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d)  MCD is an isosceles triangle,   3^^ BD bisect each other. So, MD = MC
                                                                                         The diagonals of a square
                ^y._^ 180°-ZC/WD                                                          divide the angles at the
                                                                                          corners into 45° angles.
                  = 180°-90°
                        2
                   _ 90!
                     2
                   = 45°


       Q PQRS is a rectangle. Find the following angles.

                                                                      100° ^ ^ y
           a)  APSR                           b) zlx
           c)  AQOR                           d)

                                                                        X

           Solution
           a) PQRS is a rectangle.          Each angle at the corner of a rectangle
               So, ZPS/? = 90°.             is a right angle.

           b)  QS is a straight line.
                                            Sum of angles on a straight line is 180°.
               So, ZP05= 180°-100°                                                        When two straight lines
                        = 80°.                                                            cut each other, the angles
                                                                                          opposite each other are
               PR is a straight line.
                                                                                          equal.
               So, ^x=180°-80°              Sum of angles on a straight line is 180°.
                      = 100°.                                                             AP0Q= A SOR
                                                                                          APOS=AQOR
           c)  PR is a straight line.
                                            Sum of angles on a straight line is 180°.
               So, ^Q0/?= 180°-100=                                                       Each pair of angles are
                         = 80°.                                                           vertically opposite angles.
           d) ^OQP is an isosceles triangle.
                    180°-ZOO/?                                   other.
               zx=
                                             So, OR = OS.
                   = 180°-80°
                         2
                   = 100°
                      2
                   = 50°.
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