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nirvana33 [79]
3 years ago
12

Please help what is the area

Mathematics
1 answer:
garri49 [273]3 years ago
5 0

Given:

The figure of a trapezoid.

Length of mid-segment = 5.5

Height of the trapezoid = 6\sqrt{3}

To find:

The area of the trapezoid.

Solution:

The area of a trapezoid is:

A=M\times h       ...(i)

A=\dfrac{a+b}{2}\times h

Where, a,b are bases and h is the vertical height.

We know that M=\dfrac{a+b}{2} is the length of the mid-segment.

Putting M=5.5 and h=6\sqrt{3} in (i), we get

A=5.5\times 6\sqrt{3}

A=33\sqrt{3}

Therefore, the area of the trapezoid is 33\sqrt{3} units squared.

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Identify the value of b. PLEASE HELP!!
Degger [83]

Answer:

  b = 18

Step-by-step explanation:

From an external point, the products of distances to the near circle intercept and the far circle intercept are the same. For a tangent, such as AC, point A is both the near and far intercept point, so that product is the square of the length of AC.

  (AC)² = (CG)(CV)

  b² = 12·27 = 324 . . . . substitute known values

  b = √324 . . . . . . . . . . take the square root

  b = 18

7 0
3 years ago
Calculate area and perimeter​
mina [271]

Answer:

area ≈ 12.505

perimeter ≈  16.1684

Step-by-step explanation:

We are given

- the radius of the circle (and therefore area of the circle)

- the area of the triangle

We want to find

- angle AOB/AOT. We want to find this because 360/the angle gives us how many OABs fit into the circle. For example, if AOT was 30 degrees, 360/30 = 12 (there are 360 degrees in a circle, so that's where 360 comes from). The area of the circle is equal to πr² = π6² = 36π, and because AOT is 30 degrees, there are 12 equal parts of sector OAB in the circle, so 36π/12=3π would be the area of the sector. A similar conclusion can be reached from the circumference instead of the area to find the distance between A and B along the circle, and OA + AB + BO = the perimeter of the minor sector.

First, we can say that OAT is a right triangle because a tangent line is perpendicular to the line from the center to the point on the circle, so AT is perpendicular to OA. This forms two right angles, one of which is OAT

One thing that we can start to solve is AT. We know that the area of a triangle is equal to base * height /2, and the height of this triangle is AO, with the base being AT. Therefore, we can say

15 = AO * AT / 2

15 = 6 * AT / 2

15 = 3 * AT

divide both sides by 3 to isolate AT

AT = 5

Because OAT is a right triangle, we can say that the hypotenuse ² =  the sum of the squares of the two other lengths. The hypotenuse is opposite of the largest angle (in this case, the right angle, as in a right triangle, the right angle is always the largest), so it is OT in this case. The other two sides are OA and AT, so we can say that

OA² + AT² = OT²

5²+6² = OT²

25+36=61=OT²

square root both sides

OT = √61

Next, the Law of Sines states that

sinA/a = sinB/b = sinC/c with angles A, B, and C with sides a, b, and c. Corresponding sides are opposite their corresponding angles, so in this case, AT corresponds to angle AOT, OT corresponds to angle OAT, and AO corresponds to angle ATO.

We want to find angle AOT, as stated earlier, so we have

sin(OAT)/OT = sin(ATO)/OA = sin(AOT)/AT

We know the side lengths as well as OAT/sin(OAT) and want to figure out AOT/sin(AOT), so one equation that helps us get there is

sin(OAT)/OT = sin(AOT)/AT, encompassing our 3 known values and isolating the one unknown. We thus have

sin(90)/√61 = sin(AOT) /5

plug in sin(90) = 1

1/√61 = sin(AOT)/5

multiply both sides by 5 to isolate sin(AOT)

5/√61 = sin(AOT)

we can thus say that

arcsin(5/√61) = AOT ≈39.80557

As stated previously, given ∠AOT, we can find the area and perimeter of the sector. There are 360/39.80557 ≈ 9.04396 equal parts of sector OAB in the circle. The area of the circle is πr² = 36π, so 36π / 9.04396 ≈ 12.505 as the area. The circumference is equal to π * diameter = π * 2 * radius = 12 * π, and there are 9.04396 equal parts of arc AB in the circumference, so the length of arc is 12π / 9.04396 ≈ 4.1684. Add that to OA and OB (both are equal to the radius of 6, as any point from the center to a point on the circle is equal to the radius) to get 6+6 + 4.1684 = 16.1684 as the perimeter of the sector

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All the numbers that are highlighted
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3 years ago
Math help Please!!!!
GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation  −2y=3x+7 in the form y=-\dfrac{3}{2}x-\dfrac{7}{2}. Here the slope of the given line is  m_1=-\dfrac{3}{2}. If m_2 is the slope of perpendicular line, then

m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.

Answer 1: \dfrac{2}{3}

Part B. The slope of the line y=−2x+3 is -2. Since -\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2, then lines from part A are not parallel to line a.

Since -2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1, both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line y=\dfrac{x}{4}+5 is \dfrac{1}{4}. Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then

7=-4·2+b,

b=7+8,

b=15.

Answer 4: y=-4x+15.

Part E. Consider vectors \vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a). These vectors are collinear, then

\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.

Answer 5: -\dfrac{a}{b}=-\dfrac{d}{c}.

5 0
3 years ago
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