From tech troubles to travel tips, IDNLearn.com has answers to all your questions. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.

GUY!!! PLEASE HELP it’s due in abit and I just need this final question! I’ll mark you brainliest

(geometry)
what is the perimeter of the rectangle ABCD?


GUY PLEASE HELP Its Due In Abit And I Just Need This Final Question Ill Mark You Brainliest Geometry What Is The Perimeter Of The Rectangle ABCD class=

Sagot :

Answer:

37.9 units (nearest tenth)

Step-by-step explanation:

As ABCD is a rectangle, AD = BC and AB = DC.

Find the length of AD and AB using the distance formula.

[tex]\boxed{\begin{minipage}{7.5 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}}[/tex]

From inspection of the given diagram:

  • A = (-5, -9)
  • B = (7, -5)
  • D = (-7, -3)

Therefore:

[tex]\begin{aligned}AD & =\sqrt{(x_D-x_A)^2+(y_D-y_A)^2}\\& =\sqrt{(-7-(-5))^2+(-3-(-9))^2}\\& =\sqrt{(-2)^2+(6)^2}\\& =\sqrt{4+36}\\& =\sqrt{40}\end{aligned}[/tex]

[tex]\begin{aligned}AB & =\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\& =\sqrt{(7-(-5))^2+(-5-(-9))^2}\\& =\sqrt{(12)^2+(4)^2}\\& =\sqrt{144+16}\\& =\sqrt{160}\end{aligned}[/tex]

[tex]\begin{aligned}\textsf{Perimeter of a rectangle} & = \sf 2(length + width)\\& = 2(AB+AD)\\& = 2(\sqrt{160}+\sqrt{40})\\& = 2(18.97366596...)\\& = 37.94733192...\\& = 37.9\: \sf units\:(nearest\:tenth)\end{aligned}[/tex]

Therefore, the perimeter of the rectangle ABCD is 37.9 units (nearest tenth).

Learn more about the distance formula here:

https://brainly.com/question/28144723

https://brainly.com/question/28247604