A city planner designs a park that is a quadrilateral with vertices at J(-1, 1), K (3, 3), L (5, -1), and M(-1,-1). There is an entrance to the park at the midpoint of each side of the park. A straight path
connects each entrance to the entrance on the opposite side. Assuming each unit of the coordinate plane
represents 10 meters, what is the total length of the paths to the nearest meter? Round your answer to
the nearest whole number.
The total length of the paths is approximately
meters.

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Ответ:

kristinashau

18.09.2020

Mathematics

$V =\frac{\sqrt{3}}{12}$

Step-by-step explanation:

Diagram relating to the given vertices is attached below

height of equilateral triangle is given as

h = s sin60

$h = \frac{\sqrt{3}}{2} s$

area of triangle is given as

$A = \frac{1}{2} sh$

$A = \frac{1}{2} s\times \frac{\sqrt{3}}{2} s$

$A = \frac{\sqrt{3}}{4} s^2$

equation for line of given figure below that representing diagonal is given as

y = -x + 1

integrate the from 0 to 1 we will have volume value

$V = \int_{0}^{1} \frac{\sqrt{3}}{4} s^2 dy$

$V = \int_{0}^{1} \frac{\sqrt{3}}{4} (-y+1)^2 dy$

$v = \frac{\sqrt{3}}{4} \int_{0}^{1} (-y+1)^2 dy$

$=\frac{\sqrt{3}}{4} \int_{0}^{1} (y^2-2y+1)^2 dy$

$=\frac{3}{4}[\frac{1}{3}y^3 -y^3 + y]_0^1$

$V =\frac{\sqrt{3}}{12}$

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