Page 176 - Maths Skills - 7
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174                                                                                                  Maths


        TWO SPECIAL TRIANGLES: EQUILATERAL AND ISOSCELES
        A triangle in which all the three sides are of equal lengths is called an equilateral triangle.
        Take two copies of an equilateral triangle ABC as shown Fig.              A                      A
        Keep one of them fixed. Place the second triangle on it. It fits                                 60°
        exactly into the first. Turn it round in any way and still they
        fit with one another exactly. Are you able to see that when the
        three sides of a triangle have equal lengths then the three angles   B              C   B  60°        60°  C
        are also of the same size?                                                (i)                   (ii)
        We conclude that in an equilateral triangle:
             (i)  all sides have same length                  (ii)  each angle has measure 60°


        A triangle in which two sides are of equal lengths is called an isosceles triangle.
        From a piece of paper cut out an isosceles triangle XYZ, with XY = XZ as shown in Fig. below Fold it such that
        Z lies on Y. The line XM through X is now the
        axis of symmetry. You find that ∠Y and ∠Z fit
        on each other exactly. XY and XZ are called
        equal sides; YZ is called the base; ∠Y and ∠Z
        are called base angles and these are also equal.             (i)                             M  (ii)

        Thus, in an isosceles triangle:
             (i)  two sides have same length.                 (ii)  base angles opposite to the equal sides are equal.



              Let’s Attempt


        Example 1:  In Fig., equal sides have been shown with similar markings. Find the values

                       of ∠PRQ and ∠PQR.







        Solution:      In DPQR, PR = QR                                                                      [Given]

                       \  ∠PQR = ∠QPR                       [Q Angles opposite to equal sides of a D are always equal]
                          But, ∠PRS = ∠PQR + ∠QPR
                                                      [Q Exterior angle is equal to the sum of interior opposite angles]
                       ⇒  110° = 2∠PQR

                       ⇒  ∠PQR =    110°   = 55°
                                      2
                       \  ∠PQR = 55° = ∠QPR
                          Also, ∠PRQ + ∠PRS = 180°                                            [Q Linear pair]
                       ⇒  ∠PRQ + 110° = 180°
                       ⇒  ∠PRQ = 180° – 110° = 70°

                       Hence, ∠PRQ = 70° and ∠PQR = 55°.
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