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Troyanec [42]
2 years ago
15

Help me on calculating the area of the shaded part

Mathematics
2 answers:
diamong [38]2 years ago
7 0

Answer:

1053.50 mm^2

Explanation:

first calculate the area of square ABCD

area of ABCD = AB × AB

= 70 × 70 = 4900 mm^2

if you look closely, there are 4 quater-circles with radius (AB/2 = 35 mm) and their center A,B,C and D.

now

area of any one quater circle =( pi × r^2 )/4

area of 4 quater circle =(pi × r^2)

where, pi = 3.14 and r = radius of quater circle

area of 4 quater-circle = (3.14 × 35^2)= 3846.50 mm^2

hence area of shaded region = area of square - area of 4 quater circles

= 4900 - 3846.50

=1053.50 mm^2

Ymorist [56]2 years ago
6 0

Answer:

1051 .55 {mm}^{2}

Step-by-step explanation:

Find the area of the square - area of 4 sectors

We'll use the formula to find one sector area

a =  \frac{1}{2}  \times  {r}^{2}  \times x

where r is the radius of one sector and x is the angle in radians

Since each angle is in a square, the angle of the sectors is 90 degrees.

Convert degrees to radians

90 =  \frac{1}{2} \pi \: rad

The radius of one sector is given by 70mm/2=35mm

One sector area:

\frac{1}{2}  \times {35}^{2}  \times  \frac{1}{2} \pi = 306.25\pi

Area of all 4 sectors:

4 \times 306.25\pi = 1225\pi

Area of the square:

70 \times 70  = 4900

Hence, Area of square - 4 Sectors:

4900 - 1225\pi = 1051.55 {mm}^{2} (2dp)

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A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.
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Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}

Where:

y_{o} - Initial height of the ball, measured in feet.

v_{o} - Initial speed of the ball, measured in feet per second.

g - Gravitational constant, equal to -32.174\,\frac{ft}{s^{2}}.

t - Time, measured in seconds.

Given that y_{o} = 0\,ft, v_{o} = 95\,\frac{ft}{s}, g = -32.174\,\frac{ft}{s^{2}} and y = 120\,ft, the following second-order polynomial is found:

120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}

-16.087\cdot t^{2} + 95\cdot t -120 =0

The roots of this polynomial are, respectively:

t_{1} \approx 4.075\,s and t_{2} \approx 1.831\,s.

Both roots solutions are physically reasonable, since t_{1} represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas t_{2} represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

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