A circle with radius of \green{2}2 start color green, 2, end color green sits inside a \blue{7 \times 11}7×11 start color blue,
7, times, 11, end color blue rectangle. what is the area of the shaded region
2 answers:
A circle with radius 2 sits inside a 7×11 rectangle. What is the area of the shaded region?
We draw the problem in the picture attached. The picture is not drawn to scale!
In order to find the shaded area, you need to subtract the area of the circle from the area of the rectangle.
Therefore:
A(rectangle) = b·h = 11×7
= 77 u²
A(circle) = π·r² = 3.14 · 2²
= 3.14 · 4
= 12.56 u²
A(shaded) = A(rectangle) - <span>A(circle)
= 77 - 12.56
= 64.44 u²
Hence, the shaded figure has an area of 64.44 square units.
</span>
We draw the problem in the picture attached. The picture is not drawn to scale!
In order to find the shaded area, you need to subtract the area of the circle from the area of the rectangle.
Therefore:
A(rectangle) = b·h = 11×7
= 77 u²
A(circle) = π·r² = 3.14 · 2²
= 3.14 · 4
= 12.56 u²
A(shaded) = A(rectangle) - <span>A(circle)
= 77 - 12.56
= 64.44 u²
Hence, the shaded figure has an area of 64.44 square units.
</span>
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Answer:
The coordinates of the explosion is (3300, -2750)
Step-by-step explanation:
I have plot the points and attached to thus answer for easy understanding.
Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.
Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.
Thus, 4400 = d1 = d2 = 2a
So,2a = 4400 and so, a =2200
a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;
c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000
Equation of hyperbola is given as;
(x²/a²) - (y²/b²) = 1
Plugging in the values of a² and b², we obtain ;
(x²/4,840,000) - (y²/6,050,000) = 1
Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;
(3300²/4,840,000) - (y²/6,050,000) = 1
2.25 - 1 = (y²/6,050,000)
y² = (1.25 x 6,050,000)
y² = 7562500
y = √7562500
y = ± 2750
Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.
So explosion will occur at the coordinate (3300, -2750)
