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Zina [86]
4 years ago
9

Circular garden in the middle of a square yard. The radius of the circle is 4x. The side length of the yard is 20x. What is the

area of the part of the yard that is not covered by the circle ?
Mathematics
2 answers:
Setler [38]4 years ago
7 0

answer is: 16x^2(25-pi)

100% :)

dusya [7]4 years ago
3 0
-- The area of the entire yard, before the garden was planted, was

             (side)²  =  (20x)²  =  400 x² .

-- The area of the circle where garden was put in is

             pi R²  =  (pi) · (4x)²  =  16 pi x² .

--  The remaining area of the yard ... the part not covered by
the circular garden ... is

           (400x²) - (16pi x²)  =  x² (400 - 16pi)

                                           =           349.7 x²    square units.
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Step-by-step explanation:

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Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

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