Solve the geometry problems in the image.

Steps to Solve

1. Question 1: Find the perimeter of the rectangle

The sides of the rectangular park are given as $(x+1)$ and $(x-1)$. The perimeter of a rectangle is given by $2(length + width)$. So, the perimeter is $2[(x+1) + (x-1)] = 2(2x) = 4x$.

2. Question 2: Find the side of the equilateral triangle

The side of the square PQRS is 1.2 cm. Perimeter of the square = $4 \times side = 4 \times 1.2 = 4.8$ cm. Let the side of the equilateral triangle be $a$. Perimeter of the equilateral triangle = $3a$. Since both perimeters are the same, $3a = 4.8$. Therefore, $a = \frac{4.8}{3} = 1.6$ cm.

3. Question 3: Find the length of the rectangular park

The breadth of the rectangular park is 15.4 m. The length of the wire required for fencing is 82 m, which is the perimeter. Let the length of the park be $l$. The perimeter is $2(l + breadth)$, so $2(l + 15.4) = 82$. $l + 15.4 = \frac{82}{2} = 41$. Therefore, $l = 41 - 15.4 = 25.6$ m.

4. Question 4: Find the circumference of the circle

The diameter of the circle is 7 cm. The circumference of a circle is given by $\pi d$, where $d$ is the diameter. Therefore, the circumference is $7\pi$ cm.

5. Question 5: Find the ratio of circumferences

The ratio of the radii of two circles is 2:5. Let the radii be $2r$ and $5r$ respectively. Circumference of the first circle is $2\pi(2r) = 4\pi r$. Circumference of the second circle is $2\pi(5r) = 10\pi r$. The ratio of their circumferences is $\frac{4\pi r}{10\pi r} = \frac{4}{10} = \frac{2}{5}$. Therefore, the ratio is 2:5.

6. Question 6: Find the perimeter of the region

The sides of the square are surrounded by four semicircles each of radius 10.5 cm. This means, each side of the square serves as diameter of the semicircle. Thus the side of square is $2 \times radius = 2 \times 10.5 = 21$cm. Each semicircle's arc length is half of the circle's circumference $= \frac{1}{2} \times 2 \pi r = \pi r = \frac{22}{7} \times 10.5 = 22 \times 1.5 = 33$ cm. The perimeter of the region is the sum of the lengths of the four semicircles $= 4 \times 33 = 132$ cm.

7. Question 7: Find the radius of the circular ring

An equilateral triangle-shaped wire is rebent to form a circular ring. Let the side of the equilateral triangle be $13.2$ cm (given in the figure). The perimeter of the triangle = $3 \times 13.2 = 39.6$ cm. This length is equal to the circumference of the circular ring, so $2\pi r = 39.6$. $r = \frac{39.6}{2\pi} = \frac{39.6}{2 \times \frac{22}{7}} = \frac{39.6 \times 7}{2 \times 22} = \frac{39.6 \times 7}{44} = 0.9 \times 7 = 6.3$ cm.

8. Question 8: Find the area of the parallelogram

One side of a parallelogram is 16 cm, and the distance from this side to the opposite side (height) is 7.5 cm. The area of a parallelogram is given by $base \times height = 16 \times 7.5 = 120$ cm$^2$.

9. Question 9: Find the area of the circle

The difference between the circumference and the radius of a circle is 37 cm. $2\pi r - r = 37$, so $r(2\pi - 1) = 37$. $r(2 \times \frac{22}{7} - 1) = 37$. $r(\frac{44}{7} - 1) = 37$. $r(\frac{44-7}{7}) = 37$. $r(\frac{37}{7}) = 37$. $r = 7$ cm. The area of the circle is $\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154$ cm$^2$.

10. Question 10: Find the area of the shaded region

O is the center. Two concentric circles are shown. Radius of larger circle is 14 cm and radius of smaller circle is 7 cm. The area of the shaded region = Area of larger circle - Area of smaller circle. Area of larger circle $= \pi (14)^2 = \frac{22}{7} \times 196 = 22 \times 28 = 616$ cm$^2$. Area of smaller circle $= \pi (7)^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154$ cm$^2$. Area of shaded region $= 616 - 154 = 462$ cm$^2$.