Sides of a triangular field are 15 m, 16 m and 17 m. From the three corners, a cow, a buffalo and a horse are tied with ropes of length 7 m each to graze the field. Find the area of the field which cannot be grazed by the three animals.
\(24\sqrt{21} - \dfrac{49\pi}{2}\,\text{m}^2 \;\approx\; 33.0\,\text{m}^2\)
Step 1: Find the area of the triangular field
The sides are 15 m, 16 m, and 17 m. To find the area of a triangle when all three sides are given, we use Heron’s formula:
\[ Delta = sqrt{s(s-a)(s-b)(s-c)} \]
Here, \(s\) is the semi-perimeter = half of the sum of sides.
\( s = \dfrac{15 + 16 + 17}{2} = \dfrac{48}{2} = 24 \, \text{m} \)
Now substitute into the formula:
\( Delta = sqrt{24(24-15)(24-16)(24-17)} \)
\( = sqrt{24 imes 9 imes 8 imes 7} \)
\( = sqrt{12096} \)
\( = 24\sqrt{21} \, \text{m}^2 \)
This is the total area of the triangular field.
Step 2: Find the total area grazed by the animals
Each animal is tied with a rope of length 7 m. So, each animal can graze in a circular sector of radius 7 m around the corner where it is tied.
The angle of the sector is equal to the angle of the triangle at that corner. The sum of all three angles of a triangle is \(180^\circ = \pi \, \text{radians}\).
The formula for the area of a sector is:
\( ext{Sector area} = \dfrac{1}{2} r^2 heta \)
Adding all three sectors together:
\( ext{Total grazed area} = \dfrac{1}{2} r^2 (\alpha + \beta + \gamma) \)
But \( \alpha + \beta + \gamma = \pi \).
So, \( ext{Total grazed area} = \dfrac{1}{2} imes 7^2 imes pi = \dfrac{49\pi}{2} \, \text{m}^2 \)
Step 3: Find the ungrazed area
The part of the field that cannot be grazed = (Total area of triangle) – (Total grazed area).
\( ext{Ungrazed area} = 24\sqrt{21} - \dfrac{49\pi}{2} \, \text{m}^2 \)
Now approximate the value:
\( 24\sqrt{21} \approx 109.9 \, \text{m}^2 \)
\( \dfrac{49\pi}{2} \approx 76.96 \, \text{m}^2 \)
So, \( ext{Ungrazed area} \approx 109.9 - 76.96 = 33.0 \, \text{m}^2 \)
Final Answer: The ungrazed area is about 33.0 m².