There are two ways to find or determine for the value of
c. In the first method, we can use addition and subtraction to isolate the
variable c from the other variables. In the second method, we can use the
transposition of variables to isolate the variable c from the other variables.
So solving for the value of c:
<span>Using 1st method: Addition and Subtraction</span>
We are given:
240 = 6 z + c
Simply subtract 6 z on both sides:
240 – 6 z = 6 z + c – 6 z
Cancelling 6 z – 6 z on the right side:
240 – 6 z = c
or
c = 240 – 6 z
<span>Using the 2nd method: Transposition</span>
240 = 6 z + c
What we are going to do here is to simply transpose the
variable 6 z from the right side to the left side of the equation so that we
are left with c alone on the right side. Always remember that when we
transpose, the symbol becomes opposite. That is:
240 + (- 6 z) = c
240 – 6 z = c
or
<span>c = 240 – 6 z</span>
∠BCD = 57°
∴ ∠BDR = ∠BCD = 57° (angle that meets the chord and the tangent is equi-angular to the angle at the alternate segment)
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:
Both of these examples are wrong. You cannot add/subtract integers and square roots together, however, you could add square roots together if they have the same number under the square root. For example, 2 - 2√6 will stay as 2 - 2√6 because they aren't like terms. 25 + 5√5 + 5√5 + 5 = 30 + 10√5 because 25 + 5 = 30 and 5√5 + 5√5 = 10√5. We can add 5√5 and 5√5 together because they have the same number under the square root. If we were to compute √2 + √3, we would just leave it as is because they don't have the same number under the square root.