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# A parallelogram has sides of 12cm and 8cm. The distance between the 12cm sides is 4cm, Find the distance between the 8cm sides.

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A parallelogram has sides of 12cm and 8cm. The distance between the 12cm sides is 4cm, Find the distance between the 8cm sides.

Guest Nov 29, 2014

#3
+92253
+10

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)

$$\\\frac{4-0}{13-20}=\frac{y-0}{x-20}\\\\ \frac{-4}{7}=\frac{y}{x-20}\\\\ -4x+80=7y\\\\ 4x+7y-80=0\\\\$$

Now find the perpendicular distance of this line to the point (8,0)

$$\\d=\frac{|4*8+7*0-80|}{\sqrt{16+49}}\\\\ d=\frac{48}{\sqrt{65}}\\\\ d=\frac{48\sqrt{65}}{65}\\\\ d\approx 5.95\;cm$$

Interesting - 3 mathematicians and 3 different answers.

Melody  Nov 30, 2014
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#1
+85918
+5

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.

CPhill  Nov 29, 2014
#2
+17721
+5

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

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.

geno3141  Nov 30, 2014
#3
+92253
+10

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)

$$\\\frac{4-0}{13-20}=\frac{y-0}{x-20}\\\\ \frac{-4}{7}=\frac{y}{x-20}\\\\ -4x+80=7y\\\\ 4x+7y-80=0\\\\$$

Now find the perpendicular distance of this line to the point (8,0)

$$\\d=\frac{|4*8+7*0-80|}{\sqrt{16+49}}\\\\ d=\frac{48}{\sqrt{65}}\\\\ d=\frac{48\sqrt{65}}{65}\\\\ d\approx 5.95\;cm$$

Interesting - 3 mathematicians and 3 different answers.

Melody  Nov 30, 2014
#4
+85918
0

Yeah......now that I reconsider it, I like Melody's answer, too.......

CPhill  Nov 30, 2014
#5
+26642
0

Here's yet another answer (namely 6cm), from Geogebra:

The problem with Melody's answer is that EA on Chris's diagram is not exactly 7cm (though her method is fine).

.

.

Alan  Dec 1, 2014
#6
+92253
0

Thanks Alan.  :)

Melody  Dec 1, 2014

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