| Column-I | Column-II | |
| (p) | (P) If ABCD is a parallelogram, then sum of the angles x, y and z is | (1) \[{{25}^{o}}\] |
| (q) | (Q) If ABCD is a rhombus, where \[\angle D={{130}^{o}},\]then the value of \[x\]is | (2) \[{{180}^{o}}\] |
| (r) | (R) If ABCD is a rhombus, the value of w is | (3) \[{{50}^{o}}\] |
| (s) | (S) If ABCD is a parallelogram, where\[x+y={{130}^{o}}\] then the value of B is | (4) \[{{90}^{o}}\] |
A) \[(p)\to (1),(q)\to (2),(r)\to (3),(s)\to (4)\]
B) \[(p)\to (2),(q)\to (1),(r)\to (4),(s)\to (3)\]
C) \[(p)\to (3),(q)\to (1),(r)\to (2),(s)\to (4)\]
D) \[(p)\to (4),(q)\to (3),(r)\to (1),(s)\to (2)\]
Correct Answer: C
Solution :
(P) In \[\Delta ABC,\]by angle sum property \[x+y+\angle ABC={{180}^{o}}\] \[\Rightarrow \]\[\angle ABC={{180}^{o}}-(x+y)\] ?(i) \[\therefore \]\[\angle ABC=\angle ADC\] [\[\because \]Opposite angles of a parallelogram are equal] \[\therefore \] \[z={{180}^{o}}-(x+y)\] [using (i)] \[\Rightarrow \]\[x+y+z={{180}^{o}}\] \[(Q)\angle C=2x\] [Since, diagonals bisects the angles in rhombus] Now, we have \[\angle D+\angle C={{180}^{o}}\] [Co-interior angles] \[{{130}^{o}}+2x={{180}^{o}}\] \[(\angle D={{130}^{o}})\] \[\Rightarrow \]\[2x={{180}^{o}}-{{130}^{o}}={{50}^{o}}\Rightarrow x={{25}^{o}}\] (R) Since, in a rhombus, diagonals bisect each other. \[\therefore \] \[w={{90}^{o}}\] (S) Since in a parallelogram opposite angles are equal. \[\therefore \] \[\angle B=\angle D=z\] ?(i) In\[\Delta \Alpha \Beta C,\]by angle sum property, \[\angle B+x+y={{180}^{o}}\] \[\angle B={{180}^{o}}-(x+y)\] \[\therefore \]\[z={{180}^{o}}-(x+y)\] (using (i)) \[={{180}^{o}}-{{130}^{o}}={{50}^{o}}\]
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