Given:
The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).
To find:
The area of the rectangle.
Solution:
Distance formula:
![D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=D%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Using the distance formula, we get
![AB=\sqrt{(2-0)^2+(4-1)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282-0%29%5E2%2B%284-1%29%5E2%7D)
![AB=\sqrt{(2)^2+(3)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282%29%5E2%2B%283%29%5E2%7D)
![AB=\sqrt{4+9}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B4%2B9%7D)
![AB=\sqrt{13}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B13%7D)
Similarly,
![BC=\sqrt{(6-2)^2+(0-4)^2}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%286-2%29%5E2%2B%280-4%29%5E2%7D)
![BC=\sqrt{(4)^2+(-4)^2}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%284%29%5E2%2B%28-4%29%5E2%7D)
![BC=\sqrt{16+16}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B16%2B16%7D)
![BC=\sqrt{32}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B32%7D)
![BC=4\sqrt{2}](https://tex.z-dn.net/?f=BC%3D4%5Csqrt%7B2%7D)
Now, the length of the rectangle is
and the width of the rectangle is
. So, the area of the rectangle is:
![A=length \times width](https://tex.z-dn.net/?f=A%3Dlength%20%5Ctimes%20width)
![A=\sqrt{13}\times 4\sqrt{2}](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B13%7D%5Ctimes%204%5Csqrt%7B2%7D)
![A=4\sqrt{26}](https://tex.z-dn.net/?f=A%3D4%5Csqrt%7B26%7D)
![A\approx 20](https://tex.z-dn.net/?f=A%5Capprox%2020)
Therefore, the area of the rectangle is 20 square units.
Answer:
..............
Step-by-step explanation:
Jskkdur
Sjdhrb
Zjdhbrj
Dhrbus
Djrjrvd
.......
x = 2y
1/x + 1/y = 3/10
Since we have a value for x, let's plug it into the second equation.
1/2y + 1/y = 3/10
Now, let's make the denominators equal.
Multiply the second term by 2.
1/2y + 2/2y = 3/10
Multiply the final term by 0.2y
1/2y + 2/2y = 0.6y/2y
Compare numerators after adding.
3 = 0.6y
Divide both sides by 0.6
<h3>y = 5</h3>
Now that we have the value of the second integer, we can find the first.
x = 2y
x = 2(5)
<h3>x = 10</h3>
Let's plug in these values in our equations to verify.
10 = 2(5) √ this is true
1/10 + 1/5 = 3/10 √ this is true
<h3>The first integer is equal to 10, and the second is equal to 5.</h3>