Question

A horse is tethered to one corner of a square plot of side 42cm by 35cm long rope . Find the area it can graze and the area will be left ungrazed. ​

Answer 1

Step-by-step explanation:

Diagram:- Refer the attachment

Solution:-

Area which the horse can graze = Area of sector APXQ.

$\boxed{ \sf \: Area \: of \: the \: sector = \dfrac{ \theta}{360^{ \circ } } \times\pi {r}^{2} }$

Here:-

θ = 90°

r = Length of rope = 35cm

Area of sector APXQ

$\sf= \dfrac{ 90^{ \circ} }{360^{ \circ } } \times \dfrac{22}{7} \times 35 \times 35$

$\sf= \dfrac{1}{4 } \times 22 \times 5 \times 35$

$\sf= \dfrac{3850}{4 }$

$\sf= 962.5m^2$

$\underline{\boxed{\sf \therefore Area \: it \: can \: graze = 962.5cm^2}}$

Area of the square plot = 42 × 42 = 1764m²

Area which will be left ungrazed = Area of plot - Area it can graze.

= 1764 - 962.5

= 801.5m²

$\underline{\boxed{\sf \therefore Area \: which \: will \: be \: left \: ungrazed = 801.5m^2}}$

Answer 2

$\underline{\bf{Given:-}}$

❏ Side of square plot = 42cm

❏ Length of rope = 35cm

$\underline{\bf{Find:-}}$

❏ Area the horse can graze

❏ Area left ungrazed

$\underline{\bf{Diagram:-}}$

Let, ABCD be a square field and horse is tethered to a point A

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.4mm}\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf A}\put(-0.5,4.2){\bf D}\put(4.2,4.2){\bf C}\put(4.2, - 0.5){\bf B} \qbezier(2.3,0)(2.121,2.121)(0,2.3) \put(1,1){\circle*{0.1}}\put(1,1.3){\circle*{0.1}}\put(1.4,1){\circle*{0.1}}\put(0.7,0.2){\circle*{0.1}}\put(0.2,1){\circle*{0.1}}\put(0.6,2){\circle*{0.1}}\put(0.2,2){\circle*{0.1}}\put(0.5,1.3){\circle*{0.1}}\put(0.1,0.2){\circle*{0.1}}\put(1.2,0.2){\circle*{0.1}}\put(1.6,0.2){\circle*{0.1}}\put(0.2,0.8){\circle*{0.1}}\put(0.6,0.7){\circle*{0.1}} \put(2.1,-0.5){\bf E}\put(-0.5,2.1){\bf F} \put(4.3,2.2){\vector(0, 1){1.6}} \put(4.3,1.8){\vector(0, - 1){1.6}}\put(4.1,1.9){\bf 42m} \put(1.3, - 0.2){\vector(1,0){0.8}}\put(0.8, - 0.2){\vector(-1,0){0.8}} \put(0.8, - 0.4){\bf 35m} \end{picture}$

$\underline{\bf{Solution:-}}$

Here, horse can graze the dotted part only AEF

Now,

Area horse can graze = Quadrant of circle with Radius 35cm

we, know that

$\large{\underline{\boxed{\sf Area \: of \: quadrant \: of \: circle = \dfrac{1}{4} \pi {r}^{2} }}}$

where,

• π = 22/7
• r = 35cm

So,

$\dashrightarrow\sf Area \: horse \: can \: graze = \dfrac{1}{4} \pi {r}^{2}$

$\dashrightarrow\sf Area \: horse \: can \: graze = \dfrac{1}{4} \times \dfrac{22}{7} \times {(35)}^{2}$

$\dashrightarrow\sf Area \: horse \: can \: graze = \dfrac{22}{28} \times {(35)}^{2}$

$\dashrightarrow\sf Area \: horse \: can \: graze = \dfrac{22}{28} \times 1225$

$\dashrightarrow\sf Area \: horse \: can \: graze = \dfrac{26950}{28}$

$\dashrightarrow\sf Area \: horse \: can \: graze = 962.5 {cm}^{2}$

$\purple{\small{\therefore\sf Area \: horse \: can \: graze = 962.5 {cm}^{2}}}$

$\huge{ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}$

we, know that

$\large{\underline{\boxed{\sf Area \: of \: square= (side) \times (side)}}}$

where,

• side = 42cm

So,

$\mapsto \sf Area \: of \: square \: plot= (side) \times (side)$

$\mapsto \sf Area \: of \: square \: plot= (42) \times (42)$

$\mapsto \sf Area \: of \: square \: plot= 1764 {cm}^{2}$

Now,

$\large{\underline{\boxed{\sf Area \: left \: ungrazed=Area \: of \: square \: plot - Area \: horse \: can \: graze}}}$

where,

• Area of square = 1764cm²
• Area horse can graze = 962.5cm²

So,

$\leadsto\sf Area \: left \: ungrazed=Area \: of \: square \: plot - Area \: horse \: can \: graze$

$\leadsto\sf Area \: left \: ungrazed=1764-962.5$

$\leadsto\sf Area \: left \: ungrazed=801.5 {cm}^{2}$

$\small{\therefore\sf Area \: left \: ungrazed=801.5 {cm}^{2}}$

$\huge{ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}$

Hence,

• Area horse can graze = 962.5cm²
• Area horse can not able to graze = 801.5cm²