1. Problem: Given that RO = 2x + 6 and KC = 5x - 3 in parallelogram □ROCK, find x. Step 1: State the property used: In a parallelogram, the diagonals bisect each other, so RO = KC. Step 2: Set the expressions equal: $$2x + 6 = 5x - 3$$ Step 3: Solve for x: $$2x + 6 = 5x - 3$$ $$6 + 3 = 5x - 2x$$ $$9 = 3x$$ $$x = 3$$ 2. Problem: Given $$m\angle ROC = 10x + 12$$ and $$m\angle RKC = 13x - 24$$, find $$m\angle KRO$$. Step 1: Note that in parallelogram, opposite angles are equal and adjacent angles are supplementary. Step 2: Since ROC and RKC are angles formed by diagonals intersecting, and Y is midpoint, angles ROC and RKC are vertical angles and thus equal. Step 3: Set equal and solve for x: $$10x + 12 = 13x - 24$$ $$12 + 24 = 13x - 10x$$ $$36 = 3x$$ $$x = 12$$ Step 4: Find $$m\angle KRO$$. Since KRO is adjacent to ROC and RKC, and diagonals bisect each other, angles around point Y sum to 360 degrees. Step 5: Calculate $$m\angle ROC$$: $$10(12) + 12 = 120 + 12 = 132$$ degrees. Step 6: Since ROC and RKC are vertical angles, $$m\angle RKC = 132$$ degrees. Step 7: Angles around point Y sum to 360, so $$m\angle KRO = 360 - 132 - 132 = 96$$ degrees. 3. Problem: Given $$RY = 7x + 4$$ and $$YC = 10x - 2$$, find $$RC$$. Step 1: Since Y is midpoint of diagonal RC, $$RY = YC$$. Step 2: Set equal and solve for x: $$7x + 4 = 10x - 2$$ $$4 + 2 = 10x - 7x$$ $$6 = 3x$$ $$x = 2$$ Step 3: Calculate $$RY$$: $$7(2) + 4 = 14 + 4 = 18$$ Step 4: Since Y is midpoint, $$RC = 2 \times RY = 2 \times 18 = 36$$. 4. Problem: □ROCK is a rectangle and $$m\angle KCO = 4x - 6$$, find x. Step 1: In a rectangle, all angles are 90 degrees. Step 2: Set $$4x - 6 = 90$$ and solve: $$4x = 96$$ $$x = 24$$ 5. Problem: □ROCK is a rhombus and $$m\angle RYK = 4x - 10$$, find x. Step 1: In a rhombus, diagonals bisect angles and are perpendicular. Step 2: Since diagonals are perpendicular, $$m\angle RYK = 90$$ degrees. Step 3: Set $$4x - 10 = 90$$ and solve: $$4x = 100$$ $$x = 25$$ 6. Problem: □ROCK is a square, $$RK = 6x - 8$$, $$OC = 4x - 2$$, find $$KC$$. Step 1: In a square, all sides are equal and diagonals are equal. Step 2: Since RK and OC are sides and diagonals respectively, but RK is side and OC is diagonal, use Pythagoras theorem: Step 3: Diagonal length $$KC = OC$$. Step 4: Find x by equating diagonal to $$RK \sqrt{2}$$: $$OC = RK \sqrt{2}$$ $$4x - 2 = (6x - 8) \sqrt{2}$$ Step 5: Solve for x approximately: $$4x - 2 = (6x - 8) \times 1.414$$ $$4x - 2 = 8.484x - 11.312$$ $$-2 + 11.312 = 8.484x - 4x$$ $$9.312 = 4.484x$$ $$x \approx 2.076$$ Step 6: Calculate $$KC = OC = 4x - 2 = 4(2.076) - 2 = 8.304 - 2 = 6.304$$ 7. Problem: □ROCK is a rectangle, $$RC = 3x + 7$$ and $$KO = 8x - 13$$, find $$RC$$. Step 1: In a rectangle, diagonals are equal, so $$RC = KO$$. Step 2: Set equal and solve: $$3x + 7 = 8x - 13$$ $$7 + 13 = 8x - 3x$$ $$20 = 5x$$ $$x = 4$$ Step 3: Calculate $$RC$$: $$3(4) + 7 = 12 + 7 = 19$$ 8. Problem: □ROCK is a rhombus, $$m\angle KCR = 7x - 9$$ and $$m\angle OCR = 5x + 5$$, find x. Step 1: In a rhombus, diagonals bisect angles and are perpendicular, so $$m\angle KCR + m\angle OCR = 90$$ degrees. Step 2: Set equation: $$7x - 9 + 5x + 5 = 90$$ $$12x - 4 = 90$$ $$12x = 94$$ $$x = \frac{94}{12} = 7.833$$ Final answers: 1. $$x = 3$$ 2. $$m\angle KRO = 96^\circ$$ 3. $$RC = 36$$ 4. $$x = 24$$ 5. $$x = 25$$ 6. $$KC \approx 6.304$$ 7. $$RC = 19$$ 8. $$x \approx 7.833$$