Question

The parallel sides of an isosceles trapezium are 48 cm and 24 cm. Each of its non-parallel side is 26 cm long.​

Answer 1

Answer:

Let ABCD be the given trapezium in which AB =48 cm, DC = 24 cm, BC = 26 cm, AD = 15 cm. Through C, draw CF||AD. Now, FB = AB - AF = 48 - 24 = 24 cm. So, it is an isosceles triangle.

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Let ABCD be the given trapezium in which AB =48 cm, DC = 24 cm, BC = 26 cm, AD = 15 cm.

           Through C, draw CF||AD.

            Now, FB = AB - AF = 48 - 24 = 24 cm.

            In triangle FBC, FC = BC = 26 cm

             So, it is an isosceles triangle.

             Also,CE is perpendicular to FB. So, E is the mid - point of FB.

             So,      

                     FE = 1/2FB = 1/2 *24 =12 cm

             Ina right-angled triangle CEF,

               CF^2 = FE^2 + CE^2             [by pythagoras theorem]

                26^2 = 12^2 +CE^2

               CE^2 = 676 - 144

               CE^2 = 532

               CE    = √532

Area of trapezium = 1/2(AB+DC)*CE

                            =1/2(48 + 24)*√532

                            = 1/2 * 72 *√532

                            = 36√532.

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