Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 359: 6

15.708

Given right triangle $(3\sqrt 5)^{2}$ =$x^{2}$ + $2x^{2}$ 45 = 5$x^{2}$ $x^{2}$ = $\frac{45}{5}$ = 9 x = 3 The perimeter of given triangle is x + 2x+ 3$\sqrt 5$ Put the value of x 3+6+3$\sqrt 5$ = 9 + 3$\sqrt 5$ = 3(3+$\sqrt 5$) = 15.708