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3 years ago

What is the area of a sector that has a radius of 25 mm and has an angle measure of 50

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

Answer : The area of sector is, $272.57mm^2$

Step-by-step explanation :

Formula used to calculate the area of sector in degree is:

Area of sector = $\frac{\theta}{360}\pi r^2$

where,

$\theta$ = angle = $50^o$

r = radius = 25 mm

Now put all the given values in the above formula, we get:

Area of sector = $\frac{\theta}{360}\pi r^2$

Area of sector = $\frac{50}{360}\times 3.14\times (25)^2$

Area of sector = $272.57mm^2$

Therefore, the area of sector is, $272.57mm^2$

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A 12 cm by 20 cm painting hangs in a museum. The museum curator wants to hang other paintings in the gallery that are proportion

nata0808 [166]

Answer:

A. 1.5 cm by 2.5 cm

B. 27 cm by 45 cm

C. 9 cm by 15 cm

D. 30 cm by 50 cm

Step-by-step explanation:

The dimensions of the original painting = 12 cm by 20 cm painting

Hence, the proportion is:

12/20 = 0.6

Hence, we compare the options given in the question.

A. 1.5 cm by 2.5 cm

= 1.5/2.5 = 0.6 , option A is correct

B. 27 cm by 45 cm

27cm/45 cm = 0.6 Option B is correct

C. 9 cm by 15 cm

9cm/15cm = 0.6 , Option C is correct

D. 30 cm by 50 cm

30 cm /50 cm = 0.6 Option D is correct

E. 6 cm by 14 cm

6cm/14 cm = 0.4285714286

Option E is not correct

Therefore, Options A to D are the correct options

6 0

3 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$