Question: 1.Solve for x in the diagram. \n_Diagram: Two parallel lines with a transversal cutting through. Three angles between them: 26\degree, 64\degree, and x\degree at the bottom._\n\nA. 26\degree \nB. 28\degree \nC. 38\degree \nD. 64\degree\n\n*2.* Two interior angles of a hexagon are 120\degree and 180\degree. If the remaining angles are the same, find the measure of each equal angle. \nA. 180\degree \nB. 120\degree \nC. 105\degree \nD. 90\degree\n\n*3.* A rectangle with length 11cm has a perimeter equal to 38cm. Calculate the area of the rectangle. \nA. 80cm\textsuperscript{2} \nB. 88cm\textsuperscript{2} \nC. 292cm\textsuperscript{2} \nD. 418cm\textsuperscript{2}\n\n*4.* The sum of interior angles of a regular polygon is 1260\degree. How many sides has the polygon? \nA. 6 \nB. 7 \nC. 8 \nD. 9\n\n*5.* A basic school learner decided to draw two intersecting lines and name them OX and OY. She placed a compass at O and drew an arc to intersect OX and OY at A and B respectively. With the compass point at A with a reasonable radius, she drew an arc in the space between XY and repeated same with compass point at B intersecting at C. What is the locus drawn from O through C? \nA. Perpendicular bisector. \nB. Angle bisector. \nC. Circle bisector. \nD. Intersection bisector.

1. **Problem:** Solve for $x$ in the diagram with two parallel lines cut by a transversal, angles given as $26^\circ$, $64^\circ$, and $x^\circ$.\n\n2. **Formula and rules:** Alternate interior angles and linear pairs sum to $180^\circ$.\n\n3. **Work:** The angles on a straight line sum to $180^\circ$, so $x + 64^\circ = 90^\circ$ (since $26^\circ$ and $64^\circ$ are complementary to $90^\circ$).\n\n4. Calculate $x = 90^\circ - 64^\circ = 26^\circ$.\n\n5. **Answer:** $x = 26^\circ$ (Option A).\n\n---\n\n1. **Problem:** Two interior angles of a hexagon are $120^\circ$ and $180^\circ$. The remaining four angles are equal. Find each equal angle.\n\n2. **Formula:** Sum of interior angles of a polygon = $180^\circ(n-2)$, for hexagon $n=6$, sum = $180^\circ \times 4 = 720^\circ$.\n\n3. **Work:** Sum of known angles = $120^\circ + 180^\circ = 300^\circ$. Remaining sum = $720^\circ - 300^\circ = 420^\circ$.\n\n4. Each equal angle = $\frac{420^\circ}{4} = 105^\circ$.\n\n5. **Answer:** $105^\circ$ (Option C).\n\n---\n\n1. **Problem:** Rectangle with length $11$ cm and perimeter $38$ cm. Find area.\n\n2. **Formula:** Perimeter $P = 2(l + w)$, Area $A = l \times w$.\n\n3. **Work:** $38 = 2(11 + w)$\n\n4. Divide both sides by 2: $$\cancel{2} \times (11 + w) = \frac{38}{\cancel{2}} \Rightarrow 11 + w = 19$$\n\n5. Solve for $w$: $w = 19 - 11 = 8$ cm.\n\n6. Calculate area: $A = 11 \times 8 = 88$ cm$^2$.\n\n7. **Answer:** $88$ cm$^2$ (Option B).\n\n---\n\n1. **Problem:** Sum of interior angles of a regular polygon is $1260^\circ$. Find number of sides $n$.\n\n2. **Formula:** Sum of interior angles = $180^\circ(n-2)$.\n\n3. **Work:** $1260 = 180(n-2)$\n\n4. Divide both sides by 180: $$\cancel{180} \times (n-2) = \frac{1260}{\cancel{180}} \Rightarrow n-2 = 7$$\n\n5. Solve for $n$: $n = 7 + 2 = 9$.\n\n6. **Answer:** $9$ sides (Option D).\n\n---\n\n1. **Problem:** Two intersecting lines OX and OY, arcs drawn from O intersecting OX and OY at A and B, arcs from A and B intersect at C. Find the locus drawn from O through C.\n\n2. **Explanation:** The construction describes the method to bisect the angle formed by OX and OY.\n\n3. The locus from O through C is the angle bisector.\n\n4. **Answer:** Angle bisector (Option B).