Problems on Equilateral Triangles with Detailed Solutions

Solutions to the Above Questions

1. 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

   ---

2. 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

   ---

3. 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

   ---

4. 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

   ---

5. 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
   
   \( S = \left(\dfrac{ a\dfrac{\sqrt 3}{3} } {a\dfrac{\sqrt 3}{6}} \right)^2 = 4\)

   ---

6. 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

   ---

7. 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 ? / 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 = \( \dfrac{50\sqrt 3}{6} - 50\pi / 18 \)

   ---

8. 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

   ---

9. 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

   ---

10. Solution
    
    

    

    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 ? /180) r ² = 15 ² ? /2
    
    The area of the red shape = \( ( 30^2\dfrac{\sqrt 3}{4} - 15^2 \dfrac{\pi}{2} ) \) unit ²