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  • Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three a

Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Answers (1)

\left ( 24 \sqrt{21}-72 \right )m^{2} 

Solution

            

Let their angles of triangle are \angle A,\angle B  and  \angle C  rope’s length (radius) = 7 cm

Area of sector with angle A = \frac{\pi r^{2} \times \angle A }{360}=\frac{\pi \times (7)^{2} \times \angle A}{360}

Area of sector with angle B= \frac{\pi r^{2} \times \angle B }{360}=\frac{\pi \times (7)^{2} \times \angle B}{360}

Area of sector with angle C= \frac{\pi r^{2} \times \angle C }{360}=\frac{\pi \times (7)^{2} \times \angle C}{360}

\thereforeSum of the areas are

 =\frac{\left ( \angle A +\angle B+\angle C \right )\times \pi \times (7)^{2}}{360}

 =\frac{180}{360}\times\frac{22}{7}\times 49                       {\because  Sum of angles of a triangle = 180^{\circ} }

 =77 m2

Sides of triangular field are 15m, 16m and 17m

Let a =15m, b =16m, c = 17m

     \\S=\frac{(a+b+c)}{2}\\ =\frac{(15+16+17)}{2}\\ =\frac{48}{2}\\ =24m

Area of triangular field

 \\=\sqrt{s(s-a)(s-b)(s-c)}\\ =\sqrt{24(24-15)(24-16)(24-17)}\\ =\sqrt{24 \times 9 \times 8 \times 7}\\ =\sqrt{8 \times 3 \times 9 \times 8 \times 7}\\ =8\sqrt{3 \times 3 \times 3 \times 7}\\ =24 \sqrt{21}m^{2}

So area of the field which cannot be grazed by the animals = Area of triangular field – Area of sectors of field

                                                                                              =\left (24 \sqrt{21}-77 \right )m^{2}

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