A parallelogram has sides of 12cm and 8cm. The distance between the 12cm sides is 4cm, Find the distance between the 8cm sides.
I'm going to try this too and I am going to use Chris's pic too.
Find the equation of line CD
C(20,0) D(13,4)
4−013−20=y−0x−20−47=yx−20−4x+80=7y4x+7y−80=0
Now find the perpendicular distance of this line to the point (8,0)
d=|4∗8+7∗0−80|√16+49d=48√65d=48√6565d≈5.95cm
Interesting - 3 mathematicians and 3 different answers.
The distance between the 8cm sides is just 12 cm...
Here's a picture....
Notice that the distance between BD and AC (the two 12 cm sides) is just 4. And the distance between AB and CD (the two 8 cm sides) is just 12.
I disagree with CPhill's answer because (using his diagram) the distancce from AB to CD must be measured on a perpendicular drawn from AB to CD and not just the horizontal line between the two points.
However, I don't have an easy way to find this.
I propose:
1) drawing a perpendicular line from point A to line CD;
2) finding the point where these two lines intersect; and
3) finding the distance from point A to this point of intersection.
Part 1: a perpendicular line from point A to line CD will have a slope that is the negative reciprocal of line AB.
--- Line AB has slope: -4/(4√3) = -1/√3 ---> so a line perpendicular to AB will have slope √3.
--- The equation of the lne through point A = (8,0) with slope √3 is: (use point-slope form)
y - 0 = √3(x - 8) ---> y = √3x -8√3 ---> √3x - y = 8√3
Part 2: the equation of CD is: using slope = -1/√3 and point C = (20,0):
--- y - 0 = (-1/√3)(x - 20) ---> √3y = -1(x - 20) ---> √3y = -x + 20 ---> x + √3y = 20
--- Putting these two equation together: x + √3y = 20 ---> x + √3y = 20
√3x - y = 8√3 ---> x √3 ---> 3x - √3y = 24
Adding down: 4x = 44
Dividing by 4: x = 11
Substituting 11 for x into the equation: x + √3y = 20 ---> y = 3√3
Part 3: finding the distance from the point (8,0) to the point (11, 3√3) by using the distance formula:
distance = √( (11 - 8)² + (3√3 - 0)² ) = √( (3)² + (3√3)² ) = √( 9 + 27) = √36 = 6.
I believe that the answer is 6.
I'm going to try this too and I am going to use Chris's pic too.
Find the equation of line CD
C(20,0) D(13,4)
4−013−20=y−0x−20−47=yx−20−4x+80=7y4x+7y−80=0
Now find the perpendicular distance of this line to the point (8,0)
d=|4∗8+7∗0−80|√16+49d=48√65d=48√6565d≈5.95cm
Interesting - 3 mathematicians and 3 different answers.