What is the area of the square (in cm 2) whose vertices lie on a circle of radius 5 cm?
50
Understanding the Problem: Area of an Inscribed Square
The problem asks for the area of a square whose vertices are on a circle with a radius of 5 cm. This means the square is inscribed within the circle. When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.
Connecting the Circle's Dimensions to the Square's Dimensions
We are given the radius of the circle:
- Radius ($r$) = 5 cm
The diameter of the circle ($d$) is twice the radius:
- Diameter ($d$) = $2 \times r = 2 \times 5 \text{ cm} = 10 \text{ cm}$
Since the square's vertices lie on the circle, the diagonal of the square is equal to the diameter of the circle.
- Diagonal of the square = Diameter of the circle = 10 cm
Calculating the Side Length of the Square
Let the side length of the square be $s$. In a square, the relationship between the diagonal ($d$) and the side length ($s$) is given by the Pythagorean theorem or the diagonal formula:
$d = s\sqrt{2}$
We know the diagonal is 10 cm, so we can set up the equation:
$10 = s\sqrt{2}$
To find the side length $s$, we rearrange the equation:
$s = \frac{10}{\sqrt{2}}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:
$s = \frac{10 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \text{ cm}$
Calculating the Area of the Square
The area of a square is calculated by squaring its side length:
Area = $s^2$
Substitute the value of $s$ we found:
Area = $(5\sqrt{2})^2$
Area = $(5)^2 \times (\sqrt{2})^2$
Area = $25 \times 2$
Area = $50 \text{ cm}^2$
Thus, the area of the square whose vertices lie on a circle of radius 5 cm is 50 cm$^2$.
Summary of Steps
- Find the diameter of the circle.
- Equate the diameter to the diagonal of the inscribed square.
- Use the diagonal to find the side length of the square.
- Calculate the area of the square using the side length.
| Measurement | Value |
|---|---|
| Circle Radius ($r$) | 5 cm |
| Circle Diameter ($d$) | 10 cm |
| Square Diagonal | 10 cm |
| Square Side ($s$) | $5\sqrt{2}$ cm |
| Square Area | 50 cm$^2$ |
Revision Table: Key Formulas
| Shape | Formula | Notes |
|---|---|---|
| Circle Diameter | $d = 2r$ | $r$ is radius |
| Square Diagonal | $d_{sq} = s\sqrt{2}$ | $s$ is side length |
| Square Area | Area $= s^2$ | $s$ is side length |
Additional Information: Inscribed Shapes
When a shape is inscribed in a circle, all of its vertices lie on the circle's circumference. For a square inscribed in a circle:
- The center of the circle is also the center of the square.
- The radius of the circle is half the diagonal of the square.
- The diameter of the circle is equal to the diagonal of the square.
This relationship is crucial for solving problems involving squares inscribed in circles or circles circumscribed around squares.
Important Questions from Plane Figures
If length of a rectangle is increased to its three times and breadth is decreased to its half, then the ratio of the area of given rectangle to the area of new rectangle is:
The width of the path around a square field is 4.5 m and its area is 105.75 m 2. Find the cost of fencing the field at the rate of Rs. 100 per meter.
The circumcentre of an equilateral triangle is at a distance of 3.2 cm from the base of the triangle. What is the length (in cm) of each of its altitudes?
The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m 2) of the path is:
Take \(\left(\pi=\frac{22}{7}\right)\)
If the radius of a circle is decreased by 11% then the total decrease in the area of the circle is given as:
