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Question

The ratio of the area of a circle and that of an equilateral triangle, where the diameter of the circle is equal to the sides of the equilateral tringle, is:

The correct answer is

π : \(\sqrt3\)

Understanding the Geometric Problem

The question asks us to find the ratio of the area of a circle to the area of an equilateral triangle under a specific condition: the diameter of the circle is equal to the side of the equilateral triangle. To solve this, we need the formulas for the area of both shapes and then use the given relationship to find the ratio.

Key Formulas for Area

Let's first recall the necessary area formulas:

  • Area of a Circle: If the radius of a circle is \(r\), its area is given by \(A_{circle} = \pi r^2\). If the diameter is \(d\), then \(r = \frac{d}{2}\), so the area can also be written as \(A_{circle} = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\).
  • Area of an Equilateral Triangle: If the side length of an equilateral triangle is \(s\), its area is given by \(A_{triangle} = \frac{\sqrt{3}}{4} s^2\).

Setting up the Relationship

The problem states that the diameter of the circle is equal to the side of the equilateral triangle. Let's represent the diameter of the circle by \(d\) and the side of the equilateral triangle by \(s\). The given condition is:

\(d = s\)

Calculating the Areas based on the Relationship

Now we can express the areas of both shapes using a single variable (let's use \(d\), since \(s=d\)).

  • Area of the Circle: Using the formula with diameter, \(A_{circle} = \frac{\pi d^2}{4}\).
  • Area of the Equilateral Triangle: Using the formula with side \(s\), and knowing \(s=d\), we substitute \(d\) for \(s\): \(A_{triangle} = \frac{\sqrt{3}}{4} d^2\).

Finding the Ratio of Areas

We need to find the ratio of the area of the circle to the area of the equilateral triangle, which is \(A_{circle} : A_{triangle}\) or \(\frac{A_{circle}}{A_{triangle}}\).

Ratio = \(\frac{\text{Area of Circle}}{\text{Area of Equilateral Triangle}}\) Ratio = \(\frac{\frac{\pi d^2}{4}}{\frac{\sqrt{3} d^2}{4}}\)

We can see that the term \(\frac{d^2}{4}\) appears in both the numerator and the denominator. Assuming \(d \neq 0\) (a circle and triangle must have a non-zero size), we can cancel this term out.

Ratio = \(\frac{\pi}{\sqrt{3}}\)

This ratio can be written as \(\pi : \sqrt{3}\).

Conclusion on the Ratio of Areas

The ratio of the area of the circle to that of the equilateral triangle, when the diameter equals the side, is \(\pi : \sqrt{3}\).

Shape Formula (in terms of diameter 'd' or side 's') Formula (in terms of 'd' since d=s) Calculated Area
Circle \(A_{circle} = \frac{\pi d^2}{4}\) \(A_{circle} = \frac{\pi d^2}{4}\) \(\frac{\pi d^2}{4}\)
Equilateral Triangle \(A_{triangle} = \frac{\sqrt{3}}{4} s^2\) \(A_{triangle} = \frac{\sqrt{3}}{4} d^2\) \(\frac{\sqrt{3} d^2}{4}\)

Ratio \(A_{circle} : A_{triangle} = \frac{\pi d^2}{4} : \frac{\sqrt{3} d^2}{4} = \pi : \sqrt{3}\).

Revision Table: Geometry Formulas

Shape Key Properties Area Formula
Circle Radius (r), Diameter (d = 2r) \(\pi r^2\) or \(\frac{\pi d^2}{4}\)
Equilateral Triangle Side (s), All sides equal, All angles 60° \(\frac{\sqrt{3}}{4} s^2\)

Additional Information: Understanding Ratios in Geometry

Ratios in geometry help compare the sizes of different figures or parts of figures. When calculating a ratio like \(A:B\), it means \(\frac{A}{B}\). It's important to ensure that the quantities being compared are in the same units and are related in a way that allows for simplification or comparison. In this problem, relating the diameter of the circle to the side of the triangle was the crucial step to finding a numerical (or in this case, symbolic) ratio.

Equilateral triangles have special properties due to their symmetry. Their height is \(\frac{\sqrt{3}}{2}s\), which is derived using the Pythagorean theorem on a 30-60-90 right triangle formed by the altitude. This height is used in the derivation of the area formula \( \frac{1}{2} \times base \times height = \frac{1}{2} \times s \times \frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{4} s^2 \).

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Important Questions from Plane Figures

  1. If the area of a square is 625 cm 2, then what is the perimeter of the square?

  2. The area and the perimeter of a sheet of paper are 240 cm 2and 68 cm, respectively. What would be its length and breadth?

  3. One side of rectangular field is 15 meters and one of its diagonals is 17 meters. Then find the area of the field.

  4. The bisector of ∠B in ΔABC meets AC at D. If AB = 12 cm, BC = 18 cm and AC = 15 cm, then the length of AD (in cm) is:

  5. The perimeter and the length of one of the diagonals of a rhombus is 26 cm and 5 cm respectively. Find the length of its other diagonal (in cm).

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