Question

Statement 1 For an equilateral triangle, \(\mathrm{R}=2 \mathrm{r}\) and Statement 2 \(r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)

Short answer

Question: In an equilateral triangle, the circumradius (R) is twice its inradius (r), and for any triangle, the inradius (r) can be expressed as r = 4R * sin(A/2) * sin(B/2) * sin(C/2). Explain these statements. Answer: In an equilateral triangle, the circumradius (R) refers to the radius of a circle that passes through all three vertices of the triangle, while the inradius (r) is the radius of the inscribed circle that is tangent to all three sides. It can be shown that for an equilateral triangle, R = 2r, meaning the circumradius is twice the inradius. For any triangle with angles A, B, and C, we can express its inradius (r) with the formula r = 4R * sin(A/2) * sin(B/2) * sin(C/2), where R is the circumradius. This formula helps relate the inradius of a triangle to its circumradius and angles.

Step-by-step solution

Statement 1: R = 2r in an Equilateral Triangle

Let's denote the side length of an equilateral triangle as 'a' and its circumradius as R and inradius as r. 1. Draw an equilateral triangle and its circumcenter and incenter. 2. Draw the perpendicular from the circumcenter to one side of the triangle. 3. The right triangle formed has hypotenuse R, one leg as r, and the other leg as (a/2). 4. Apply the Pythagorean theorem in the created right triangle: R^2 = r^2 + (a/2)^2 5. For an equilateral triangle, a = r * sqrt(3), so substitute this relation in Step 4: R^2 = r^2 + (r * sqrt(3)/2)^2 6. Simplify the equation and solve for R: R = 2r So, for an equilateral triangle, the circumradius (R) is twice its inradius (r).

Statement 2: Proof of r = 4R sin(A/2) sin(B/2) sin(C/2) for any triangle

Let's prove the given statement for any triangle with side lengths a, b, c and angles A, B, and C: 1. Recall the area formula for a triangle using semiperimeter and angle measures: Area = r * s, where s is the semiperimeter (s = (a + b + c)/2) 2. Use the sine formula for the area of a triangle: Area = (1/2)ab * sin(C) 3. From Steps 1 and 2, we have: r * s = (1/2)ab * sin(C) 4. Divide both sides of the equation by R and use the definition of the circumradius (R) in terms of the triangle's sides and angles: r/R = ab * sin(C) / 2sR 5. Use the sine rule in Step 4 to relate a, b, and C to the angles A, B, and R: r/R = (2R * sin(A) * 2R * sin(B) * sin(C)) / (4sR^2) 6. In the given exercise, we have A + B + C = π, so sin(C) = sin(π - (A + B)) = sin(A + B). Apply the sine addition formula to simplify sin(A + B) to sin(A/2) * cos(B/2) + sin(B/2) * cos(A/2) 7. Replace sin(C) in Step 5 and simplify: r = 4R * sin(A/2) * cos(B/2) * sin(B/2) * cos(A/2) r = 4R * sin(A/2) * sin(B/2) * sin(π - (A + B)/2) r = 4R * sin(A/2) * sin(B/2) * sin(C/2) So, for any triangle, the inradius (r) can be expressed as r = 4R * sin(A/2) * sin(B/2) * sin(C/2).

Key concepts

Circumradius and Inradius Relationship

Understanding the relationship between the circumradius and inradius of an equilateral triangle is essential to dive into the geometry of triangles. With an equilateral triangle, every side and angle is congruent, which simplifies our approach to many geometric problems. Let's consider the circumradius (R), the radius of the circumscribed circle that passes through all three vertices of the triangle. On the other hand, we have the inradius (r), the radius of the inscribed circle that touches all three sides.

Now, many students might find it intriguing that in an equilateral triangle, the relationship between the circumradius and inradius can be summarized by the formula \( R = 2r \). This is a special property of equilateral triangles, and it comes from exploring the geometry within the triangle. By drawing an altitude from the center of the circumscribed circle to one side of the triangle, you can form a right triangle where you can apply the Pythagorean theorem. This theorem helps us set up an equation relating the circumradius, inradius, and the triangle's side.

The synergy between the circumradius and inradius in an equilateral triangle not only showcases the harmony of triangle properties but also can be applied in various mathematical problems and proofs.

Pythagorean Theorem

At the heart of many geometric proofs and computations is the Pythagorean theorem. This principle is a cornerstone in mathematics that states, for a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In mathematical terms, if we label these sides as 'a', 'b', and 'c', where 'c' is the hypotenuse, the theorem gives us \( c^2 = a^2 + b^2 \).

This theorem is not just a formula; it's a gateway to understanding relationships between different geometric entities. For instance, when exploring equilateral triangles, we use the Pythagorean theorem to relate the side of the triangle to its circumradius and inradius. By seeing the triangle split into two right-angled triangles through an altitude, students can discover how the circumradius and the inradius are connected with each other through the theorem. This use of the theorem illuminates a deeper understanding of geometric properties and how they interlink.

Triangle Area Formulas

The area of a triangle is a measure of the space enclosed within its three sides. There are several formulas to calculate the area, depending on the information available. For a right-angled triangle, the area can be determined by the simple formula \( Area = \frac{1}{2} \times base \times height \). However, for non-right-angled triangles, particularly when dealing with sides and angles, other formulas come into play.

One such formula involves the semiperimeter 's' and the inradius 'r'; it states \( Area = r \times s \), where \( s = \frac{a + b + c}{2} \). On the other hand, the area can be derived using trigonometry with \( Area = \frac{1}{2}ab \times \sin(C) \), capitalizing on the relationship between sides and angles. These formulas are not just tools for calculation; they reveal the intrinsic link between linear dimensions and angles of triangles. By maneuvering through these formulas, students can learn to select the most efficient path to solve area-related problems, deepening their understanding of triangular geometry and trigonometry.