Problems on Equilateral Triangles with Detailed Solutions

Question: What is the area of an equilateral triangle of perimeter 45 cm?

Solution:

If the side of the equilateral triangle is \(a\), its perimeter \(P\) is given by:
\(P = 3a\)
\(a = P / 3 = 45 / 3 = 15 \, \text{cm}\)

Area \(A = a^2 \frac{\sqrt{3}}{4} = 15^2 \frac{\sqrt{3}}{4} = \frac{225}{4} \sqrt{3} \, \text{cm}^2\)

Question: What is the area of an equilateral triangle of height 20 units?

Solution:

If \(h\) is the height of the equilateral triangle and \(a\) its side, we have the relationship:
\(h = a \frac{\sqrt{3}}{2}\)
Given \(h = 20\), we set up the equation: \(20 = a \frac{\sqrt{3}}{2}\)
Solve for \(a\): \(a = \frac{40}{\sqrt{3}}\)

Area \(A = a^2 \frac{\sqrt{3}}{4} = \left(\frac{40}{\sqrt{3}}\right)^2 \frac{\sqrt{3}}{4} = \frac{400}{3} \sqrt{3} \, \text{unit}^2\)

Question: What is the area of the circumscribed circle of an equilateral triangle of side \(a = 5\) inches?

Solution:

Let \(R\) be the radius of the circumscribed circle to an equilateral triangle of side \(a\):
\(R = a \frac{\sqrt{3}}{3} = 5 \frac{\sqrt{3}}{3}\) (since \(a = 5\))

Area of the circle of radius \(R\) is:
\(A = \pi R^2 = \pi \left(5 \frac{\sqrt{3}}{3}\right)^2 = \frac{25}{3}\pi \, \text{inches}^2\)

Question: What is the radius of the inscribed circle of an equilateral triangle with an area of 100 cm²?

Solution:

Let \(r\) be the radius of the inscribed circle to an equilateral triangle of side \(a\):
\(r = a \frac{\sqrt{3}}{6}\)
Squaring both sides gives: \(r^2 = \frac{a^2}{12}\)

The area \(A\) of the circle of radius \(r\) is given by: \(A = \pi r^2 = \pi \frac{a^2}{12}\)

The area of the equilateral triangle is given as 100:
\(100 = a^2 \frac{\sqrt{3}}{4} \implies a^2 = \frac{400}{\sqrt{3}}\)

Substitute \(a^2\) into the circle area formula:
\(A = \pi \frac{\frac{400}{\sqrt{3}}}{12} = \pi \frac{100\sqrt{3}}{9} \, \text{cm}^2\)

Question: What is the ratio of the area of the circumscribed circle to the area of the inscribed circle of an equilateral triangle?

Solution:

If \(R\) is the radius of the circumscribed circle and \(r\) the radius of the inscribed circle, the ratio \(S\) is:
\(S = \frac{\pi R^2}{\pi r^2} = \frac{R^2}{r^2} = \left(\frac{R}{r}\right)^2\)

Using the formulas for \(R\) and \(r\):
\(S = \left(\frac{a\frac{\sqrt{3}}{3}}{a\frac{\sqrt{3}}{6}}\right)^2 = 4\)

Question: What is the area of triangle \(CA'B'\) if \(ABC\) is an equilateral triangle and \(A'B'\) is parallel to \(AB\)?

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 \(AB'C' = 4^2 \frac{\sqrt{3}}{4} = 4\sqrt{3} \, \text{unit}^2\)

Question: What is the area of the shaded (in green) shape shown below if \(ABC\) is an equilateral triangle of side \(a = 10\) with an inscribed circle, center \(O\), tangent at \(P\) and \(M\) to sides \(AC\) and \(AB\) respectively?

Solution:

Area of green shape = Area of quadrilateral \(AMOP\) - Area of sector \(MOP\)

Quadrilateral \(AMOP\) is made up of two congruent right triangles (\(AC \perp PO\) and \(AB \perp MO\) since \(P\) and \(M\) are points of tangency, with \(AP = a/2\)).

Area of triangle \(APO = \frac{1}{2} AP \times PO = \frac{1}{2} (10 / 2) \times \left(10 \frac{\sqrt{3}}{6}\right) = \frac{25\sqrt{3}}{6}\)

Angle of sector \(MOP = 360^\circ / 3 = 120^\circ\)

Area of sector \(MOP = \frac{1}{2} \left(\frac{120\pi}{180}\right) r^2 = \frac{\pi}{3} \left(10 \frac{\sqrt{3}}{6}\right)^2 = \frac{50\pi}{18}\)

Area of green shape = \(2 \times \text{Area of triangle } APO - \text{Area of sector } MOP = \frac{50\sqrt{3}}{6} - \frac{50\pi}{18}\)

Question: Find the coordinates of point \(C\) such that triangle \(ABC\) is equilateral of side 12 units.

Solution:

\(AB = \sqrt{b^2} = 12 \implies b = 12\)

\(AC = \sqrt{x^2 + y^2} = 12 \implies x^2 + y^2 = 12^2\)

\(BC = \sqrt{(x - 12)^2 + y^2} = 12 \implies (x - 12)^2 + y^2 = 12^2\)

Expand the last equation: \(x^2 - 24x + 12^2 + y^2 = 12^2\)

Substitute \(x^2 + y^2 = 12^2\) into the equation to obtain:
\(24x = 12^2 \implies x = 6\)

Substitute \(x = 6\) into \(x^2 + y^2 = 12^2\) to find \(y = 6\sqrt{3}\).

Question: 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 \(AB\)?

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 \(30^\circ\) since the size of angle \(B'BB''\) is \(60^\circ\).

\(\sin(30^\circ) = \frac{BB''}{BB'} \implies BB'' = 3\), and use Pythagoras' theorem to get \(B'B'' = 3\sqrt{3}\).

Area of triangle \(BB'B'' = \frac{1}{2} B'B'' \times BB'' = \frac{1}{2} \times 3\sqrt{3} \times 3 = 4.5\sqrt{3} \, \text{unit}^2\)

Question: 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?

Solution:

The figure above shows that the centers of the circles form 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 a \(60^\circ\) angle).

Area of the equilateral triangle = \((2r)^2 \frac{\sqrt{3}}{4} = 30^2 \frac{\sqrt{3}}{4}\)

Areas of three congruent sectors = \(3 \times \text{Area of one sector} = 3 \times \frac{1}{2} \left(\frac{60\pi}{180}\right) r^2 = 15^2 \frac{\pi}{2}\)

The area of the red shape = \(\left(30^2 \frac{\sqrt{3}}{4} - 15^2 \frac{\pi}{2}\right) \text{unit}^2\)