Respuesta :
Hiya,
So I did this quiz just now and I have the answers. c:
1.) What is the area of a rectangle with vertices at
4.) What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
Answer: 21.6 Units
5.) What is the perimeter of a polygon with vertices at
So I did this quiz just now and I have the answers. c:
1.) What is the area of a rectangle with vertices at
(−3, −1)(−3, −1) , (1, 3)(1, 3) , (3, 1)(3, 1) , and (−1, −3)(−1, −3) ?
Enter your answer in the box. Do not round any side lengths.
Answer: 16 Units
2.) What is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?
Answer: 26.8 Units
3.) What is the area of a triangle with vertices at
(0, −2) , (8, −2) , and (9, 1) ?
Enter your answer in the box.
Answer: 12 Units4.) What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
Answer: 21.6 Units
5.) What is the perimeter of a polygon with vertices at
(−2, 1) , (−2, 4) , (2, 7) , (6, 4) , and (6, 1) ?
Enter your answer in the box. Do not round any side lengths.
Answer: 24 Units
I really hope this helps! Let me know if it did... or anyone else that used this info!
The length of the side joining points (-3, -1) and (1, 3) is given by
[tex]l= \sqrt{(1+3)^2+(3+1)^2} \\ \\ = \sqrt{4^2+4^2} = \sqrt{16+16} \\ \\ = \sqrt{32} [/tex]
The length of the side joining points (1, 3) and (3, 1) is given by
[tex]w= \sqrt{(3-1)^2+(1-3)^2} \\ \\ = \sqrt{2^2+(-2)^2} = \sqrt{4+4} \\ \\ = \sqrt{8}[/tex]
The area of a rectangle is given by length times width.
Thus, the area of the given rectangle is given by
[tex]Area=l\times w \\ \\ = \sqrt{32} \times \sqrt{8} \\ \\ = \sqrt{32\times8} = \sqrt{256} \\ \\ =\bold{16} \ units^2[/tex]
[tex]l= \sqrt{(1+3)^2+(3+1)^2} \\ \\ = \sqrt{4^2+4^2} = \sqrt{16+16} \\ \\ = \sqrt{32} [/tex]
The length of the side joining points (1, 3) and (3, 1) is given by
[tex]w= \sqrt{(3-1)^2+(1-3)^2} \\ \\ = \sqrt{2^2+(-2)^2} = \sqrt{4+4} \\ \\ = \sqrt{8}[/tex]
The area of a rectangle is given by length times width.
Thus, the area of the given rectangle is given by
[tex]Area=l\times w \\ \\ = \sqrt{32} \times \sqrt{8} \\ \\ = \sqrt{32\times8} = \sqrt{256} \\ \\ =\bold{16} \ units^2[/tex]