WebOct 10, 2024 · Solution: EF and CD are parallel ; CE is a transversal pass through the parallel lines EF and CD. Sum of angles in same side of transversal which passes through parallel lines is 180°. ∠ DCE + angle CEF = 180°. ∠ DCE + 130° = 180°. ∠ DCE = 180° − 130°. ∠ DCE = 50°. AB is parallel to CD. WebThe area of the figure formed by joining the mid points If x4+1x4=194 then x3+1x3= 76 52 64 none of these x4+1x In the figure if l m n p and ∠1=850 find ∠2 In th
Class 9 RD Sharma Solutions - Chapter 8 Introduction to Lines and ...
WebGiven: In the figure, if AB CD and CD EF, ∠ B A C = 70 0, ∠ C E F = 130 0 ∵ E F C D ∴ ∠ E C D + ∠ C E F = 180 0 (Co-interior angles) ⇒ ∠ E C D + 130 0 = 180 0 ∴ ∠ E C D = 180 0 − 130 … WebAC = AB, AB // DF, BC // ED, AC // EF and ∠ CAB = 70°. Find ∠ x, ∠ y and ∠ z. Δ ABC is an isosceles triangle. ∠ ACB = ∠ ABC = (180° − 70°) ÷ 2 = 55° DF // AE and AD // EF So, AEFD is a parallelogram . ∠ ... AB = AC, ∠ AEC = 53° and ... ∠ t = ∠ FCE − ∠ ACE = ∠ FCE − ∠ ACB = 127° − 38 ° = 89 ... terry goodkind book covers
In the given figure AB CD and CD EF . Also EA AB . If BEF
WebMar 22, 2024 · Transcript. Ex 6.2, 3 In the given figure, If AB CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. Now, AB CD & GE is transversal, Hence, ∠ AGE = ∠ GED ∠ … WebAug 9, 2024 · In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that (i) ∆ACE ≅ ∆DBF (ii) AE = DF. Solution: Question 10. In the given figure, AB = AC and D is mid-point of BC. Use SSS rule of congruency to show that (i) ∆ABD ≅ ∆ACD (ii) AD is bisector of ∠A (iii) AD is perpendicular to BC. Solution: Question 11. WebMar 22, 2024 · If ∠ BEF = 55°, find the values of x, y and z. Given AB CD and BE is the transversal So, x = y Also, CD EF and BE is transversal So, y + 55° = 180° y = 180° – 55° y … trigyn technologies salary