 Krishna
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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