After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.
The area of a triangle with vertices known is given by the matrix
M =
![\left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20x_%7B1%7D%26y_%7B1%7D%261%5C%5Cx_%7B2%7D%26y_%7B2%7D%261%5C%5Cx_%7B3%7D%26y_%7B3%7D%261%5Cend%7Barray%7D%5Cright%5D%20)
Area = 1/2· | det(M) |
= 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
= 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |
Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
= 1/2· | -5·(1) - 4·(-13) - 1·(12) |
= 1/2 | 35 |
= 35/2
Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
= 1/2· | -5·(11) + 4·(-13) - 1·(2) |
= 1/2 | -109 |
<span> = 109/2</span>
The total area of the quadrilateral will be the sum of the areas of the two triangles:
A(ABCD) = A(ABC) + A(ADC)
= 35/2 + 109/2
= 72
Answer:
≈ 30.63 cm²
Step-by-step explanation:
The shaded area is calculated by subtracting the area of the inner circle from the area of the outer circle.
outer circle has radius = 8 ÷ 2 = 4 and inner circle has radius = 5 ÷ 2 = 2.5
shaded area = πr₁² - πr₂² (r₁ is outer and r₂ is inner )
A = π (4² - 2.5²)
= π(16 - 6.25) = 9.75π ≈ 30.63 cm²
Answer: 168 inches.
Explanation:
First thing you must always do when solving word problems is read closely, and take note of any number.
In this case, the first thing we must do is find the scale factor.
For this, we use one of the dimensions. We will use the width of the photo.
We have then:
k = 132/11
k = 12
Then, we look for the value of the height of the new photo. To do this, we multiply the scale factor by the original dimension.
We have then:
14k = 14 (12) = 168
Thus, 168 inches is your answer! :)
Hope this helps! :)