Answer:
\large\underline{\sf{Given- }}
Given−
A cow is tied with a rope of length 14m at the corner of rectangular field of dimesions 20m x 16m.
\begin{gathered} \\ \\ \large\underline{\sf{To\:Find - }}\end{gathered}
ToFind−
A of field in which cow can graze.
\begin{gathered} \\ \\ \large\underline{\sf{Solution-}}\end{gathered}
Solution−
Given that,
Since, the cow is tied with a rope of 14 m at one of the corner of rectangular field.
So, Area grazed by cow is quadrant of a circle of radius, r = 14 m.
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = Area_{(quadrant)} \\ \\ \end{gathered}
Area
(grazedbycow)
=Area
(quadrant)
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = \frac{1}{4} \pi \: {r}^{2} \\ \\ \end{gathered}
=
4
1
πr
2
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \\ \\ \end{gathered}
×
7
22
×14×14
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = 22 \times 7 \\ \\ \end{gathered}
=22×7
\begin{gathered}\bf\implies \: Area_{(grazed \: by \: cow)} = 154 \: {m}^{2} \\ \\ \\ \end{gathered}
⟹Area
=154m
\rule{190pt}{2pt}
Additional information
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}
†
FormulasofAreas:−
⋆Square=(side)
⋆Rectangle=Length×Breadth
⋆Triangle=
×Base×Height
⋆Scalene△=
s(s−a)(s−b)(s−c)
⋆Rhombus=
×d
d
4a
−d
⋆Parallelogram=Base×Height
⋆Trapezium=
(a+b)×Height
⋆EquilateralTriangle=
3
(side)
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Answers & Comments
Verified answer
Answer:
\large\underline{\sf{Given- }}
Given−
A cow is tied with a rope of length 14m at the corner of rectangular field of dimesions 20m x 16m.
\begin{gathered} \\ \\ \large\underline{\sf{To\:Find - }}\end{gathered}
ToFind−
A of field in which cow can graze.
\begin{gathered} \\ \\ \large\underline{\sf{Solution-}}\end{gathered}
Solution−
Given that,
A cow is tied with a rope of length 14m at the corner of rectangular field of dimesions 20m x 16m.
Since, the cow is tied with a rope of 14 m at one of the corner of rectangular field.
So, Area grazed by cow is quadrant of a circle of radius, r = 14 m.
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = Area_{(quadrant)} \\ \\ \end{gathered}
Area
(grazedbycow)
=Area
(quadrant)
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = \frac{1}{4} \pi \: {r}^{2} \\ \\ \end{gathered}
Area
(grazedbycow)
=
4
1
πr
2
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \\ \\ \end{gathered}
Area
(grazedbycow)
=
4
1
×
7
22
×14×14
\begin{gathered}\sf \: Area_{(grazed \: by \: cow)} = 22 \times 7 \\ \\ \end{gathered}
Area
(grazedbycow)
=22×7
\begin{gathered}\bf\implies \: Area_{(grazed \: by \: cow)} = 154 \: {m}^{2} \\ \\ \\ \end{gathered}
⟹Area
(grazedbycow)
=154m
2
\rule{190pt}{2pt}
Additional information
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}
†
FormulasofAreas:−
⋆Square=(side)
2
⋆Rectangle=Length×Breadth
⋆Triangle=
2
1
×Base×Height
⋆Scalene△=
s(s−a)(s−b)(s−c)
⋆Rhombus=
2
1
×d
1
×d
2
⋆Rhombus=
2
1
d
4a
2
−d
2
⋆Parallelogram=Base×Height
⋆Trapezium=
2
1
(a+b)×Height
⋆EquilateralTriangle=
4
3
(side)
2