Krishna
0

Step 1: Remember the area formula and name the areas of similar isosceles triangles.

          NOTE: Let A_1 and   A_2 be the areas of the triangle with

          lateral sides   a_1 = 2 and a_2 = 10 respectively

           A_1 = \frac{1}{2} (a_1)^2 sin(α)

                     A_2 = \frac{1}{2} (a_2)^2 sin(α)


Step 2: Find the ratio of the areas of the similar isosceles triangles.

           \frac{A_1}{A_2} = \frac{\frac{1}{2} (a_1)^2 sin(α)}{\frac{1}{2} (a_2)^2 sin(α)}

               \frac{A_1}{A_2} = \frac{(a_1)^2}{(a_2)^2} ...........(1)


Step 3: Find the height of the second isosceles triangle

            NOTE: Use Pythagoras theorem in the right triangle

                         a^2 = (\frac{b}{2})^2 + h^2

                         h = \sqrt{a^2 - (\frac{b}{2})^2}

                         h = \sqrt{10^2 - (\frac{12}{2})^2}

                         h = \sqrt{64} = 8


Step 4: Find the area of the second isosceles triangle A_2

                     A_2 = \frac{1}{2}bh

                     A_2 =   \frac{1}{2} (12 * 8)

                    {A_2} = 48 unit^2


Step 5: Find the area of the first isosceles triangle by using the ratio of the areas

                     \frac{A_1}{A_2} = \frac{a_1}{a_2}  (Since equation (1) see step 2 )

        A_2 = 48 unit^2,   a_1 = 2 , a_2 = 10

                       \frac{A_1}{48} = \frac{2^2}{10^2}  

                       A_1 = \frac{4}{100} *48

                       A_1 = 1.92 units^2