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Do this ASAP and put answers in the "Answer" Tab. Any Absurd answers shall be reported and deleted. Remember when your answer gets deleted you loose the points you received. [tex]\boxed{ Perimeter\; and\; Area\; on \;the \;Coordinate \;Plane}[/tex]

Do this ASAP and put answers in the Answer Tab Any Absurd answers shall be reported and deleted Remember when your answer gets deleted you loose the points you class=
Do this ASAP and put answers in the Answer Tab Any Absurd answers shall be reported and deleted Remember when your answer gets deleted you loose the points you class=
Do this ASAP and put answers in the Answer Tab Any Absurd answers shall be reported and deleted Remember when your answer gets deleted you loose the points you class=

Respuesta :

[tex].....................part \: 1. \: a.......................... \\\boxed{ Perimeter} \\ the \: perimeter \: of \: a \: triangle \: is \: given \: by \to\\ \boxed{p} = a + b + c........ \\ were \: \\ a \: b \: and \: c \: are \: the \: sides \: \: of \:the \: triangle \: when : \\ a = su = \: ?\\ b = ut = 8 \: units\\ c = ts = 14 \: units\\ to \: find \: su \: we \: apply \: pythagorean \: rule\: \to \: \\ (su) {}^{2} = {(ut)}^{2} + {(ts)}^{2} \\ su = \sqrt{ {8}^{2} + {14}^{2} } \\ su = 2 \sqrt{65} \\ hence : \\ \boxed{p} = 8 + 14 + 2 \sqrt{65} \\ p = 22 + 2 \sqrt{65} \\\boxed{p= 38 .12\: units}\\ \\ .......................part \: 1. \: b....................... \\ \boxed{Area} \\ the \: area \: of \: a \: triangle \: is \: given \: by : \\ A = \frac{1}{2} bc \\ hence : \\ A = \frac{1}{2} \times 8 \times 14 \\ \boxed{ A = 56\: {units}^{2}}[/tex]

Note.......I took care of part 1 only......cos handling both questions on here will be much on me

or.......you can set part 2 as a different question......but applying the same methods for the triangle will definitly work for the rectangle.

♨Rage♨

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