Question 26 of 42advanced🔍 AnalyzeLong Answer5 marks

An equilateral triangle is inscribed in a circle of radius rr. Show that the ratio of the area of the triangle to the area of the circle is equal to 334π≈0.413\dfrac{3\sqrt{3}}{4\pi} \approx 0.413.

Correct Answer

The ratio of the area of the triangle to the area of the circle is 334π≈0.413\dfrac{3\sqrt{3}}{4\pi} \approx 0.413.

Exercise: Exercise Set 6.3 | Q: 8 | (Page 31)
For More Understanding

Explanation

The textbook context provides a hint that the answer for the hexagon is exactly twice the answer for the equilateral triangle. This implies the area of the inscribed equilateral triangle is exactly half the area of the inscribed regular hexagon.

By calculating the area of the hexagon and subtracting the area of the three isosceles triangles formed at the alternate vertices, we can find the area of the inscribed equilateral triangle and subsequently the required ratio.

Solution Steps

  1. Step 1: The area of the circle with radius rr is πr2\pi r^2.

  2. Step 2: Consider a regular hexagon inscribed in the same circle. It has side length rr and is composed of 6 equilateral triangles of side rr. The area of the hexagon is:

    6×34r2=332r26 \times \frac{\sqrt{3}}{4}r^2 = \frac{3\sqrt{3}}{2}r^2

  3. Step 3: The inscribed equilateral triangle is formed by joining alternate vertices of the hexagon. This divides the hexagon into the inscribed equilateral triangle and 3 smaller isosceles triangles at the remaining vertices. Each isosceles triangle has sides rr, rr and an included angle of 120°120°. Dropping a perpendicular from the 120°120° angle to the base (length rr) gives a height of r32\dfrac{r\sqrt{3}}{2}.

    The area of one isosceles triangle is:

    12×r×r32=34r2\frac{1}{2} \times r \times \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{4}r^2

    The total area of the 3 isosceles triangles is:

    3×34r2=334r23 \times \frac{\sqrt{3}}{4}r^2 = \frac{3\sqrt{3}}{4}r^2

  4. Step 4: The area of the inscribed equilateral triangle is the area of the hexagon minus the area of the 3 isosceles triangles:

    332r2−334r2=334r2\frac{3\sqrt{3}}{2}r^2 - \frac{3\sqrt{3}}{4}r^2 = \frac{3\sqrt{3}}{4}r^2

    This confirms the triangle's area is exactly half the hexagon's area.

  5. Step 5: The ratio of the area of the triangle to the area of the circle is:

    334r2πr2=334π≈0.413\frac{\dfrac{3\sqrt{3}}{4}r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi} \approx 0.413
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