Seating 7 students in a row in a classroom
2 answers:
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One way of solving this problem can be Heron's formula. It is a formula that allows us to compute the area of a triangle, knowing the length of its three sides.
For the sake of clarity, let's assume that the point in the top-left lies in the origin, and let's call it point A. Then, the "middle" point is point B, and the bottom-right point is point C.
If we fix the coordinate axis with the origin in A, we have the following coordinates for the three points:
We can compute the length of any side using the formula for the distance between two points:
Plugging the approriate values, we get the following lenghts:
Now that we have the lengths, we can use Heron's formula: given the side lenghts and the semiperimeter
, the area is given by
If you plug our values, you will get an area of 6.5. So, unless I'm mistaken, none of the answers seem to mach, whereas 7 seems the best approximation.
1.
In any trapezoid, the length of the midsegment is
substituting the known values:
2.
Notice that since D and have the same y-coordinate, then DE is horizontal, and since F and D have the same x-coordinate, FD is vertical.
Thus FD and DE are perpendicular, so the triangle FED is a right triangle.
The median drawn from the right angle, is equal to half the hypotenuse.
That is, |DO|=1/2 |FE|, thus |OE|=|OF|=|OD|, are all radii of the circle centered at O.
O is the midpoint of EF, and is found by the Midpoint formula:
Answer:
1. 14
2. (4, 3)
In any trapezoid, the length of the midsegment is
substituting the known values:
2.
Notice that since D and have the same y-coordinate, then DE is horizontal, and since F and D have the same x-coordinate, FD is vertical.
Thus FD and DE are perpendicular, so the triangle FED is a right triangle.
The median drawn from the right angle, is equal to half the hypotenuse.
That is, |DO|=1/2 |FE|, thus |OE|=|OF|=|OD|, are all radii of the circle centered at O.
O is the midpoint of EF, and is found by the Midpoint formula:
Answer:
1. 14
2. (4, 3)