Question

The sides of a triangular field are 35m, 50m and 110m. Find the
length of the wire required to fence the field.​

Answer 1

Answer:

Length of wire = perimeter of triangular field

i.e

35+50+110 = 195 m

Answer 2

$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Question}}}$

The sides of a triangular field are 35m, 50m and 110m. Find the length of the wire required to fence the field.

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Answer}}}$

The sides of a triangular field are 35m, 50m and 110m. 195m the length of the wire required to fence the field .

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★ Formula Used :-

$\begin{gathered}\begin{gathered}\\\;\boxed{\sf{\red{perimeter\;of\; triangle\;=\;\bf{sum \: of \: all \: sides}}}}\end{gathered} \end{gathered}$

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Solution}}}$

Given,

» side of triangular field

• 35 m

• 50 m

• 110 m

To find ,

» length of the wire required to fence the field

~ Let's solve it ,

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we know that ,

$\begin{gathered}\begin{gathered}\\\;{\sf{{perimeter\;of\; triangle\;=\;\bf{sum \: of \: all \: sides}}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\\;\large{ \longrightarrow{\sf perimeter \: of \: triangle = 35 + 50 + 110}}\end{gathered}\end{gathered} \end{gathered}\end{gathered} \end{gathered} \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\\;\large{ \longrightarrow{\sf perimeter \: of \: triangle = 195m}}\end{gathered}\end{gathered} \end{gathered}\end{gathered} \end{gathered} \end{gathered}\end{gathered}$

$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: hence , length \: of \: wire \: required \: for \: fencing = \underline{195 \: m}}}}$

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★ More to know :-

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; perimeter\;of\;rectangle\;=\;2(length\;\times\;breadth)}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; perimeter\;of\;square\;=\;4a}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; perimeter\;of\;parallelogram\;=\;2(sum \: of \: two \: adjacent \: sides)}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; perimeter\;of\;circle\;=\;2\pi r}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; perimeter\;of\;triangle\;=\;sum \: of \: three \: sides}}\end{gathered} \end{gathered}$