C is the center of the circle. Find the area of the segment shown in the figure,

Mathematics

1 answer:

Semmy [17]3 years ago

6 0

Answer: A 28.50 sq. u

Step-by-step explanation:

In the given figure , radius of the circle : r= 10 u

Central angle made by Arc ED: $\theta$ = 90° ( right angle )

Now , the area of the segment is given by :-

Area of segment = Area of sector (ECD) - Area of triangle(ECD) (i)

Formula : Area of right triangle = 0.5 x Base x Height

Area of sector = $\dfrac{\theta}{360^{\circ}}\times\pi r^2$

Using above formulas in (i) and substituting corresponding values , we get

Area of segment =

$\dfrac{90^{\circ}}{360^{\circ}}\times\pi (10)^2-0.5\times10\times10\\\\=\frac{1}{4}(3.14)(100)-50\\\\=78.5-50=28.5\ sq.u$

Hence, the correct answer is A 28.50 sq. u

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A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate