Question 1

Mathematics Area Medium

The area of the equilateral triangle is \(49\sqrt{3}cm^{2}\). Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. (Take \(\sqrt{3}=1.73\))

Question Image
(A) \(7.77~cm^{2}\)
(B) \(6.77~cm^{2}\)
(C) \(5.77~cm^{2}\)
(D) \(8.77~cm^{2}\)
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

To find the required area, we subtract the area of the three circular sectors from the triangle's total area.

1. Find the side of the triangle (a):
The area of an equilateral triangle is \(A = \frac{\sqrt{3}}{4}a^2\).
Given \(A = 49\sqrt{3} \text{ cm}^2\).
\[\begin{aligned} 49\sqrt{3} &= \frac{\sqrt{3}}{4}a^2 \\ 49 \times 4 &= a^2 \\ a &= \sqrt{196} = 14 \text{ cm} \end{aligned}\]
2. Find the radius of the circles (r):
The radius is half the side length.
\[r = \frac{a}{2} = \frac{14}{2} = 7 \text{ cm}\]
3. Find the area of the three sectors:
The three \(60^\circ\) sectors at the vertices combine to form a semicircle (\(3 \times 60^\circ = 180^\circ\)).
Area of three sectors = Area of a semicircle.
\[A_{\text{sectors}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (7)^2 = 77 \text{ cm}^2\]
4. Calculate the final area:
First, find the numerical area of the triangle using \(\sqrt{3} = 1.73\).
\[A_{\text{triangle}} = 49\sqrt{3} = 49 \times 1.73 = 84.77 \text{ cm}^2\]
Now, subtract the area of the sectors from the area of the triangle.
\[A_{\text{required}} = A_{\text{triangle}} - A_{\text{sectors}} = 84.77 - 77 = 7.77 \text{ cm}^2\]
Result:
The area of the triangle not included in the circles is \(7.77 \text{ cm}^2\).