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Olenka [21]
3 years ago
11

The cup, on the 9th hole of a golf course, is located, dead center, in the middle of a circular green, which is 35 feet in radiu

s. The ball follows a straight-line path, and exits the green at the right-most edge. Assume the ball travels 11 ft/sec. Introduce coordinates, so that the cup is the origin of an xy-coordinate system. Provide numerical answers below, with two decimal places of accuracy. Ball starts at (-40,-50). Suppose that L is a line, tangent to the boundary of the golf green, and parallel to the path of the ball. Let Q be the point where the line is tangent to the circle. Notice that there are two possible positions for Q. Find the possible x-coordinates of Q.
Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

± 27.33 ft

Step-by-step explanation:

For the given problem, we can estimate the initial and final coordinates of the line of the ball path as (-40,-50) and (0,0). Therefore, the slope is:

(-50-0)/(-40-0) = 50/40 = 1.25

Similarly, we can estimate the slope of a perpendicular line to the line of the ball path as: -1*(1/1.25) = -0.8.

Therefore, using (0,0) and the slope -0.8, the equation of the perpendicular line is: -0.8 = (y-0)/(x-0);

-0.8 = y/x

y = -0.8x

Furthermore, we are given the circle radius as 35 ft and we can use the distance formula to find the two points 35 ft far from the origin:

35^2 = x^2 + y^2

y = -0.8x

35^2 = x^2 + (-0.8x)^2

1225 = (x^2 + 0.64x^2)

1225 = 1.64x^2

x^2 = 1225/1.64 = 746.95

x = sqrt(746.95) = ± 27.33 ft

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Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

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