Segment QR of length r is a tangent at Q to a circle of radius r with centre at P. What is the area of the part of the triangle PQR, which is outside the circular region?
r 2/2 - πr 2/8
Understanding the Geometry Problem
The problem asks for the area of a specific part of triangle PQR: the part that lies outside a circular region. We are given that QR is a segment of length \(r\) which is tangent to a circle at point Q. This circle has its centre at P and a radius of \(r\).
Let's break down the given information:
- The circle has centre P and radius \(r\).
- QR is a tangent to the circle at Q.
- The length of QR is \(r\).
A key property of tangents is that the radius drawn to the point of tangency is perpendicular to the tangent at that point. In this case, PQ is the radius to the point of tangency Q. Therefore, the radius PQ is perpendicular to the tangent QR at Q. This means that the angle ∠PQR is 90 degrees.
So, triangle PQR is a right-angled triangle with the right angle at Q.
Calculating the Area of Triangle PQR
For a right-angled triangle, the area is given by half the product of its perpendicular sides (legs). In triangle PQR, the legs are PQ and QR.
We know the following lengths:
- PQ = radius of the circle = \(r\).
- QR = length of the tangent segment = \(r\).
The area of triangle PQR is:
Area\((PQR)\) = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area\((PQR)\) = \(\frac{1}{2} \times PQ \times QR\)
Area\((PQR)\) = \(\frac{1}{2} \times r \times r\)
Area\((PQR)\) = \(\frac{r^2}{2}\)
Identifying the Circular Region Inside the Triangle
The circular region we are concerned with has its centre at P and radius \(r\). The part of triangle PQR that is inside this circular region is a sector of the circle with centre P. The boundaries of this sector are defined by the lines PQ and PR, and the arc of the circle between Q and the point where PR intersects the circle (let's call this point S).
To find the area of this sector (the part of the circle inside the triangle), we need the angle at the centre P. This angle is ∠QPR.
Calculating the Angle of the Sector (∠QPR)
Triangle PQR is a right-angled triangle with ∠PQR = 90°. We also know that PQ = \(r\) and QR = \(r\). Since the two sides forming the right angle are equal, triangle PQR is an isosceles right-angled triangle.
In an isosceles right-angled triangle, the angles opposite the equal sides are equal. So, ∠QPR = ∠QRP.
The sum of angles in a triangle is 180°.
\(\angle PQR + \angle QPR + \angle QRP = 180^\circ\)
\(90^\circ + \angle QPR + \angle QPR = 180^\circ\)
\(90^\circ + 2 \angle QPR = 180^\circ\)
\(2 \angle QPR = 180^\circ - 90^\circ\)
\(2 \angle QPR = 90^\circ\)
\(\angle QPR = \frac{90^\circ}{2}\)
\(\angle QPR = 45^\circ\)
The angle of the sector of the circle inside the triangle PQR, with centre P, is 45°.
Calculating the Area of the Sector
The area of a sector of a circle with radius \(r\) and angle \(\theta\) (in degrees) is given by:
Area\((Sector)\) = \(\frac{\theta}{360^\circ} \times \pi r^2\)
In our case, the angle \(\theta = 45^\circ\) and the radius is \(r\).
Area\((Sector)\) = \(\frac{45^\circ}{360^\circ} \times \pi r^2\)
Area\((Sector)\) = \(\frac{1}{8} \times \pi r^2\)
Area\((Sector)\) = \(\frac{\pi r^2}{8}\)
This is the area of the part of the triangle PQR that is inside the circular region.
Finding the Area Outside the Circular Region
The area of the part of triangle PQR that is outside the circular region is the total area of the triangle minus the area of the part that is inside the circle (which is the sector calculated above).
Area\((Outside Circular Region)\) = Area\((PQR)\) - Area\((Sector)\)
Area\((Outside Circular Region)\) = \(\frac{r^2}{2} - \frac{\pi r^2}{8}\)
Summary of Area Calculation
We found the area of triangle PQR to be \(\frac{r^2}{2}\) and the area of the circular sector within the triangle to be \(\frac{\pi r^2}{8}\). The required area is the difference between these two values.
\(\text{Area} = \frac{r^2}{2} - \frac{\pi r^2}{8}\)
This matches one of the given options.
| Item | Formula / Value | Calculation |
|---|---|---|
| Triangle PQR Type | Right-angled at Q, Isosceles (PQ=QR=r) | ∠PQR = 90°, PQ = r, QR = r |
| Area of Triangle PQR | \(\frac{1}{2} \times \text{base} \times \text{height}\) | \(\frac{1}{2} \times r \times r = \frac{r^2}{2}\) |
| Angle of Sector at P | \(\angle QPR\) | From triangle properties: 45° |
| Area of Sector | \(\frac{\theta}{360^\circ} \times \pi r^2\) | \(\frac{45^\circ}{360^\circ} \times \pi r^2 = \frac{1}{8} \times \pi r^2 = \frac{\pi r^2}{8}\) |
| Area Outside Circle | Area(Triangle) - Area(Sector) | \(\frac{r^2}{2} - \frac{\pi r^2}{8}\) |
Revision Table: Triangle and Circle Geometry
| Concept | Description | Relevance to Problem |
|---|---|---|
| Tangent Property | A tangent to a circle is perpendicular to the radius at the point of tangency. | Used to determine ∠PQR = 90°. |
| Area of Right Triangle | \(\frac{1}{2} \times \text{product of perpendicular sides}\). | Used to calculate Area(PQR). |
| Isosceles Triangle Properties | If two sides are equal, the angles opposite them are equal. | Used with right angle to find ∠QPR. |
| Sum of Angles in Triangle | Angles add up to 180°. | Used to find ∠QPR. |
| Area of Circle Sector | \(\frac{\theta}{360^\circ} \times \pi r^2\). | Used to calculate the area of the circular part inside the triangle. |
Additional Information: Related Geometry Concepts
This problem combines concepts from triangles and circles. Understanding how these shapes interact is crucial in geometry.
- Radius: A line segment from the center of a circle to any point on its circumference.
- Tangent: A line that touches a circle at exactly one point. The point of contact is called the point of tangency.
- Sector: A part of a circle enclosed by two radii and the arc between them. Its area is proportional to the central angle.
- Right Triangle: A triangle with one angle measuring 90 degrees.
- Isosceles Triangle: A triangle with two sides of equal length.
Problems like this often require breaking down a complex shape (like the area outside a region) into simpler, calculable shapes (like a triangle and a sector) and then combining or subtracting their areas.
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