Explanation:
a. The line joining the midpoints of the parallel bases is perpendicular to both of them. It is the line of symmetry for the trapezoid. This means the angles and sides on one side of that line of symmetry are congruent to the corresponding angles and sides on the other side of the line. The diagonals are the same length.
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b. We observe that adjacent pairs of points have the same x-coordinate, so are on vertical lines, which have undefined slope. KN is a segment of the line x=1; LM is a segment of the line x=3. If the trapezoid is isosceles, the midpoints of these segments will be on a horizontal line. The midpoint of KN is at y=(3-2)/2 = 1/2. The midpoint of LM is at y=(1+0)/2 = 1/2. These points are on the same horizontal line, so the trapezoid <em>is isosceles</em>.
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c. We observed in part (b) that the parallel sides are KN and LM. The coordinate difference between K and L is (1, 3) -(3, 1) = (-2, 2). That is, segment KL is the hypotenuse of an isosceles right triangle with side lengths 2, so the lengths of KL and MN are both 2√2.
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For part (c), we used the shortcut that the hypotenuse of an isosceles right triangle is √2 times the leg length.
Answer:
2.82843
Step-by-step explanation:
it was what i got
We know that
<span>the regular hexagon can be divided into 6 equilateral triangles
</span>
area of one equilateral triangle=s²*√3/4
for s=3 in
area of one equilateral triangle=9*√3/4 in²
area of a circle=pi*r²
in this problem the radius is equal to the side of a regular hexagon
r=3 in
area of the circle=pi*3²-----> 9*pi in²
we divide that area into 6 equal parts------> 9*pi/6----> 3*pi/2 in²
the area of a segment formed by a side of the hexagon and the circle is equal to <span>1/6 of the area of the circle minus the area of 1 equilateral triangle
</span>so
[ (3/2)*pi in²-(9/4)*√3 in²]
the answer is
[ (3/2)*pi in²-(9/4)*√3 in²]
Answer:
46
Step-by-step explanation:
(2/5)*115
230/5
46 students