Geometry Problem?

Question

Find the length of the median of the trapedzoid in terms of j:

The trapezoid looks kinda like a rhombus, with the top and bottom parallel; the upper right angle is 45 degrees; the lower left angle is 60 degrees; the left side of the trapedzoid is j and the bottom side of the trapezoid is j.

The answer is 3j plus j root 3 divided 4

How would you get this answer?

Answer 1

It's hard to explain without a diagram but here goes.

Look at the 60° angle in the bottom left corner. If you drop a perpendicular from the top left corner to the base of the trapezoid you get a 30° 60° 90° triangle. The hypotenuse is the left side which has length j. The bottom leg L is therefore

cos 60° = 1/2 = L/j
L = j/2

Splitting the bottom base into two equal halves. Drop a second perpendicular from the bottom right corner to the top. Now we have marked off a section on the top of length j/2 also.

Looking again at the 30° 60° 90° triangle. The second leg h is the altitude of the trapezoid.

sin 60° = √3/2 = h/j
h = j√3/2

Now look at the top right angle of 45°. That means you can mark off a section equal to h.

So the full length of the top base is L + h.

The median is the average of the bases. The length of the median m is:

m = ½(j + L + h) = ½(j + j/2 + j√3/2) = ½(3j/2 + j√3/2)
m = (3 + √3)j / 4

Answer 2

First, draw a perpendicular segment from the upper left vertex to the bottom of the trapezoid. You have a 30-60-90 right triangle with hypotenuse j, so the shorter leg (which lies on the bottom of the trapezoid) equals j/2. Also, the longer leg equals (j sqrt(3))/2. Because the bottom of the trapezoid is j, the segment from the base of the perpendicular to the bottom right vertex equals j/2. Now draw a perpendicular segment from the bottom right vertex to the top of the trapezoid. Because its a trapezoid, the two perpendiculars are equal, so its length is also (j sqrt(3))/2. Because the upper right angle is 45 degrees, the other leg of this right triangle also equals (j sqrt(3))/2. Just by looking at it (it can be proven), the segment from the base of this perpendicular to the upper left vertex equals j/2, because it is equal in length to the corresponding segment on the bottom of the trapezoid. So, adding the two segments together, it follows that the length of the top of the trapezoid is (j+j sqrt(3))/2. The length of the median is the average of the two bases, and so the median equals ((j+j sqrt(3))/2 + j)/2=(3j+j sqrt(3))/4

Answer 3

Just to make sure I understand you correctly: the interior angles of the trapezoid are (clockwise from bottom left) 60, 120, 45, 135. Meanwhile, the bottom and left edges of the trapezoid both have length j.

Assuming this is correct, you might find it helpful to glue two triangles onto your trapezoid in order to form a rectangle (you'll need an upside-down 30-60-90 triangle on the left side and a 45-45-90 triangle on the right side). If the 30-60-90 triangle has hypotenuse j, then it has short side (width) j/2 and long leg (height) j*Sqrt(3)/2.

Meanwhile, the 45-45-90 triangle at right must have the same height, j*Sqrt(3)/2. This means its base (width) is also j*Sqrt(3)/2.

Looking at our trapezoid, it has bottom base j. The rectangle has base j+j*Sqrt(3)/2, on top and bottom. If we look to the top edge of the rectangle and subtract off the j/2 triangle leg, we are left with a length of j/2 + j*Sqrt(3)/2 for the trapezoid's top base.

The length of the trapezoid's median is the average of its bases:
[j + (j/2 + j*Sqrt(3)/2)]/2 = [3j + j*Sqrt(3)]/4.

Answer 4

Draw the figure ABCD with A being in the lower left hand side, B being on the lower right hand side, C being on the upper right side, and D being on the upper left hand side.

Let E be the midpoint of AD and F be the midpoint of BC. Then EF is the median.

Now drop a perpendicular from D to AB at H, and intersecting EF at K. Also draw a line fromB perpendicular to EF at L.

So if you have drawn the diagram correctly, you should see that the median ED = EK + KL + LF.

Now the triangle is a 30-60-90 triangle so AH = AD/2 =j/2.
Therefore EK = (j/2)/2 = j/4 [line connecting midpoints of two sides of a triangle is = 1/2 the third side.]

Now since AH = j/2, BH must = j/2 since AB = j.
Therefore KL = BH = j/2.

HD = (j/2) sqrt(3)) so HK = (j/4) sqrt(3)
HK = BL = FL because triangle BLF is isosceles.
Therefore LF = (j/4) sqrt(3).

So ED = j/4 + j/2 + (j/4) sqrt(3)
= 3j/4 +(jsqrt(3))/4 = [3j + jsqrt(3)]/4