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 =
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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:
17.7 cm^2
Step-by-step explanation:
Use trig to find the height of the triangle. Then the area is bh/2.
Extend side BC to the right until it is vertically below point A. Draw a segment from point A vertically down until it intersects the extension of side BC. Call the point of intersection D. <D is a right angle.
Use triangle ABD to find the height, AD, of triangle ABC.
For <B of 37 deg, AD is the opposite leg. AB is the hypotenuse. The trig ratio that relates the opposite lefg to the hypotenuse is the sine.
sin B = opp/hyp
sin 37 deg = AD/13.1
AD = 13.1 * sin 37 deg
AD = 7.9
AD is the height of triangle ABC. BC is the base. We can find the area of triangle ABC.
area = bh/2
area = (4.5 cm)(7.9 cm)/2
area = 17.7 cm^2
The question ask to determine what is a regular polygon that contains an interior angle of 9000 degree, base on my research, according to the geometrical research, I would say that the said polygon is a 52 sided convex polygon. I hope this would help
Answer:
answer is in the picture with work, just know your formulas and plug the numbers in and you'll be fine
Answer:
45
Step-by-step explanation:
AEF is a similar triangle to ABC. that means it has the same angles, and the sides (and all other lines in the triangle) are scaled from the ABC length to the AEF length by the same factor f.
now, what is f ?
we know this from the relation of AC to FA.
FA = 12 mm
AC = 12 + 28 = 40 mm
so, going from AC to FA we multiply AC by f so that
AC × f = FA
40 × f = 12
f = 12/40 = 3/10
all other sides, heights, ... if ABC translate to their smaller counterparts in AEF by that multiplication with f (= 3/10).
the area of a triangle is
baseline × height / 2
aABC = 500
and because of the similarity we don't need to calculate the side and height in absolute numbers. we can use the relative sizes by referring to the original dimensions and the scaling factor f.
baseline small = baseline large × f
height small = height large × f
we know that
baseline large × height large / 2 = 500
baseline large × height large = 1000
aAEF = baseline small × height small / 2 =
= baseline large × f × height large × f / 2 =
= baseline large × height large × f² / 2 =
= 1000 × f² / 2 = 500 × f² = 500 ×(3/10)² =
= 500 × 9/100 = 5 × 9 = 45 mm²