you didn't need a, you could have solved it with just the areas. Area of the triangle is 6sqrt(3)*4sqrt(3)*sin(60)/2=r*(6sqrt(3)*4sqrt(3))/2. Also you don't have a value for alfa angle you have an approximation. But you have the exact value for sinus and then you can calculate the tangent from that
Getting an exact value for d (no need to compute an approximate value for α or tan(α)): At 3:55, sin(α) = (√21)/7. Off to the side, construct a right triangle with one angle equal to α and the side opposite α having length equal to √21. Then, the hypotenuse has length 7 to make sin(α) = (√21)/7. Using the Pythagorean theorem, the side adjacent to angle α must have length 2√7 ( (2√7)² + (√21)² = 28 + 21 = 49, which is the square of the hypotenuse 7). Then, tan(α) = (opposite side)/(adjacent side) = (√21)/(2√7) = (√3)/2. Skip ahead to 7:05, where r = (18tan(α))/(3tan(α) + √3). Replace tan(α) with (√3)/2: r = (18((√3)/2))/((3(√3)/2) + √3) = (9√3)/((√3)(3/2 + 1)) = 9/(5/2) = 18/5 = 3.6. The diameter is twice the radius, or d = 7.2, as Mr Talbot Maths also found by approximation.
Once we have the lengths of all 3 sides (at 3:05, the length of the third side has been found to be 2√21), we can use Heron's formula to find the large triangle's area. Then, we find the large triangle's area a different way. The large triangle has been divided into 2 triangles with height r, one with base 6√3 and the other with base 4√3. The area of the large triangle is the sum of these two areas: (1/2)(6√3)(r) + (1/2)(4√3)(r) = (5√3)(r). Set (5√3)(r) equal to the area found by Heron's formula and solve for r. Multiply r by 2 to get the diameter. To simplify the mathematics for Heron's formula, first compute the area for the similar triangle with sides a = 6, b = 4, and c = 2√7 and then multiply the computed area by (√3)(√3) = 3. The semi-perimeter s = (a + b + c)/2 = (6 + 4+ 2√7)/2 = 5 + √7, (s - a) = s - 6 = 5 + √7 - 6 = -1 + √7, (s - b) = s - 4 = 5 + √7 - 4 = 1 + √7, (s - c) = s - 2√7 = 5 + √7 - 2√7 = 5 - √7. According to Heron's formula, area = √(s(s - a)(s - b)(s - c)) = √((5 + √7)(-1 + √7)(1 + √7)(5 - √7)). Note that (-1 + √7)(1 + √7) = -1 + 7 = 6 and (5 + √7)(5 - √7) = 25 - 7 = 18. So, area = √((6)(18)) = √((3)(36)) = 6√3. This is the area of the similar triangle. We multiply by 3 to get the original triangle's area = 18√3. We set (5√3)(r) = 18√3 and find r = 18/5 = 3.6. The diameter d is twice the radius length, so d = 7.2, as Mr Talbot Maths also found.
you didn't need a, you could have solved it with just the areas. Area of the triangle is 6sqrt(3)*4sqrt(3)*sin(60)/2=r*(6sqrt(3)*4sqrt(3))/2. Also you don't have a value for alfa angle you have an approximation. But you have the exact value for sinus and then you can calculate the tangent from that
Great idea. You meant 6sqrt(3)*4sqrt(3)*sin(60)/2=r*(6sqrt(3)+4sqrt(3))/2.
Getting an exact value for d (no need to compute an approximate value for α or tan(α)): At 3:55, sin(α) = (√21)/7. Off to the side, construct a right triangle with one angle equal to α and the side opposite α having length equal to √21. Then, the hypotenuse has length 7 to make sin(α) = (√21)/7. Using the Pythagorean theorem, the side adjacent to angle α must have length 2√7 ( (2√7)² + (√21)² = 28 + 21 = 49, which is the square of the hypotenuse 7). Then, tan(α) = (opposite side)/(adjacent side) = (√21)/(2√7) = (√3)/2. Skip ahead to 7:05, where r = (18tan(α))/(3tan(α) + √3). Replace tan(α) with (√3)/2: r = (18((√3)/2))/((3(√3)/2) + √3) = (9√3)/((√3)(3/2 + 1)) = 9/(5/2) = 18/5 = 3.6. The diameter is twice the radius, or d = 7.2, as Mr Talbot Maths also found by approximation.
Once we have the lengths of all 3 sides (at 3:05, the length of the third side has been found to be 2√21), we can use Heron's formula to find the large triangle's area. Then, we find the large triangle's area a different way. The large triangle has been divided into 2 triangles with height r, one with base 6√3 and the other with base 4√3. The area of the large triangle is the sum of these two areas: (1/2)(6√3)(r) + (1/2)(4√3)(r) = (5√3)(r). Set (5√3)(r) equal to the area found by Heron's formula and solve for r. Multiply r by 2 to get the diameter.
To simplify the mathematics for Heron's formula, first compute the area for the similar triangle with sides a = 6, b = 4, and c = 2√7 and then multiply the computed area by (√3)(√3) = 3. The semi-perimeter s = (a + b + c)/2 = (6 + 4+ 2√7)/2 = 5 + √7, (s - a) = s - 6 = 5 + √7 - 6 = -1 + √7, (s - b) = s - 4 = 5 + √7 - 4 = 1 + √7, (s - c) = s - 2√7 = 5 + √7 - 2√7 = 5 - √7. According to Heron's formula, area = √(s(s - a)(s - b)(s - c)) = √((5 + √7)(-1 + √7)(1 + √7)(5 - √7)). Note that (-1 + √7)(1 + √7) = -1 + 7 = 6 and (5 + √7)(5 - √7) = 25 - 7 = 18. So, area = √((6)(18)) = √((3)(36)) = 6√3. This is the area of the similar triangle. We multiply by 3 to get the original triangle's area = 18√3. We set (5√3)(r) = 18√3 and find r = 18/5 = 3.6. The diameter d is twice the radius length, so d = 7.2, as Mr Talbot Maths also found.