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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.

 Dec 9, 2018
 #1
avatar+94501 
+1

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

 

 

cool cool cool

 Dec 9, 2018
 #2
avatar+94501 
+1

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

 

 

cool cool cool

 Dec 9, 2018
 #3
avatar+94501 
+1

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  

 

 

cool cool cool

 Dec 9, 2018
edited by CPhill  Dec 9, 2018
 #4
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0

i don't think 108 is correct or 54 for number 4

Guest Dec 9, 2018
 #6
avatar+94501 
0

Sorry...slight math error....see my corrected answer....!!!

 

 

cool cool cool

CPhill  Dec 9, 2018
 #7
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0

it is not 54

Guest Dec 9, 2018
 #10
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0

For number 4 the answer is not 54? so what will it be? CPhill

Guest Dec 9, 2018
edited by Guest  Dec 9, 2018
 #5
avatar+94501 
+1

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

 

 

cool cool cool

 Dec 9, 2018
 #8
avatar+94501 
+1

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

 

 

cool cool cool

 Dec 9, 2018
 #9
avatar+94501 
+1

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

 

 

cool cool cool

 Dec 9, 2018

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