Solve the geometry problems in the image.

Steps to Solve

1. Area of quadrilateral PQRS

The quadrilateral $PQRS$ can be divided into two triangles: $\triangle PQS$ and $\triangle QRS$. The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

Area of $\triangle PQS = \frac{1}{2} \times 10 \times 4 = 20$ Area of $\triangle QRS = \frac{1}{2} \times 6 \times 3 = 9$

Area of quadrilateral $PQRS = $ Area of $\triangle PQS + $ Area of $\triangle QRS = 20 + 9 = 29$.

2. Area of trapezoid ABCD

Given vertices $A(2, 2)$, $B(4, 6)$, $C(4, -3)$, and $D(2, -1)$. The bases of the trapezoid are the vertical segments $AD$ and $BC$. The height is the horizontal distance between the parallel sides. Length of $AD = |2 - (-1)| = 3$. Length of $BC = |6 - (-3)| = 9$. Height $= |4 - 2| = 2$. Area of trapezoid $= \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (3 + 9) \times 2 = 12$.

3. Area of a circle

Circumference $C = 20\pi$. Since $C = 2\pi r$, we have $2\pi r = 20\pi$, which gives $r = 10$. Area of a circle $A = \pi r^2 = \pi (10)^2 = 100\pi$.

4. Area of the quadrilateral

The quadrilateral can be divided into a rectangle and a right triangle. The base of the right triangle is $7.8 - 5.5 = 2.3$ inches. The height of the right triangle can be found using trigonometry: $\tan(45^\circ) = \frac{\text{height}}{2.3}$. Since $\tan(45^\circ) = 1$, the height is $2.3$ inches. The height of the rectangle is also $2.3$ inches. Area of the rectangle $= 5.5 \times 2.3 = 12.65$ square inches. Area of the triangle $= \frac{1}{2} \times 2.3 \times 2.3 = \frac{1}{2} \times 5.29 = 2.645$ square inches. Area of the quadrilateral $= 12.65 + 2.645 = 15.295 \approx 15$ square inches (to the nearest square inch).

5. Perimeter of the rectangular garden

Area of the rectangle $= 128$ square yards. Let the width be $w$ and the length be $l$. We are given that $l = 2w$. The area is $A = lw = (2w)w = 2w^2 = 128$. So, $w^2 = 64$, which gives $w = 8$ yards. The length $l = 2w = 2(8) = 16$ yards. Perimeter $= 2(l + w) = 2(16 + 8) = 2(24) = 48$ yards.

6. Perimeter of the walking track

The track consists of a square with side 12 yards and two semicircles. Two semicircles make up one full circle. The diameter of the circle is equal to the side of the square, so the diameter is 12 yards and the radius is 6 yards. The perimeter of the two semicircles (one circle) is $2\pi r = 2\pi(6) = 12\pi$. The perimeter of the square is normally $4 \times 12$, but two of the sides are replaced by the semicircles, so we only have two sides of the square in the walking track perimeter. So, the perimeter of the two sides of the square is $2 \times 12 = 24$ yards. Total perimeter of the track $= 24 + 12\pi$ yards.