Calculator Techniques for Circles and Triangles in Plane Geometry
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 about 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.
Calculator Techniques for Circles and Triangles
Calculator techniques for problems related to circles and triangles are more on algebra, trigonometry, and geometry. Memorization of formulas is what is needed. Here are the contents of the article.
 A circle inscribed in a triangle
 Circle circumscribing a triangle
 Three circles mutually tangent to each other
 Area of a sector
 A circle inscribed in a sector
 The bisector of a triangle
 The side measures of a triangle
 Chord of a circle
 A point outside a triangle
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.
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 centimeters
b. Solve for the perimeter of the triangle.
Perimeter (P) = 8 + 10 + 14 Perimeter (P) = 32 centimeters
c. 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 centimeters
Final 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?
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 meters
b. 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 meters
c. 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 meters
Final 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?
Calculator Technique
a. Form three equations for three unknowns.
x + y = 10 x + z = 12 y + z = 14
b. Arrange the equations in a system form.
x + y + 0z = 10 x + 0y + z = 12 0x + y + z = 14 x = 4 y = 6 z = 8
c. 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 units
Final 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 3rd side.
Calculator Technique
a. Solve for the third side C.
a = 8 centimeters b = 10 centimeters Area of triangle = 39.19 square centimeters radius = abc / 4A 7.14 centimeters = (8) (10) (c) / 4 (39.19 square centimeters) 7.14 centimeters = (8) (10) (c) / 156.76 square centimeters 7.14 (156.76) = (8) (10) (c) c = 14.00 centimeters
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.
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 degrees
b. 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 centimeters
Final 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?
Calculator Technique
a. Solve for the radius of the sector.
Radius of sector = 8 + (8 / sin40°) Radius of sector = 20.45 centimeters
b. 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 = 818
c. 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 centimeters
Final 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.
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 centimeters
b. Solve for the radius of the circumscribed circle.
Radius (R) = ABC / 4A Radius (R) = (30) (36) (48) / 4 (539.32) Radius (R) = 24.03 centimeters
c. 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 centimeters
Final Answer: The distance from the point of intersection of perpendicular bisectors to side BC is 15.92 centimeters.
Problem 8: Measure of Sides of a Triangle
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.
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.
X
 Y
 C


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 centimeters
Final 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. Find the ratio of the area of the smaller part to the larger one?
Calculator Technique
a. Solve for the radius of the circle.
Area of circle = (pi) (r)^2 1018 square centimeters = (pi) (r)^2 r = 18 centimeters
b. Solve for θ.
cosine (θ/2) = 8 / 18 θ = 127.22 degrees
c. 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 centimeters
d. Solve for the larger segment.
Area of larger segment = (pi) (18)^2  230.70 Area of larger segment = 787.176 square centimeters
e. Solve for the ratio between the two segments.
Ratio = 230.70 / 787.16 Ratio = 0.293
Final Answer: The ratio between the two segments is 0.293.
Problem 10: A Point Outside a Triangle
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?
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 centimeters
Final answer: The length of one side of the triangle is 19.95 centimeters.
Other Calculator Techniques
 Calculator Techniques for Polygons in Plane Geometry
Solving problems related to plane geometry especially polygons can be easily solved using a calculator. Here is a comprehensive set of problems about polygons solved using calculators.
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