**Additional Materials**

**Materials Required**

**Estimated Time Needed**

*(Times are approximate and will depend on the needs of the students.)*

Homework

- Review Individual Solutions to Circles and Triangles (10 minutes)
**Ask students to look again at their original individual solutions to the problem.***Read through your original solution to the Circles and Triangles problem.**Write what you have learned during the lesson.**Suppose a friend began work on this task tomorrow. What advice would you give your friend to help him or her produce a good solution?*

- Solutions: Circles and Triangles
Bill's Method *Question 1*cos 30° = b = √3 ⇒ b = r√3 r 2 2 sin 30° = c = 1 ⇒ c = r r 2 2 tan 60° = a = √3 ⇒ a = r√3 r Area of small equilateral triangle:

6 × ½ × b × c = 3 × r√3 × r = 3√3r ^{2}2 2 4 Area of large equilateral triangle:

6 × ½ × a × r = 3 × r√3 × r = 3√3r

^{2}Ratio of area of the outer to the area of the inner triangle:

3√3r ^{2}:3√3 r ^{2}= 4 : 14 *Question 2*c is the radius of the inscribed circle.

c = r 2 Ratio of area of circles:

= πr

^{2}: πc^{2}= πr

^{2}: π^{r2}⁄_{4}_{= 4 : 1}Carla's Method *Question 1*The small equilateral triangle is rotated through 60° about O, the center of the circle. The arm of the rotation is the radius of the circle. Therefore points D, E, and F are all points on the circumference of the circle.

These points bisect the sides of △ABC.

△CFE is isosceles (CF = CE because the lengths of two tangents to a circle from a point are equal), so ∠CFE = ∠FEC = (180 − 60) ÷ 2 = 60°. Therefore △CFE is equilateral.

It follows by symmetry that all four small triangles are equilateral and congruent. Hence the ratio of the area of the outer to the area of the inner triangle = 4 : 1.

*Question 2*Ratio of the area of the outer to the area of the inner triangle:

= (3 × area △OCB) : (3 × area △FDO)

= (3 × ½ × r × CB) : (3 × ½ × h × ½ × CB)

= 2r : h

Since we know from Q1 that this ratio is 4 : 1 ⇒ h =

^{r}⁄_{2}Ratio of area of circles:

= πr

^{2}: πh^{2}= πr

^{2}: π^{r2}⁄_{4}= 4 : 1

Darren's Method *Question 1*Area of △DEF = 3 × area of △OEF

⇒ ½ × 2n × (h + r) = 3 × ½ × 2n × h

⇒ n(h + r) = 3nh

⇒ h + r = 3h

⇒ h =

^{r}⁄_{2}Triangle OQE is similar to triangle OPB:

△POB is common to both triangles and OQE = OPB = 90° (altitudes of an equilateral triangle).

Therefore PB = 2n and so CB = 4n (altitudes of an equilateral triangle bisect a side).

Area △ABC = 3 × area △OBC = 3 × ½ × 4n × r = 6nr.

Area △DEF= 3 × area △OEF = 3 × ½ × 2n ×

^{r}⁄_{2}=^{3nr}⁄_{2}.Ratio of areas of triangles = 6nr :

^{3nr}⁄_{2}= 4 : 1.*Question 2*h is the radius of the inscribed circle.

h =

^{r}⁄_{2}.The ratio of the area of the outer circle to the area of the inner circle:

= πr

^{2}: πh^{2}= πr

^{2}: π^{r2}⁄_{4}= 4 : 1