1. In the diagram below, ABCE is an isosceles trapezoid. Point D lies on line CE so that AE is parallel to BD and angle CBD=28 degrees Find angle BAE in degrees.
2. Trapezoid ABCD has bases line AB and line CD. The extensions of the two legs of the trapezoid intersect at P. If the area of PAB=10 and CD=2AB what is the area of PCD?
3.
ABCD is a trapezoid with bases line AB and line DC. We know that Angle C=60 degrees and Angle D=45 degrees. If AB=5-2sqrt(3) and BC=4, what is DC?
4.
The ratio of the four sides of a trapezoid is 2:1:1:1 The area of the trapezoid is 27sqrt(3) Find the length of the median of the trapezoid.
5. Trapezoid ABCD has bases line AB and line CD. The extensions of the two legs of the trapezoid intersect at P. If the area of ABD=3 and the area of BCD=7, then what is the area of PAB?
6.
Line NM is the median of trapezoid ABCD. If the area of NMCD=2* the area of ABMN what is CD/AB?
7.
The diagonals of a trapezoid are perpendicular and have lengths 3 and 4. Find the length of the median of the trapezoid.
+130099 2. Triangles PBA and PCD are similar
Since CD = 2AB, then the scale factor of triangle PCD to triangle PBA = 2
So....the area of triangle PCD =
area of triangle PBA * scale factor^2 =
10 * 2^2 =
10 * 4 =
40 units^2
+130099 1.
Angle EAB = Angle CBA = y
And because ABCE is a parallelogram, we have that
And EAB + DBA = 180
And DBA = y - 28
So
y + ( y - 28) = 180
2y - 28 = 180
2y = 208
y = 104° = EAB = BAE
+130099 4.
The ratio of the four sides of a trapezoid is 2:1:1:1 The area of the trapezoid is 27sqrt(3) Find the length of the median of the trapezoid.
Let the bottom base be the longest side = 2S
Let the top base = S
Using the Pythagorean Theorem, the height, h, can be found as
sqrt [1 - (1/2)^2] = sqrt [ 3/4 ] = sqrt(3)/2
So....using the formula for the area of a trapezoid, we have
27sqrt (3) = (1/2)h ( 2S + S)
27sqrt (3) = (1/2) (sqrt(3)/2 ) (3S)
27 = (3/4)S
S = 36
The length of the median = (sum of bases)/2 = (3/2)S = (3/2)*36 = 54
+130099 3.
ABCD is a trapezoid with bases line AB and line DC. We know that Angle C=60 degrees and Angle D=45 degrees. If AB=5-2sqrt(3) and BC=4, what is DC?
A B
D C
Draw BE perpendicular to DC
Angle BEC = 90°
Angle EBC = 30°
Then, because BEC s a 30-60-90 right triangle, then EC = (1/2)BC = 2
And BE = 2sqrt(3) = height of trapezoid
Draw AF perpendicular to DC
AF = 2sqrt(3)
And angle DAF = 45°
So...we have a 45 - 45 - 90 right triangle with AF = DF = 2sqrt(3)
So...DC = EC + EF + DF ....and EF = AB....so...
DC = [ 2] + [ 5 - 2 sqrt (3) ] + 2sqrt(3) = 2 + 5 = 7
+130099 5. Trapezoid ABCD has bases line AB and line CD. The extensions of the two legs of the trapezoid intersect at P. If the area of ABD=3 and the area of BCD=7, then what is the area of PAB?
Triangles ABD and BDC are under the same heights
Area of BDC 7 (1/2) h *DC
__________ = ___ = _________
Area of ABD 3 (1/2) h * AB
So.....the ratio of base DC to base AB is 7 : 3
But triangle PAB is similar to triangle PDC
So...Area of triangle PDC : Area of triangle PAB = 7^2 / 3^2 = 49 / 9
So area PDC - area of ABCD = area of triangle PAB
(49/9)[area of PAB] - 10 = area of triangle PAB
(49/9) [area of PAB] -area of PAB = 10
(40./9)[area of PAB] = 10
area of PAB = 10 (9/40) = 9/4 units^2 = 2.25 units^2
+130099 6.
Line NM is the median of trapezoid ABCD. If the area of NMCD=2* the area of ABMN what is CD/AB?
Area of NMCD = 2 * Area of ABMN = (1/2)h ( DC + NM) ⇒ Area of ABMN =(1/4)h (DC + MN) (1)
Area of ABMN = (1/2) h (AB + MN) (2)
Equate (1), (2)
(1/4)h (DC + MN) = (1/2)h (AB + MN)
DC + MN = 2 (AB + MN)
DC + MN = 2AB + 2MN
DC + ( AB + DC)/2 = 2AB + 2 [ (AB + DC)/ 2 ]
(3/2)DC + (1/2)AB = 2AB + AB + DC
(3/2)DC + (1/2)AB = 3AB + DC
(1/2)DC = (5/2)AB
DC = 5AB
CD / AB = 5
