# Can you find area of the Blue shaded region? | (Semicircle) | #math #maths | #geometry

TLDRIn this geometry-focused video, the presenter tackles the problem of finding the area of a blue shaded region within a semicircle. Given the lengths of two vertical segments and the length of segment EF, the process involves applying the Pythagorean theorem to find the radii of the semicircles and the value of x. The area of the blue region is then deduced by subtracting the area of triangle OED from the sector AOD, resulting in an approximate area of 1.77 square units. The video concludes with a reminder to subscribe for more educational content.

### Takeaways

- 📏 The problem involves finding the area of a blue shaded region within a semicircle confined by two vertical segments, EF and FC.
- 🔍 The lengths of the segments are given: DE is 2 units, CF is 1 unit, and EF is 3 units.
- 📐 The first step is to find the radius of the semicircle to calculate the area of the blue region.
- 📝 The Pythagorean theorem is used to establish two equations involving the radii R and r, and the unknown segment EO.
- ✂️ By applying the identity for the square of a difference, the equations are simplified and the unknown x is found to be 1 unit.
- 🔢 Substituting x = 1 into the equations yields the radii R and r, both being the square root of 5.
- 📐 An angle θ is identified in the right triangle OED, and its tangent is calculated to be 2, leading to an angle of approximately 63.435°.
- 🔄 The area of the blue shaded region is the difference between the area of sector AOD and the area of triangle OED.
- 📐 The area of triangle OED is calculated using the formula for the area of a triangle, resulting in 1 square unit.
- 📐 The area of sector AOD is calculated using the sector area formula, yielding approximately 2.77 square units.
- 📝 The final step is to subtract the area of triangle OED from the area of sector AOD, resulting in an area of approximately 1.77 square units for the blue shaded region.

### Q & A

### What is the main task of the video?

-The main task of the video is to calculate the area of the blue shaded region in a semicircle with specific given dimensions.

### What are the lengths of the vertical segments DE and FC?

-The length of segment DE is 2 units, and the length of segment FC is 1 unit.

### What is the length of segment EF?

-The length of segment EF is 3 units.

### Why is it necessary to find the radius before calculating the area of the blue shaded region?

-The radius is necessary because the area of the sector of the semicircle and the right triangle within it are dependent on the radius for their calculations.

### What theorem is used to relate the sides of the right triangles in the video?

-The Pythagorean theorem is used to relate the sides of the right triangles in the video.

### What identity is applied to simplify the equation involving the binomial (3 - x)² + 1²?

-The identity (a - b)² = a² - 2ab + b² is applied to simplify the equation.

### How is the value of x determined in the script?

-The value of x is determined by solving the equation x² - 6x + 10 = x² + 4, which simplifies to x = 1 unit.

### What is the radius of the semicircle after solving for x?

-After solving for x, the radius of the semicircle is found to be √5 units.

### How is the angle Theta calculated in the video?

-Angle Theta is calculated using the tangent of the angle, which is the ratio of the opposite side (2 units) to the adjacent side (1 unit), resulting in an angle of approximately 63.435°.

### What is the formula used to calculate the area of a sector of a circle?

-The formula used to calculate the area of a sector is (angle in degrees / 360) * π * r², where r is the radius of the circle.

### What is the final calculated area of the blue shaded region?

-The final calculated area of the blue shaded region is approximately 1.77 square units.

### Outlines

### 📐 Introduction to the Geometry Problem

The video begins with an introduction to a geometry problem involving two vertical segments, E and FC, confined within a semicircle. The segments have specific lengths, with D being 2 units and CF being 1 unit. The task is to calculate the area of a blue shaded region. The presenter emphasizes the importance of finding the radius before proceeding with the area calculation. The diagram provided is noted to not be perfectly to scale, and the audience is encouraged to like, subscribe, and follow along.

### 🔍 Applying the Pythagorean Theorem to Find the Radius

The presenter connects the center of the semicircle, O, with points C and D to define the radii OC and OD, labeled as R and r respectively. A segment EO is introduced, with a length of x units, and the remaining segment OD is calculated as 3 - x. The Pythagorean theorem is applied to two right triangles, OEC and OFC, resulting in two equations involving x, R, and r. By using the identity for the square of a binomial, the presenter simplifies the equations and equates the right-hand sides to find the value of x, which is determined to be 1 unit.

### 📏 Calculating the Radius and Angle for the Geometry Problem

With x found to be 1 unit, the presenter substitutes this value into one of the equations to find the radius R, which is calculated to be √5 units. The radius r is also inferred to be √5 units. The presenter then focuses on the right triangle OED, introduces an angle Theta, and uses the tangent function to find the angle's measure, which is approximately 63.435°. The goal now is to calculate the area of the blue shaded region, which is the difference between the area of a sector AOD and the area of triangle OED.

### 📐 Final Calculations for the Area of the Shaded Region

The presenter calculates the area of triangle OED using the formula for the area of a triangle (1/2 * base * height), resulting in an area of one square unit. Next, the area of sector AOD is calculated using the sector area formula (angle in degrees / 360 * π * radius squared), yielding an approximate area of 2.77 square units. Finally, the area of the blue shaded region is determined by subtracting the area of triangle OED from the area of sector AOD, resulting in an area of approximately 1.77 square units. The video concludes with a reminder to subscribe for more content.

### Mindmap

### Keywords

### 💡Semicircle

### 💡Vertical segments

### 💡Pythagorean theorem

### 💡Radius

### 💡Area of a triangle

### 💡Sector

### 💡Tangent

### 💡Angle Theta

### 💡Area of the blue shaded region

### 💡Right triangle

### Highlights

Introduction to the problem of finding the area of the blue shaded region within a semicircle.

Identification of the two vertical segments, DE and FC, confined within the semicircle.

Lengths of segments DE and CF are given as 2 units and 1 unit, respectively.

Length of segment EF is determined to be 3 units.

The importance of finding the radius before calculating the area of the blue shaded region.

Drawing radii from the center of the semicircle to points C and D, labeling them as R and r.

Segment EO is labeled as X units, with the remaining segment OD being 3 - X units.

Application of the Pythagorean theorem to the right triangle OED.

Equation 1: x² + 2 = R², derived from the Pythagorean theorem.

Application of the Pythagorean theorem to the right triangle OFC.

Equation 2: (3 - x)² + 1² = r², derived from the Pythagorean theorem.

Use of the identity for (a - b)² to simplify Equation 2.

Equating Equation 1 and Equation 2 to find the value of x.

Solving for x, which is determined to be 1 unit.

Substituting x = 1 into Equation 1 to find the radius R.

Calculation of the radius R, which is found to be √5 units.

Observation that the area of the blue shaded region is the sector AOD minus the triangle OED.

Calculation of the area of triangle OED using the formula for the area of a triangle.

Determination of the angle Theta using the tangent function.

Calculation of the area of sector AOD using the sector area formula.

Final calculation of the area of the blue shaded region as approximately 1.77 square units.

Conclusion of the video with a reminder to subscribe for more educational content.