3) Calculate the area and perimeter of an isosceles trapezoid whose bases measure \( 2 / 5 \) and \( 4 / 5 \), and its height is \( 3 / 2 \).

To calculate the area and perimeter of the isosceles trapezoid, we can use the following formulas: 1. **Area of a trapezoid**: \[ A = \frac{1}{2} \times (b_1 + b_2) \times h \] where \( b_1 \) and \( b_2 \) are the lengths of the two bases, and \( h \) is the height. 2. **Perimeter of a trapezoid**: \[ P = b_1 + b_2 + 2s \] where \( s \) is the length of the non-parallel sides (the legs) of the trapezoid. ### Given: - \( b_1 = \frac{2}{5} \) - \( b_2 = \frac{4}{5} \) - \( h = \frac{3}{2} \) ### Step 1: Calculate the Area Substituting the values into the area formula: \[ A = \frac{1}{2} \times \left(\frac{2}{5} + \frac{4}{5}\right) \times \frac{3}{2} \] \[ A = \frac{1}{2} \times \left(\frac{6}{5}\right) \times \frac{3}{2} \] \[ A = \frac{1}{2} \times \frac{18}{10} = \frac{9}{10} \] ### Step 2: Calculate the Length of the Legs To find the length of the legs \( s \), we can use the Pythagorean theorem. The legs of the trapezoid can be found by considering the right triangle formed by the height and half the difference of the bases. 1. Calculate the difference of the bases: \[ \text{Difference} = b_2 - b_1 = \frac{4}{5} - \frac{2}{5} = \frac{2}{5} \] 2. Half of the difference: \[ \text{Half Difference} = \frac{1}{2} \times \frac{2}{5} = \frac{1}{5} \] 3. Now, we can use the Pythagorean theorem to find \( s \): \[ s = \sqrt{h^2 + \left(\frac{1}{5}\right)^2} \] \[ s = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{1}{5}\right)^2} \] \[ s = \sqrt{\frac{9}{4} + \frac{1}{25}} \] To add these fractions, we need a common denominator: \[ \frac{9}{4} = \frac{225}{100}, \quad \frac{1}{25} = \frac{4}{100} \] \[ s = \sqrt{\frac{225 + 4}{100}} = \sqrt{\frac{229}{100}} = \frac{\sqrt{229}}{10} \] ### Step 3: Calculate the Perimeter Now we can calculate the perimeter: \[ P = b_1 + b_2 + 2s \] \[ P = \frac{2}{5} + \frac{4}{5} + 2 \times \frac{\sqrt{229}}{10} \] \[ P = \frac{6}{5} + \frac{2\sqrt{229}}{10} \] \[ P = \frac{6}{5} + \frac{\sqrt{229}}{5} = \frac{6 + \sqrt{229}}{5} \] ### Final Results - **Area**: \( \frac{9}{10} \) - **Perimeter**: \( \frac{6 + \sqrt{229}}{5} \)