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Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.
Calculator techniques for circles and triangles in plane geometry
John Ray Cuevas
Calculator Techniques for Circles and Triangles
Calculator techniques for problems related to circles and triangles are more about algebra, trigonometry, and geometry. Memorization of formulas is what is needed. Here are the contents of the article:
- A circle inscribed in a triangle
- A circle circumscribing a triangle
- Three circles mutually tangent to each other
- The area of a sector
- A circle inscribed in a sector
- The bisector of a triangle
- The side measures of a triangle
- The chord of a circle
- A point outside a triangle
Inscribed and Circumscribed Circles
A circle can either be inscribed or circumscribed. A circle circumscribing a triangle passes through the vertices of the triangle, while a circle inscribed in a triangle is tangent to the three sides of the triangle. The third connection linking circles and triangles is a circle escribed outside a triangle. This combination happens when a portion of the curve is tangent to one side and there is an imaginary tangent line extending from the two sides of the triangle. Given A, B, and C as the sides of the triangle and A as the area, the formula for the radius of a circle circumscribing a triangle is r = ABC / 4A, and for a circle inscribed in a triangle is r = A / S where S = (A + B + C) / 2.
Problem 1: Circle Inscribed in a Triangle
The sides of a triangle are 8 cm, 10 cm, and 14 cm. Determine the radius of the inscribed circle.
John Ray Cuevas
Calculator Technique
a. Using Heron's formula, solve for the area of the triangle.
A = 8 centimeters B = 10 centimeters C = 14 centimeters X = (A + B + C) / 2 X = (8 + 10 +14) / 2 X = 16 centimeters Area of triangle (A) = √X (X - A) (X - B) (X - C) Area of triangle (A) = √16(16 - 8) (16 - 10) (16 - 14) Area of triangle (A) = 16√6 square centimetersb. Solve for the perimeter of the triangle.
Perimeter (P) = 8 + 10 + 14 Perimeter (P) = 32 centimetersc. Solve for the radius of the inscribed circle.
Radius of inscribed circle = 2A / P Radius of inscribed circle = 2 (16√6) / (32) Radius of inscribed circle = 2.45 centimetersFinal Answer: The radius of the inscribed circle is 2.45 centimeters.
Problem 2: Circle Circumscribing an Equilateral Triangle
The area of a circle circumscribing about an equilateral triangle is 250.45 square meters. What is the area of the triangle?
John Ray Cuevas
Calculator Technique
a. Given the area of the circle, solve for the radius.
Area of circle = (pi)(r)^2 250.45 square meters = (pi) (r)^2 r = 8.93 metersb. Solve for a side of the equilateral triangle.
A = B = C = Side of the triangle (S) radius = ABC / 4A radius = S^3 / 4A 8.93 meters = S^3 / 4 (S^2) (√3 / 4) S = 15.47 metersc. Solve for the area of the equilateral triangle.
Area of triangle = S^2 (√3 / 4) Area of triangle = (15.47)^2 (√3 / 4) Area of triangle = 103.59 square metersFinal Answer: The area of the equilateral triangle inscribed in a circle is 103.59 square meters.
Problem 3: Three Circles Mutually Tangent
The distance between the centers of the three circles which are mutually tangent to each other externally is 10, 12 and 14 units. What is the area of the largest circle?
John Ray Cuevas
Calculator Technique
a. Form three equations for three unknowns.
x + y = 10 x + z = 12 y + z = 14b. Arrange the equations in a system form.
x + y + 0z = 10 x + 0y + z = 12 0x + y + z = 14 x = 4 y = 6 z = 8c. Solve the area of the largest circle using the largest radius from Step 2.
Area = (pi) (8)^2 Area = 64 (pi) Area = 201.06 square unitsFinal Answer: The area of the largest circle is 201.06 square units.
Problem 4: Triangle Inscribed in a Circle
The area of the triangle inscribed in a circle is 39.19 square centimeters, and the radius of the circumscribed circle is 7.14 centimeters. If the two sides of the inscribed triangle are 8 centimeters and 10 centimeters respectively, find the third side.
Calculator Technique
a. Solve for the third side C.
Final Answer: The length of the third side is 14.00 centimeters.
Problem 5: Area of a Sector
The angle of a circle's sector is 300 degrees, and the radius is 15 centimeters. Find the area in square centimeters.
John Ray Cuevas
Calculator Technique
a. Use the formula for finding the area of the sector.
Area of a sector = 1/2 (r)^2 (θ) Area of a sector = 1/2 (15)^2 (300°) Area of a sector = 33750 in degreesb. Go to radian mode. Once in radian mode, enter "Shift Answer" and choose the degree sign.
Area of sector = 33750 Area of sector = 589.05 square centimetersFinal Answer: The area of the sector is 598.05 square centimeters.
Problem 6: Circle Inscribed in a Sector
A sector circumscribes a circle with a radius of 8.00 centimeters. If the sector's central angle is 80 degrees, what is the area?
John Ray Cuevas
Calculator Technique
a. Solve for the radius of the sector.
Radius of sector = 8 + (8 / sin40°) Radius of sector = 20.45 centimetersb. Solve for the area of the sector.
Area of sector = 1/2 (r)^2 (θ) Area of sector = 1/2 (20.45) (80°) Area of sector = 818c. Go to radian mode. Once in radian mode, enter "Shift Answer" and choose the degree sign.
Area of sector = 818 Area of sector = 291.83 square centimetersFinal Answer: The area of the sector is 291.83 square centimeters.
Problem 7: Bisector of a Triangle
Given a triangle ABC with sides AB = 30 centimeters, BC = 36 centimeters and AC = 48 centimeters. Find the distance of the point of intersection of perpendicular bisectors to side BC.
John Ray Cuevas
Calculator Technique
a. The point of intersection of the perpendicular bisectors is the radius of the circumscribed circle. Solve for the area of the triangle using Heron's Formula.
A = 30 centimeters B = 36 centimeters C = 48 centimeters X = A + B + C X = 30 + 36 + 48 / 2 X = 57 centimeters Area of triangle (A) = √X (X - A) (X - B) (X - C) Area of triangle (A) = √57(57 - 30) (57 - 36) (57 - 48) Area of triangle (A) = 539.32 square centimetersb. Solve for the radius of the circumscribed circle.
Radius (R) = ABC / 4A Radius (R) = (30) (36) (48) / 4 (539.32) Radius (R) = 24.03 centimetersc. Using the Pythagorean theorem, solve for the missing value x.
X^2 + 18^2 = R^2 X^2 + 18^2 = (24.03)^2 X = 15.92 centimetersFinal Answer: The distance from the point of intersection of perpendicular bisectors to side BC is 15.92 centimeters.
The perimeter of a triangle ABC is 400 centimeters. If angle A is 30 degrees and angle B is 58 degrees, find the measure of side AC.
John Ray Cuevas
Calculator Technique
a. Assume that AC = 1. Then use sine law technique in solving for the sides AB and BC. Set your calculator in equation mode and input the following values:
cos (30) | cos (58) | 1 |
sin (30) | - sin (58) | 0 |
b. Solve for the length of side AC.
AB = 0.85 BC = 0.5 0.85X + 0.5X + X = 400 centimeters X = AC = 170.31 centimetersFinal Answer: The length of side AC is 170.31 centimeters.
Problem 9: Chord of a Circle
A circle giving an area of 1018 square centimeters is cut into two segments by a chord 8 centimeters from the center. Can you find the ratio of the area of the smaller part to the area of the larger one?
John Ray Cuevas
Calculator Technique
a. Solve for the radius of the circle.
Area of circle = (pi) (r)^2 1018 square centimeters = (pi) (r)^2 r = 18 centimetersb. Solve for θ.
cosine (θ/2) = 8 / 18 θ = 127.22 degreesc. Find the area of the small segment.
Area of small segment = 1/2 (R)^2 (θ - sin (θ)) Area = 1/2 (18)^2 (127.22 - sin (127.22)) Area = 230.70 square centimetersd. Solve for the larger segment.
Area of larger segment = (pi) (18)^2 - 230.70 Area of larger segment = 787.176 square centimeterse. Solve for the ratio between the two segments.
Ratio = 230.70 / 787.16 Ratio = 0.293Final Answer: The ratio between the two segments is 0.293.
From a point outside an equilateral triangle, the distances to the vertices are 10 centimeters, 18 centimeters, and 10 centimeters, respectively. What is the length of one side of the triangle?
John Ray Cuevas
Calculator Technique
a. Solve for the length of one side X using the Cosine law.
a^2 = b^2 + c^2 -2bc cosine (α) 10^2 = X^2 + 18^2 - 2 (X) (18) (cosine(30)) X = 19.95 centimetersFinal answer: The length of one side of the triangle is 19.95 centimeters.
© 2018 Ray
Ray (author) from Philippines on April 11, 2019:
You're very welcome, Ali Hassan. There are a lot more articles like this. You can check it in my profile. Happy hubbing!
Ali Hassan on April 11, 2019:
It's a very helpful and informative article thanks for making this
Ray (author) from Philippines on July 17, 2018:
Thanks, Syed Aabis! I appreciate it. Have a great day pal! Enjoy reading my articles.
Syed Aabis from Pakistan on July 17, 2018:
It is very informative article i will like to read more like this topic