Question

Find the area of each regular polygon. Round your answer to the nearest tenth if necessary​

Answer 1

Answer:

Step-by-step explanation:

13). Area of a square = (Side)²

                                   = (BC)²

     Since, diagonals of a square bisect each other at 90°,

     ΔBOC is a right triangle.

     By applying Pythagoras theorem in the given triangle,

     BC² = OB² + OC²

     BC² = 2(OB)²

     BC² = 2(7√2)²

     BC = $\sqrt{196}$

     Area of square ABCD = (BC)²

                                          = (√196)²

                                          = 196 units²

14). Measure of interior angles of the regular hexagon = 120°

     Area of the regular hexagon = $\frac{3\sqrt{3}}{2}(\text{side})^2$

      From the given picture,

      m∠BAC = m∠ABC = m∠ACB = 60°

      Therefore, ΔABC is an isosceles triangle.

      And all sides of this triangle will be equal in measure.

      AB = AC = BC = 9 units

      Area of the given regular hexagon = $\frac{3\sqrt{3}}{2}(9)^2$

                                                                 = 210.44 square units

                                                                 ≈ 210.4 square units