Question

Sides of a right triangular field are 25m, 24m and 7m. At the three corners of the field, a cow, a buffalo
and a horse are tied separately with ropes of 3.5 m each to graze in the field. Find the area of the field
that cannot be grazed by these animals.
​

Answer 1

AnswEr :

⋆ Sides of Right Triangular Field :

• Hypotenuse = 25 m
• Perpendicular = 24 m
• Base = 7 m

⋆ A Cow, A Buffalo, and A Horse are tied separately with ropes of 3.5 m at three corners.

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{A}}\put(7.7,1){\large{B}}\put(10.6,1){\large{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\mathsf{\large{24 m}}}\put(9,0.7){\matsf{\large{7 m}}}\put(9.4,1.9){\mathsf{\large{25 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

• Area of Right Triangular Field ABC :

$\longrightarrow \tt Area_{\tiny ABC} =\dfrac{1}{2} \times Base \times Height \\ \\\longrightarrow \tt Area_{\tiny ABC} =\dfrac{1}{2} \times7 \times 24 \\ \\\longrightarrow \tt Area_{\tiny ABC} =\dfrac{1}{ \cancel2} \times\cancel{168} \\ \\\longrightarrow \blue{\tt Area_{\tiny ABC} = 84 \: {m}^{2}}$

∴ Total Area of Grass Field is 84 m².

$\rule{300}{1}$

Now, if these animals are tied at the corners and will make Sector i.e. (3 Sectors of Radius 3.5 m), So will Find the Area that Animals can actually Graze.

• Area that Animals can Graze are :

$\longrightarrow \tt Area = \dfrac{\angle A}{360\degree}\pi {r}^{2} + \dfrac{\angle B}{360\degree}\pi {r}^{2} + \dfrac{\angle C}{360\degree}\pi {r}^{2} \\ \\\longrightarrow \tt Area = \pi {r}^{2} \bigg(\dfrac{\angle A}{360\degree}+ \dfrac{\angle B}{360\degree} + \dfrac{\angle C}{360\degree} \bigg)\\ \\\longrightarrow \tt Area = \pi {r}^{2} \bigg(\dfrac{\angle A +\angle B +\angle C}{360\degree}\bigg) \\ \\\longrightarrow \tt Area = \pi {r}^{2} \bigg( \cancel\dfrac{180 \degree}{360\degree}\bigg) \\ \\\longrightarrow \tt Area = \dfrac{22}{7} \times 3.5 \times 3.5 \times \dfrac{1}{2} \\ \\\longrightarrow \blue{\tt Area =19.25 \: {m}^{2}}$

∴ Total Area that can Graze is 19.25 m²

$\rule{300}{2}$

• AREA THAT CAN'T BE GRAZE IS :

⇒ Area Can't Graze = Total Area of Field – Area Can Graze

⇒ Area Can't Graze = (84 – 19.25) m²

⇒ Area Can't Graze = 64.75 m²

⠀

∴ Area of Field that can't graze is 64.75 m²

#answerwithquality #BAL

Answer 2

Answer:

$\underline{\underline{\mathfrak{Required \ area\ =\ 64.75\ {m}^{2}}}}$

$\huge{\sf{Solution}} :$

Sides of the triangle = 25 m, 24 m & 7m

⇒ Perimeter = (25+24+7) m

⇒ Perimeter = 56 m

⇒ Semi perimeter = 56/2 = 28 m

__________________________

Area of the triangle :-

⇒ √[s (s-a) (s-b) (s-c)]

⇒ √[28 (28-25)(28-24)(28-7)]

⇒ √[28 * 3 * 4 * 21]

⇒ √[7056]

⇒ 84 m²

___________________________

Three animals are tied with ropes of 3.5 m.

So total area that can be grazed :-

$\longmapsto\bold{3 \times \dfrac{\theta}{360\degree}\times\pi{r}^{2}}$

$\longmapsto\sf\frac{(Angle \ A + Angle\ B+Angle \ C)}{360 ^{\circ}}\times 3.1416\times{3.5}^{2}$

Here, A + B + C = 180°

⇒ 180°/360° × 3.1416 × 12.25

⇒ 19.25 m²

__________________________

$\longmapsto\bold{Area \: that \: can \: be \: grazed = 19.25 {m}^{2}}$

Now, area that cannot be grazed :-

$\longmapsto\bold{Total \ area - Area \ that \ can \ be \ grazed}$

$\longmapsto\sf(84 - 19.25) \ {m}^{2}$

$\longmapsto{\underline{\boxed{\large\bold{ 64.75 \ {m}^{2}}}} \ \star$

Therefore,

Area that cannot be grazed by these animals = 64.75 m²