1. **Problem 21: Find the measures of angles in rectangle ABCD.** A rectangle has four right angles, so each angle measures 90 degrees. - m\angle1 = 90 - m\angle2 = 90 - m\angle3 = 90 - m\angle4 = 90 Angles 5, 6, and 7 are not specified in the problem, but if they are angles formed by diagonals or other lines inside the rectangle, we need more information. Since no further data is given, we assume they are also right angles or related to the rectangle's properties. Typically, diagonals in a rectangle are equal and bisect each other, forming two congruent triangles. 2. **Problem 22: Rhombus-shaped parallelogram with diagonals intersecting at O.** (a) The best name is **Rhombus** because all sides are equal and diagonals bisect each other at right angles. (b) Given AO = x, CO = y, and side AD = 17. In a rhombus, diagonals bisect each other at right angles, so by the Pythagorean theorem: $$AD^2 = AO^2 + CO^2$$ $$17^2 = x^2 + y^2$$ $$289 = x^2 + y^2$$ Without more info, we cannot find exact values of x and y. 3. **Problem 23: Rectangle ABCD with diagonals intersecting. AB = 7, DC = 24.** (a) The best name is **Rectangle**. (b) Diagonals of a rectangle are equal and bisect each other. Length AB = 7 (width), DC = 24 (length). Diagonal length $d$ is: $$d = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$ Since diagonals bisect each other, each segment is half the diagonal: $$x = y = z = \frac{25}{2} = 12.5$$ 4. **Problem 24: Quadrilateral MATH with MT and AH bisecting each other at R and MR = HR.** Since the diagonals bisect each other, MATH must be a parallelogram. If MR = HR, then the diagonals bisect each other equally, which is true for parallelograms. No information about right angles or equal sides to confirm rectangle or square. Answer: A. I only (Parallelogram only). 5. **Problem 25: Wire length for rectangle with diagonals.** Rectangle dimensions: length = 40 in, width = 9 in. Wire needed = perimeter + 2 diagonals. Perimeter: $$P = 2(40 + 9) = 2(49) = 98$$ Diagonal length: $$d = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41$$ Two diagonals: $$2d = 2 \times 41 = 82$$ Total wire length: $$98 + 82 = 180$$ **Final answers:** 1. m\angle1 = 90, m\angle2 = 90, m\angle3 = 90, m\angle4 = 90, m\angle5 = unknown, m\angle6 = unknown, m\angle7 = unknown 2. (a) Rhombus, (b) $x^2 + y^2 = 289$ 3. (a) Rectangle, (b) $x = y = z = 12.5$ 4. A. I only (Parallelogram) 5. 180 units of wire needed