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2 years ago

Find the areas of the trapezoids

Mathematics

1 answer:

Gennadij [26K]2 years ago

4 0

Answer:

figure 1 - 10.5 unit^2

figure 2 - 12 unit^2

Step-by-step explanation:

figure 1

1. find the area of the rectangle

3*2 = 6 unit^2

2. find the area of the triangle

(3*3)/2 = 4.5 unit^2

3. add the area of the rectangle and the area of the triangle together. The sum would be the area of the trapezoid.

6 + 4.5 = 10.5 unit^2

figure 2

1. find the area of the rectangle

2*4 = 8 unit^2

2. find the area of both triangles

(1*4)/2 = 2 unit^2

3. add the area of the rectangle and the area of both triangles together. The sum would be the area of the trapezoid.

8 + 2 + 2 = 12 unit^2

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Delicious77 [7]

The second one it's double.

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2 years ago

The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$

$sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}$

$cos\angle AOB = \dfrac{Base}{Hypotenuse}$

$\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}$

$tan\angle AOB = \dfrac{Perpendicular}{Base}$

$\Rightarrow tan\angle AOB = \dfrac{3}{4}$

$cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}$

$\Rightarrow cosec\angle AOB = \dfrac{5}{3}$

$sec\angle AOB = \dfrac{Hypotenuse}{Base}$

$\Rightarrow sec\angle AOB = \dfrac{5}{4}$

$cot\angle AOB = \dfrac{Base}{Perpendicular}$

$\Rightarrow cot\angle AOB = \dfrac{4}{3}$

3 0

2 years ago

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GREYUIT [131]

Answer:

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Step-by-step explanation:

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2 years ago

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ycow [4]

Answer:

4986 ounces/m³

Step-by-step explanation:

1 kilogram = 35.274 ounces

1 cubic foot = 0.0283 cubic metre

We are converting kg/ft³ to ounces/m³

Hence:

4kg/ft³ × 35.274 ounces/ 1 kg × 1 ft³/0.0283m³

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Approximately to the nearest whole number = 4986 ounces/m³

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2 years ago

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Oksanka [162]

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2 years ago