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anyanavicka [17]
3 years ago
14

Find the are of the blue shaded region.

Mathematics
2 answers:
ICE Princess25 [194]3 years ago
7 0

Answer:

area = 73.37 cm²

Step-by-step explanation:

area of circle:

A = π5² = 78.5 cm²

area of 80° segment:

(80/360)(78.5) = 17.44 cm²

area of isosceles triangle w/ 5 cm legs and 80° angle:

sin 40° = b/5

0.6428 = b/5

b = 3.214

base = 2(3.21) = 6.43   (we only figured for one-half of the base)

cos 40° = h/5

0.7660 = h/5

height = 3.83

A = 1/2(6.43)(3.83) = 12.31 cm²

area of unshaded part:

17.44 - 12.31 = 5.13 cm²

area of total shaded part:

78.5 - 5.13 = 73.37 cm²

Firdavs [7]3 years ago
6 0

Answer:

25π

Step-by-step explanation:

i hope u hav a gud day ^_^

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PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of
Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cone volume)

V = \dfrac{ \pi *r^2*h}{3}

V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}

V = 3.14*16*4

\boxed{V = 200.96\:cm^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4* \pi * \dfrac{r^3}{3} , but we have a hemisphere, so the formula will be half the volume of the hemisphere V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}

Formula: (Volume of the hemisphere)

V = 2* \pi * \dfrac{r^3}{3}

Solving:

V = 2* \pi * \dfrac{r^3}{3}

V = 2*3.14 * \dfrac{4^3}{3}

V = 2*3.14 * \dfrac{64}{3}

V = \dfrac{401.92}{3}

\boxed{ V_{hemisphere} \approx 133.97\:cm^3}

What is the approximate volume of this figure?

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

\boxed{\boxed{\boxed{V = 334.93\:cm^3 \to Volume\:of\:the\:figure \approx 335\:cm^3 }}}\end{array}}\qquad\quad\checkmark

Answer:

The volume of the figure is approximately 335 cm³

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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