Three horse are tethered at $3$ corners of a triangular plot having sides $20 m, 30 m$ and $40 m$ with ropes of $7 m$ length each. Find the area of this plot which can be grazed by the horses.

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Solution

Area of $△ A B C$
$= \sqrt{B (s - a) (3 - b) (3 - c)}$ sq.cm
Now, $s = \frac{20 + 40 + 30}{2} = 45$ m
$∴$ Area of $△ A B C = \sqrt{45 (45 - 20) (45 - 30) (45 - 40)}$ sq.cm
$\sqrt{45 \times 25 \times 15 \times 5} = 270.47$ sq.cm
Area of plot $= \frac{π r^{2}}{360} \times 180^{0} = \frac{1}{2} π r^{2}$
$= \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77$ sq.m
$∴$ AZrea of plot cannot be grazed $= 90.47 - 77$ sq.cm
$= 213.47$ sq.cm

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Three horse are tethered at 3 corners of a triangular plot having sides 20 m,30 m and 40 m with ropes of 7 m length each. Find the area of this plot which can be grazed by the horses.

Area of △ABC=√B(s−a)(3−b)(3−c) sq.cmNow, s=20+40+302=45 m∴ Area of △ABC=√45(45−20)(45−30)(45−40) sq.cm√45×25×15×5=270.47 sq.cmArea of plot=πr2360×1800=12πr2=12×227×7×7=77sq.m∴ AZrea of plot cannot be grazed =90.47−77 sq.cm=213.47 sq.cm

Three horse are tethered at $3$ corners of a triangular plot having sides $20 m, 30 m$ and $40 m$ with ropes of $7 m$ length each. Find the area of this plot which can be grazed by the horses.