1. **Problem statement:** Given a logo composed of three circles each with radius $6$ cm intersecting to form a shaded equilateral triangle inside, find: a. The area of the shaded equilateral triangle, rounded to 0.1 cm². b. The total area of the composite logo (the union of the three circles), rounded to 0.1 cm². Use $\pi \approx 3.14$. 2. **Formulas and rules:** - Area of an equilateral triangle with side length $a$ is: $$\text{Area} = \frac{\sqrt{3}}{4} a^2$$ - The side length $a$ of the equilateral triangle formed by the centers of the circles equals the distance between circle centers, which is $2 \times$ radius = $2 \times 6 = 12$ cm. - Area of a circle is: $$A = \pi r^2$$ - The total area of the logo is the union of three circles minus the overlapping parts. The overlapping area of two circles of radius $r$ with centers $a$ apart is given by: $$A_{overlap} = 2r^2 \arccos\left(\frac{a}{2r}\right) - \frac{a}{2} \sqrt{4r^2 - a^2}$$ Since the circles intersect such that the centers form an equilateral triangle of side $12$ cm, the overlapping areas are symmetric. 3. **Step a: Area of the shaded equilateral triangle** - Side length $a = 12$ cm. - Calculate area: $$\text{Area} = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36 \sqrt{3}$$ - Approximate $\sqrt{3} \approx 1.732$: $$36 \times 1.732 = 62.352$$ - Rounded to 0.1: $$62.4 \text{ cm}^2$$ 4. **Step b: Total area of the logo** - Area of one circle: $$A_{circle} = 3.14 \times 6^2 = 3.14 \times 36 = 113.04$$ - Total area of three circles without overlap: $$3 \times 113.04 = 339.12$$ - Calculate overlap area between two circles: Distance between centers $a = 12$ cm, radius $r=6$ cm. Calculate: $$\arccos\left(\frac{12}{2 \times 6}\right) = \arccos(1) = 0$$ Since $a = 2r$, the circles touch at exactly one point, so overlap area between two circles is 0. - However, the problem states the circles intersect forming a triangle, so the distance between centers must be less than $2r$ for overlap. Given the problem's figure and description, the side length of the triangle formed by the centers is $6$ cm (radius), not $12$ cm. Recalculate side length: - The triangle inside is equilateral with side equal to radius $6$ cm. - Area of triangle: $$\frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9 \sqrt{3} \approx 9 \times 1.732 = 15.588$$ - Rounded to 0.1: $$15.6 \text{ cm}^2$$ - Area of one circle: $$3.14 \times 36 = 113.04$$ - Distance between centers $a = 6$ cm. Calculate overlap area between two circles: $$A_{overlap} = 2 \times 6^2 \times \arccos\left(\frac{6}{2 \times 6}\right) - \frac{6}{2} \times \sqrt{4 \times 6^2 - 6^2}$$ Calculate each term: $$2 \times 36 \times \arccos(0.5) = 72 \times \frac{\pi}{3} = 72 \times 1.047 = 75.384$$ $$\frac{6}{2} = 3$$ $$\sqrt{4 \times 36 - 36} = \sqrt{144 - 36} = \sqrt{108} = 10.392$$ So: $$A_{overlap} = 75.384 - 3 \times 10.392 = 75.384 - 31.176 = 44.208$$ - There are 3 pairs of overlapping circles, so total overlap area: $$3 \times 44.208 = 132.624$$ - The triple overlap area (the shaded triangle) is counted three times in overlaps, so add it back twice: Total area: $$3 \times 113.04 - 132.624 + 15.588 = 339.12 - 132.624 + 15.588 = 222.084$$ - Rounded to 0.1: $$222.1 \text{ cm}^2$$ **Final answers:** - a) Area of shaded triangle: $15.6$ cm² - b) Total logo area: $222.1$ cm²