Question

What is the radius of the smallest circle that encloses an equilateral triangle with 12 inch sides? What is the radius of the largest circle that will fit inside the same triangle?

Answer 1

This is geometry through and through.  Plus a little trig thrown in for fun.  If you inscribe an equilateral triangle inside a circle and the triangle has side lengths of 12, you have part of what you need to use Pythagorean's Theorem to find the hypotenuse of the triangle which is also the radius of the circle. First, use the formula 360/3 to find that the central angle measure of each angle INSIDE a triangle is 120. So you have 3 triangles within the large one, each with a top angle of 120 and a base of 12. If you extract that one triangle and then split it in half, you have a right triangle with a base of 6. This is a 30-60-90 triangle and this is important so you can check your work. Now use the apothem formula for a right triangle as it relates to a side in an equilateral triangle, which is a = sqrt3/6 * s. Our values are a = sqrt3/6(12) which simplifies to 2sqrt3. That's our apothem. If you're familiar with a 30-60-0 triangle, you could check this to see it's correct. Now you have the base leg of 6 and the height of 2sqrt3, now use Pythagorean's Theorem to find the hypotenuse, which is also the radius of the circle. This was really a difficult one to explain.

Answer 2

Side of equilateral Triangle =12 inch

Let r be the radius of Smallest circle which fits inside the triangle.

Theorem Which will be Used

→The Point of intersection of end point of radius and tangent through that end point makes an angle of 90°.

→Angle Subtended by an arc at the center is twice the angle subtended at any point on the circle.

→Perpendicular from the center of the circle to the chord bisects the chord.

→ Area of Equilateral Triangle

                                  $=\frac{\sqrt{3}}{4} \times \text{side}^2$

→Area of Triangle

                     $=\frac{\sqrt{1}}{2} \times \text{Base}\times \text{Height}$

⇒ Radius of the smallest circle

→ Area of Equilateral Triangle having side 12 inches

                                  $=\frac{\sqrt{3}}{4} \times (12)^2\\\\=\frac{\sqrt{3}}{4} \times 144\\\\=36\sqrt{3}$

                                 ------------------------------(1)

→Area of 3 Triangle having base =12 inches and height =r inches

                     $=3 \times \frac{\sqrt{1}}{2} \times 12 \times r\\\\=18 r$

                                                    -----------------------------(2)

Equating (1) and (2), that is both the Areas are equal.

$\rightarrow 18r=36 \sqrt{3}\\\\\rightarrow r=\frac{36 \sqrt{3}}{18}\\\\\rightarrow r=2 \sqrt{3}\\\\\rightarrow r=2 \times 1.732\\\\\rightarrow r=3.464 \text{inch}$

⇒Radius of the largest circle

        Each Interior angle of equilateral Triangle=60°

∠AOC=2×60°=120°-----Angle Subtended by an arc at the center is twice the angle subtended at any point on the circle.

r=Radius of circle

→∠AOC+∠CAO+∠ACO=180°---Angle sum property of triangle.

→ 120°+2∠CAO=180°------[OA=OC, →∠CAO=∠OCA------Angle Opposite to equal side of a triangle are equal.]

→2∠CAO=180°-120°

→2∠CAO=60°

→∠CAO=30°-------Dividing both sides by, 2

→∠MAO+∠AMO+∠AOM=180°---Angle sum property of triangle.

→∠MAO+90°+60°=180°

→∠MAO=180°-150°

→∠MAO=30°

In ΔOMA

AC=12 inches

AM=MC=6 inches--Perpendicular from the center of the circle to the chord bisects the chord.

$\cos 30 ^{\circ}=\frac{AM}{AO}\\\\\cos 30 ^{\circ}=\frac{6}{r}\\\\r=\frac{6 \times 2}{\sqrt{3}}\\\\r=4\sqrt{3}\\\\r=4 \times 1.732\\\\r=6.928 \text{inches}$