Anyone solve this......xd

1.	Anyone solve this......xd
Answer» $for \: finding \: area \: of \: the \: park \\ \\ we \: have \\ \\ 2s = 50m + 120m + 80m = 250m \\ \\ ie \: \: s = 125m \\ \\ therefore \: area \: of \: the \: park \\ \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{125(125 - 50)(125 - 80)(125 - 120)} \\ \\ = \sqrt{125 \times 75 \times 45 \times 5} \\ \\ = 375 \sqrt{15} \: \: {m}^{2} \\ \\ also \: primeter \: of \: the \: park \: \\ \\ = ab + bc + ca \\ \\ = 250m \\ \\ therefore \\ \: length \: of \: the \: wire \: needed \\ \: for \: fencing \\ \\ = (250 - 3)m \\ \\ = 247m \\ \\ cost \: of \: fencing \: \\ \\ = 20 \times 247 \\ \\ = rs \: 4940$ *THANK* *and* *MARK* as *BRAINLIEST*... ❤️

Answer»

$for \: finding \: area \: of \: the \: park \\ \\ we \: have \\ \\ 2s = 50m + 120m + 80m = 250m \\ \\ ie \: \: s = 125m \\ \\ therefore \: area \: of \: the \: park \\ \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{125(125 - 50)(125 - 80)(125 - 120)} \\ \\ = \sqrt{125 \times 75 \times 45 \times 5} \\ \\ = 375 \sqrt{15} \: \: {m}^{2} \\ \\ also \: primeter \: of \: the \: park \: \\ \\ = ab + bc + ca \\ \\ = 250m \\ \\ therefore \\ \: length \: of \: the \: wire \: needed \\ \: for \: fencing \\ \\ = (250 - 3)m \\ \\ = 247m \\ \\ cost \: of \: fencing \: \\ \\ = 20 \times 247 \\ \\ = rs \: 4940$

THANK and MARK as BRAINLIEST...

❤️

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