The size of the perimeter of the square ABCD is equal to 100 cm. The length of the segment MN is equal to 5 cm and the triangle MNC is isosceles. Find the area of the pentagon ABNMC. (The picture on that site isn't to scale. This is.)The square's perimeter is 100 cm, so each side of the square is 25 cm. The area of the square is (25 cm)² or 625 cm² We need to subtract the area of the triangle. It is a right triangle, and it isosceles, so MC = CN. We use the Pythagorean theorem to find MC and CN. MC² + CN² = MN² MC² + MC² = 5² 2MC² = 25 MC² = 12.5 MC = = CN Area of the triangle = BASE×HEIGHT = MC×CN = × = = (12.5) = 6.25 cm² Area of pentagon = Area of square - Area of triangle = 625 cm² - 6.25 cm² = 618.75 cm² ----------------------------------------------------------- I won't draw the other one. Initially the rectangular prism on the left was full of water. It measures 2cm×4cm×10cm Volume = length × width × height = 2cm × 4cm × 10cm = 80cm³ so the volume of water in the original container is 80cm³ The first container into which part of it was poured to a height of h is a rectangular prism like the original except that it is only filled to a height of h cm. Volume of water = length × width × height = 2cm × 4cm × h cm = 8h cm³ The second container into which the other part of it was poured to a height of h cm is a cylinder with radius 1 cm. Volume of water = p × (radius)² × (height) = p(1)²h = ph = (3.14)h so the volume of water in the cylinder is 3.14h cm³. water in original = water in 1st + water in 2nd 80 = 8h + 3.14h 80 = 11.14h = h 7.2 = h So the height is 7.2 cm to the nearest tenth of a centimeter. Edwin