24. Rectangle A is 16 feet wide and is 8 feet longer than Rectangle B which is 14 feet wide.
1 answer:
Answer:
The Area of rectangle A is 576 feet² .
Step-by-step explanation:
Given as :
The width of rectangle A = 16 feet
The Length of rectangle A = 8 feet + The length of rectangle B
The width of rectangle B = 14 feet
Let The length of rectangle B = L feet
So, The Length of rectangle A = 8 feet + L feet
The perimeter of rectangle A + The perimeter of rectangle B = 156 feet
So, 2 × ( Length A + width A ) + 2 × ( Length B + width B ) = 156 feet
Or, 2 × ( 8 + L + 16 ) + 2 × ( L + 14 ) = 156 feet
Or, 2 ×( 24 + L ) + 2 × ( L + 14 ) = 156 feet
Or, 48 + 2 L + 2 L + 28 = 156
Or, 76 L + 4 L = 156
So, 4 L = 156 - 76
Or, 4 L = 80
∴ L = = 20 feet
So , The Length of rectangle A = 8 feet + 28 feet = 36 feet
And The width of rectangle A = 16 feet
So, Area of rectangle A = Length of rectangle A × width of rectangle A
I.e Area of rectangle A = 36 × 16 = 576 feet²
Hence The Area of rectangle A is 576 feet² . Answer
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Answer:
- resultant force: 93.946∠-10.62° N
- line of action: 17.314x +92.337y = 809.433
Step-by-step explanation:
We can use the notation a∠b to represent the (x, y) components (a·cos(b), a·sin(b)), where angle b is measured CCW from the +x direction. If we label the forces a, b, c, d clockwise from A, then we have ...
a = 80∠0° = (80, 0)
b = 60∠90° = (0, 60)
c = 90∠45° = (63.640, 63.640)
d = 150∠-110° = (-51.303, -140.954)
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If we label point A the origin, then the clockwise torque on point A is the sum of products of the force x-component and its y location, and its y-component and the negative of its x location.
T = (0, 0)·(80, 0) +(3, 0)·(0, 60) +(3, -8)·(63.640, 63.640) +(0, -8)·(-51.303, -140.954)
T = 809.433 . . . . n·m, the CW torque on point A
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The sum of forces is ...
F = a +b +c +d = (92.337, -17.314) = 93.946∠-10.62° . . . N
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This force, applied to the point of application, must generate the same torque as the given forces. That is ...
F·(y, -x) = 809.433
Then the equation of the line of action is ...
17.314x +92.337y = 809.433 . . . . . x and y in meters measured from A
Any point (x, y) on this line will serve as a point of application of the force. Unfortunately, this line of action does not pass through the rectangular plate. The attachment shows the point (D) on the line of action that is closest to point A.
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<em>Additional comment</em>
The resultant force could be decomposed into two forces acting <em>on the rectangular plate</em>. One could be of much larger magnitude, operating at the corner opposite point A. This force would provide the necessary torque. Another would be acting on point A, providing no torque, but with components such that the resultant has the correct magnitude and direction.
The legs are defined by . The Hypotenuse if defined by
. The perimeter is defined by:
D. P(x) = 2x √2 + 4x