1)
A pair of parallel lines divide a parallelogram as shown in the diagram into three regions with the same area. If one of the three regions is an equilateral triangle with perimeter 1, then what is the perimeter of the original parallelogram?
2)
An isosceles trapezoid with one base twice as long as the other has a height of 12 cm. The shorter base is 18 cm. In centimeters, what is the perimeter of this trapezoid?
3)
Two long strips of paper are each 2 inches wide, and they overlap so that the angle between the strips is 30°. What is the number of square inches in the area of the parallelogram formed by the overlap?
4)
The median and the height of an isosceles trapezoid are equal in length. One diagonal of the trapezoid has length 12. What is the area of the trapezoid?
Thanks!
+532 For number one, the parallelogram is a rhombus as one angle is 60 degrees (because equaliteral triangle), and since the triangle's perimeter is 1, the sides are 1/3 and the area is sqrt3 / 36. So, the area of the rhombus is sqrt3 / 12. We can cut it into two equaliteral triangle with area sqrt3 / 24 each, so each side is sqrt6 / 6, and the perimeter is 2sqrt6 / 3.
For number 2, we can draw a rectangle with the shorter base as one of its sides that is in the trapezoid. We know that the longer base is 18 x 2 = 36, so the part of it not contained by the rectangle's side is (36 - 18) / 2 = 9. Now we use the pythagorean theorem to get that the two congruent sides are 15, so the perimeter is 15 + 15 + 18 + 36 = 54 + 30 = 84 cm.
For number 3, if the angle between the two strips is 30 degrees, then one of the angles of the parallelogram is 30. The parallelogram is obviously a rhombus as all four sides are equal, and its area is found by multiplying the two diagonals and dividing by 2. A simple way to solve this (with some trigonometry), we noting that sine 15 and cosine 15 are (sqrt6 plus or minus sqrt2)/4, so the area of one of the triangles that the diagonals split the rhombus into is 1/2. Multiply this by 4 to get 2 sq in.
For number 4, you should define which median, as there are several medians. :)
Hoping this helped,
asdf334
+129849 4)
The median and the height of an isosceles trapezoid are equal in length. One diagonal of the trapezoid has length 12. What is the area of the trapezoid?
See the following image ( not to scale )
Let AD be the diagonal......call the median GJ and the height AE , M
Let GH , IJ = X
And let the top base = B
Then GJ - HI = GH + IJ = 2IJ
So....M - B = GH + IJ = 2X = 2IJ
And since triangles BFD and BIJ are similar...and 2BI = BF and 2BJ = BD....then 2IJ = FD
But 2IJ = 2X
So.....
EF + FD =
B + (2IJ ) =
B + (M - B) =
M =
ED
And because AED is a right angle.. and AED a right triangle......we have that
AE^2 + ED^2 = AD^2
M^2 + M^2 =12^2
2M^2 = 144
M^2 = 72
Note that the area of the trapezoid is (1/2) (height ( sum of the bases) =
(height) (sum of the bases) / 2 =
And ...(sum of the bases) / 2 = the median M ....so....
M * M = M^2 = 72 = area of the trapezoid
