Area of second figure is 56.6 square meter
Solution:
Given that,
Two given figures are similar
Perimeter of first figure = 20 m
Perimeter of second figure = 34 m
Area of first figure = 19.6 square meter
Area of second figure = ?
Find the ratio of perimeter of first figure to second
[tex]\frac{\text{perimeter of first figure}}{\text{perimeter of second figure}} = \frac{20}{34} = \frac{10}{17}[/tex]
In two similar triangles: The perimeters of the two triangles are in the same ratio as the sides
Therefore,
[tex]\frac{\text{perimeter of first figure}}{\text{perimeter of second figure}} = \frac{10}{17}[/tex]
If two figures are similar, then ratio of their perimeters is equal to ratio of their sides
[tex]\frac{\text{side of first figure}}{\text{side of second figure}} = \frac{10}{17}[/tex]
Taking square on both sides, then it becomes ratio of their areas
Because,
[tex]area = side^2[/tex]
Therefore,
[tex](\frac{\text{side of first figure}}{\text{side of second figure}})^2 = (\frac{10}{17})^2\\\\\frac{\text{area of first figure}}{\text{area of second figure}} = (\frac{10}{17})^2\\\\\frac{19.6}{\text{area of second figure}} = \frac{100}{289}\\\\\text{area of second figure } = 19.6 \times \frac{289}{100}\\\\\text{area of second figure } = 56.644 \approx 56.6[/tex]
Thus area of second figure is 56.6 square meter
