Respuesta :
A(2, 3); B(4, 9); C(7, 8); D(5, 2)
1. How long is the longer edge of the table?
[tex]AB= \sqrt{(x_b-x_a)^2+(y_b-y_a)^2} = \sqrt{(4-2)^2+(9-3)^2}= \\ = \sqrt{2^2+6^2}= \sqrt{4+36}= \sqrt{40} \ \text{feet} [/tex]
2. What is the area of the tabletop?
[tex]BC= \sqrt{(x_c-x_b)^2+(y_c-y_b)^2} = \sqrt{(7-4)^2+(8-9)^2}= \\ = \sqrt{3^2+(-1)^2}= \sqrt{9+1}= \sqrt{10} \ \text{feet} \\ \\ Area=AB*BC= \sqrt{40}* \sqrt{10}= \sqrt{400}= 20 \ \text{square feet}[/tex]
Answer: √40; 20.
1. How long is the longer edge of the table?
[tex]AB= \sqrt{(x_b-x_a)^2+(y_b-y_a)^2} = \sqrt{(4-2)^2+(9-3)^2}= \\ = \sqrt{2^2+6^2}= \sqrt{4+36}= \sqrt{40} \ \text{feet} [/tex]
2. What is the area of the tabletop?
[tex]BC= \sqrt{(x_c-x_b)^2+(y_c-y_b)^2} = \sqrt{(7-4)^2+(8-9)^2}= \\ = \sqrt{3^2+(-1)^2}= \sqrt{9+1}= \sqrt{10} \ \text{feet} \\ \\ Area=AB*BC= \sqrt{40}* \sqrt{10}= \sqrt{400}= 20 \ \text{square feet}[/tex]
Answer: √40; 20.
Answer:
The answer is 40 and 20 for the second one
Step-by-step explanation: