD
WebHint: ab+ bc +ac = 2(a+b+c)2−(a2+b2+c2). The number of ordered triples (a,b,c) of positive integers which satisfy the simultaneous equations ab +bc = 44, ac +bc = 33. Your solution is correct. Noe that a = 1,b −c = 11 and a = 11,b −c = 1 both lead to a+b = 12+ c. Then 33 = ac + bc = (12+ c)c indeed has no solution. WebMay 15, 2024 · Best answer Given : D is any point on side AB of ∆ABC and E and F are two points on side BC. line segment DF, DE and AE are drawn. To prove : BF/FE= BE/EC Proof : In ∆BCA, DE AC (given) BE/EC = BD/DA ….. (i) (By Basic Prop. Theorem) Again In ∆BEA, DF AE (given) BF/FE = BD/DA ….. (ii) (By Basic Prop. Theorem) From equation (i) and (ii) WebOct 12, 2024 · To find: The value of EC. Solution: In the given figure, the triangles ABC and ADE have the ∠A common. Since BC and DE are parallel to one another, the angles ∠E and ∠C are equal. Also, angles ∠D and ∠B are equal. This is … random character quirks generator