The length of an arc of a circle is given by the formula L = (theta)/(360^) x 2pi r. Calculate the angle theta if the arc length is 1 (19)/(36) cm and the radius is 3.5 cm. Use pi = (22)/(7).

Here are the solutions to the questions.

Question 31:
The length of an arc of a circle is given by the formula $L = \frac{\theta}{360^\circ} \times 2\pi r$.

Step 1: Convert the given arc length and radius to fractions.

$$L = 1 \frac{19}{36} \text{ cm} = \frac{36 \times 1 + 19}{36} \text{ cm} = \frac{55}{36} \text{ cm}$$ $$r = 3.5 \text{ cm} = \frac{7}{2} \text{ cm}$$

We are given $\pi = \frac{22}{7}$.

Step 2: Substitute the values into the arc length formula.

$$\frac{55}{36} = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times \frac{7}{2}$$

Step 3: Simplify the right side of the equation.

$$\frac{55}{36} = \frac{\theta}{360^\circ} \times \left(2 \times \frac{22}{7} \times \frac{7}{2}\right)$$ $$\frac{55}{36} = \frac{\theta}{360^\circ} \times 22$$

Step 4: Solve for $\theta$.

$$\theta = \frac{55}{36} \times \frac{360^\circ}{22}$$ $$\theta = \frac{55 \times 360^\circ}{36 \times 22}$$ $$\theta = \frac{55 \times 10^\circ}{22}$$ $$\theta = \frac{5 \times 10^\circ}{2}$$ $$\theta = 5 \times 5^\circ$$ $$\theta = 25^\circ$$

The final answer is $\boxed{\text{C. } 25^\circ}$.

Question 32:
In the diagram, $PU \parallel SR$ and $PS \parallel TR$. This means that $PTRS$ is a parallelogram.
We are given that the area of $\triangle SUR = 24 \text{ cm}^2$ and $SR = 8 \text{ cm}$.

Step 1: Find the height of $\triangle SUR$.
Let $h$ be the perpendicular height from $U$ to the base $SR$.
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

$$24 \text{ cm}^2 = \frac{1}{2} \times SR \times h$$ $$24 \text{ cm}^2 = \frac{1}{2} \times 8 \text{ cm} \times h$$ $$24 \text{ cm}^2 = 4 \text{ cm} \times h$$ $$h = \frac{24 \text{ cm}^2}{4 \text{ cm}} = 6 \text{ cm}$$

Step 2: Calculate the area of parallelogram $PTRS$.
Since $PU \parallel SR$, the perpendicular distance between the parallel lines $PU$ and $SR$ is $h$.
The parallelogram $PTRS$ has base $SR = 8 \text{ cm}$ and height $h = 6 \text{ cm}$.
The area of a parallelogram is given by $\text{base} \times \text{height}$.

$$\text{Area of } PTRS = SR \times h$$ $$\text{Area of } PTRS = 8 \text{ cm} \times 6 \text{ cm}$$ $$\text{Area of } PTRS = 48 \text{ cm}^2$$

The final answer is $\boxed{\text{B. } 48 \text{ cm}^2}$.

Question 33:
In the diagram, $PQR$ is a circle with center $O$. $OP$, $OQ$, and $OR$ are radii of the circle.
We are given $\angle OPQ = 48^\circ$.

Step 1: Find $\angle POQ$.
Since $OP$ and $OQ$ are radii, $\triangle OPQ$ is an isosceles triangle with $OP = OQ$.
Therefore, the base angles are equal: $\angle OQP = \angle OPQ = 48^\circ$.
The sum of angles in a triangle is $180^\circ$.

$$\angle POQ = 180^\circ - (\angle OPQ + \angle OQP)$$ $$\angle POQ = 180^\circ - (48^\circ + 48^\circ)$$ $$\angle POQ = 180^\circ - 96^\circ$$ $$\angle POQ = 84^\circ$$

Step 2: Find the value of $m^\circ$, which is $\angle ROQ$.
From the diagram, $R, O, P$ lie on a straight line, which means $RP$ is a diameter.
Angles on a straight line sum to $180^\circ$.

$$\angle ROQ + \angle POQ = 180^\circ$$ $$$$