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An equilateral triangle has all three sides equal and and all three angles equal to 60° The relationship between the side \( a \) of the equilateral triangle and its area A, height h, radius R of the circumscribed and radius r of the inscribed circle are give by:
Formulas for Equilateral Triangles
![equilateral triangle formulas](http://www.analyzemath.com/Geometry/triangles/equilateral_triangle.gif)
Problem 1
What is the area of an equilateral triangle of perimeter 45 cm?
Problem 2
What is the area of an equilateral triangle of height 20 units?
Problem 3
What is the area of the circumscribed circle of an equilateral triangle of side a = 5 inches?
Problem 4
What is the radius of the inscribed circle of an equilateral triangle with an area of 100 cm 2?
Problem 5
What is the ratio of the area of the circumscribed circle to the area of the inscribed circle of an equilateral triangle?
Problem 6
What is the area of triangle CA'B' if ABC is an equilateral triangle and A'B' is parallel to AB?
![two equilateral triangle](http://www.analyzemath.com/Geometry/triangles/equilateral_triangle2.gif)
Problem 7
What is is the area of the shaded (in green) shape shown below if ABC is an equilateral triangle of side a = 10 has an inscribed circle, with center O, and is tangent at P and M to the sides AC and AB respectively?
![equilateral triangle with inscribed circle](http://www.analyzemath.com/Geometry/triangles/equilateral_triangle3.gif)
Problem 8
Find the coordinates of point C such that triangle ABC is equilateral of side 12 units.
![equilateral triangle with inscribed circle](http://www.analyzemath.com/Geometry/triangles/equilateral_triangle4.gif)
Problem 9
What is the area of the triangle BB'B" if ABC is an equilateral triangle of side 10 units and A'B' is parallel to AB?
![equilateral triangle and a right triangle](http://www.analyzemath.com/Geometry/triangles/equilateral_triangle5.gif)
Problem 10
What is the area of the shaded shape (in red) if all three circles have equal radii of 15 units and are tangent to each other?
![equilateral triangle made with centers of three tangent congruent circles](http://www.analyzemath.com/Geometry/triangles/equilateral_triangle6.gif)
Solutions to the Above Questions
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Solution
If the side of the equilateral triangle is \( a \), its perimeter P is given by
P = 3 \( a \)
\( a \) = P / 3 = 45 / 3 = 15 cm
Area A = \( a^2\dfrac{\sqrt 3}{4} \) = = \( 15^2\dfrac{\sqrt 3}{4} \) = \( \dfrac{225} {4} {\sqrt 3} \) cm 2
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Solution
If h is the height of the equilateral triangle and \( a \) its side, we have the relationship (see formula above)
h = \( a \dfrac{\sqrt 3}{2} \)
h = 20, hence the equation: 20 = \( a\dfrac{\sqrt 3}{2} \)
Solve for \( a \) to get : \( a = \dfrac{40}{\sqrt 3} \)
Area A = \( a^2\dfrac{\sqrt 3}{4} \) = = \( (\dfrac{40}{\sqrt 3})^2\dfrac{\sqrt 3}{4} = \dfrac{400}{3} \sqrt 3 \) unit 2
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Solution
Let R be the radius the circumscribed circle to an equilateral triangle of side a, then (see formula above)
R = \( a\dfrac{\sqrt 3}{3} \) = \( 5 \dfrac{\sqrt 3}{3} \) , \( a = 5\) given
Area of circle of radius R = \( \pi R^2 = \pi (5\dfrac{\sqrt 3}{3})^2 = \dfrac{25}{3}\pi \) inches 2
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Solution
Let r be the radius the inscribed circle to an equilateral triangle of side a, then (see formula above)
r = \( a\dfrac{\sqrt 3}{6} \)
Square both sides of the above to get : r 2 = \( \dfrac{a^2}{12} \)
Area A of circle of radius r is given by: A = \( \pi r^2 = \pi \dfrac{a^2}{12} \)
The area of the equilateral triangle is given, hence: 100 = \( a^2\dfrac{\sqrt 3}{4} \)
Solve the above to find: \( a^2 = \dfrac{400} {\sqrt 3}\)
Substitute a 2 found above into the expression of the area A = \( = \pi \dfrac{a^2}{12} \) found above to find
A = \( = \pi \dfrac{\dfrac{400} {\sqrt 3}}{12} = \pi \dfrac{100\sqrt 3}{9} \) cm 2
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Solution
If R is the radius of the circumscribed circle and r the radius the inscribed circle to an equilateral triangle of side a, then the ratio S is given by
\( S = \dfrac{\pi R^2}{\pi r^2} = \dfrac{ R^2} {r^2} = (\dfrac{ R} {r})^2 \)
We now use the formulas for R and r given above and simplify
\( S = \left(\dfrac{ a\dfrac{\sqrt 3}{3} } {a\dfrac{\sqrt 3}{6}} \right)^2 = 4\)
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Solution
Since A'B' is parallel to AB, triangles ABC and AB'C' are similar and therefore triangle AB'C' is also equilateral and has side equal to 4.
Area of AB'C' = \( 4^2\dfrac{\sqrt 3}{4} = 4 \sqrt 3 \) unit 2
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Solution
Area of green shape = area of quadrilateral AMOP - area of sector MOP
Quadrilateral AMOP is made up of two congruent right triangles (AC perpendicular to PO and AB perpendicular to MO) since P and M are points of tangency. (Note: AP = a/2)
area of triangle APO = (1/2) AP × PO = (1/2) AP × r = (1/2) (10 / 2) × \( 10 \dfrac{\sqrt 3}{6} = \dfrac{25\sqrt 3}{6}\)
Angle of sector MOP = 360 / 3 = 120°
area of sector MOP = (1/2) (120 π / 180 ) r 2 = \( \dfrac{\pi}{3} (10\dfrac{\sqrt 3}{6})^2 = 50\pi / 18 \)
Area of green shape = 2 × area of triangle APO - area of sector MOP = \( \dfrac{50\sqrt 3}{6} - 50\pi / 18 \)
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Solution
AB = \( \sqrt{b^2} \) = 12
Solve for b
b = 12
AC = \( \sqrt{(x^2 +y^2)} \) = 12 gives \( x^2 +y^2 = 12^2\)
BC = \( \sqrt{((x-12)^2 +y^2)} \) = 12 gives \( (x-12)^2 +y^2 = 12^2 \)
Expand the last equation: \( x^2 -24x + 12^2 + y^2 = 12^2 \)
Use \( x^2 +y^2 = 12^2\) in the last equation to obtain
\( 24 x = 12^2 \) , solve for x: x = 6
Substitute x = 6 in the equation \( x^2 +y^2 = 12^2 \) to find y = 6 √3
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Solution
Since A'B' is parallel to AB, triangles ABC and AB'C' are similar and both equilateral. Hence CB' = 4 and B'B = 10 - 4 = 6
Angle BB'B'' has a size of 30° since the size of angle B'BB" is 60°.
sin(30°) = BB" / BB' , hence BB" = 3 and use Pythagora's to get B'B" = 3√3
area of triangle BB'B" = (1/2) B'B" × BB" = (1/2) 3√3 × 3 = 4.5√3 unit 2
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Solution
![equilateral triangle made with centers of three tangent congruent circles](http://www.analyzemath.com/Geometry/triangles/equ_sol10.gif)
Figure above shows that the centers of the circles make an equilateral triangle of side 2r where r = 15 (given) is the radius of one circle.
The area of the red shape = area of the equilateral triangle - areas of three congruent sectors (each sector has 60° angle)
area of the equilateral triangle = \( (2r)^2\dfrac{\sqrt 3}{4} = 30^2\dfrac{\sqrt 3}{4} \)
areas of three congruent sectors = 3 × area of one sector = 3 × (1/2) ( 60 π /180) r 2 = 15 2 π /2
The area of the red shape = \( ( 30^2\dfrac{\sqrt 3}{4} - 15^2 \dfrac{\pi}{2} ) \) unit 2
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