Page 178 - ICSE Math 7
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•  As all the three sides of an equilateral triangle are equal, all the three             A
                       angles are also equal and each angle is of magnitude 60°.                              60°
                    •  As all the three sides of a scalene triangle are unequal, all the angles of a
                       scalene triangle are also unequal.                                                  60°    60°
                    •  If all the angles of a triangle are equal, then it must be an equilateral triangle.  B          C

                    Property 3: In any triangle, the sum of any two sides is always greater than the third side.

                    Example 1:  In a DPQR, ∠P = 50° and ∠Q = 55°. Find ∠R.
                    Solution:     As the sum of all the angles of a triangle is 180°, therefore
                                  ∠P + ∠Q + ∠R = 180°
                                  ⇒ 50° + 55° + ∠R = 180°
                                  ⇒ ∠R = 180° – 105° = 75°

                    Example 2:  Find angle x in each of the following.
                                               P                                            C
                                                                       A                 35°
                                               46°

                                  (a)                          (b)            60°  E

                                       Q                R                                  x   D
                                          x                              B
                                        S                 T                                     F


                    Solution:     (a)  In DPQR, PQ = PR
                                      \ ∠PQR = ∠PRQ
                                      Also,  ∠PQR + ∠PRQ + ∠RPQ  = 180°    (Sum of the angles of a triangle)
                                      ⇒ ∠PQR + ∠PQR + 46°  = 180°
                                      ⇒ 2∠PQR = 180° – 46° = 134°
                                                   134°
                                      ⇒ ∠PQR =           = 67°
                                                     2
                                      ∠PQR + ∠RQS = 180°                           (Linear pair)
                                      ⇒ 67° + x = 180°
                                      ⇒ x = 180° – 67° = 113°
                                      or, ∠RQS  = ∠QPR + ∠PRQ                      (Exterior angle is equal to the sum of
                                                 = 113°                            its two interior opposite angles)
                                      ⇒ x = 46° + ∠PQR
                                      ⇒ x = 46° + 67° = 113°
                                  (b)  ∠AEB = ∠CED                                 (Vertically opposite angles)
                                      ⇒ ∠CED = 60°
                                      ∠EDF = ∠CED + ∠DCE                           (Exterior angle is equal to the sum of
                                                                                   its two interior opposite angles)
                                      ⇒ x = 60° + 35° = 95°

                                       Try This                                           P

                                      In an isosceles DPQR, PQ = PR and the bisectors of   86°
                                      ∠Q and ∠R intersect at S. Find the angle x.          S
                                                                                          x
                                                                                Q                   R

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