Calculator Techniques for Circles and Triangles in Plane Geometry

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.

Calculator Formula

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.

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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 Method

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 third 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 Method

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 Method

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:

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. Can you find the ratio of the area of the smaller part to the area of 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 Method

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.

© 2018 Ray