A) 6 cm
B) 3 cm
C) \[6\sqrt{2}\]cm
D) \[3\sqrt{3}\]cm
Correct Answer: A
Solution :
(a): \[\angle PQR={{60}^{{}^\circ }}\] \[PQ=QR\] \[\therefore \]\[\angle QPR=\angle QRP={{60}^{{}^\circ }}\] \[\therefore \]\[\Delta PQR\] is an equilateral triangle.You need to login to perform this action.
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