The height of isosceles trapezoid ABDC is 12 units. The : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 Feb 2017, 05:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The height of isosceles trapezoid ABDC is 12 units. The

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 75

Kudos [?]: 3077 [2] , given: 217

The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

07 Feb 2012, 16:02
2
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

61% (03:18) correct 39% (01:49) wrong based on 257 sessions

### HideShow timer Statistics

Attachment:

Trapezoid ABCD.PNG [ 4.37 KiB | Viewed 12387 times ]
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?

(A) 72
(B) 90
(C) 96
(D) 108
(E) 180
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 75

Kudos [?]: 3077 [0], given: 217

Re: Area of Trapezoid ABCD? [#permalink]

### Show Tags

07 Feb 2012, 16:06
Sorry guys - I should have said how I am trying to solve.

I draw the two perpendiculars from vertex A and B and called them E and F. So that I have a rectangle called ABEF. Now as we know its an isosceles trapezoid AC = BD and therefore angle C is equal to angle D. Height is 12 and diagonal is 15. Therefore, ED = 9. But, I am struggling to find CE and FD? Can someone please help?
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
Joined: 02 Sep 2009
Posts: 37153
Followers: 7275

Kudos [?]: 96850 [6] , given: 10808

Re: Area of Trapezoid ABCD? [#permalink]

### Show Tags

07 Feb 2012, 16:22
6
KUDOS
Expert's post
3
This post was
BOOKMARKED
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?
(A) 72
(B) 90
(C) 96
(D) 108
(E) 180
Attachment:

Trapezoid-area.PNG [ 5.88 KiB | Viewed 14023 times ]
ED^2+AE^2=AD^2 -->ED^2+12^2=15^2 --> ED=9. Now, as the trapezoid isosceles then CE=FD=x --> AB=9-x and CD=9+x.

Area of trapezoid $$are=a*\frac{b_1+b_2}{2}$$, where b1, b2 are the lengths of the two bases a is the altitude of the trapezoid. Hence, the are of trapezoid ABCD is $$area=AE*\frac{AB+CD}{2}=12*\frac{(9-x)+(9+x)}{2}=12*9=108$$.

_________________
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 75

Kudos [?]: 3077 [0], given: 217

Re: Area of Trapezoid ABCD? [#permalink]

### Show Tags

07 Feb 2012, 16:26
Bunuel - thanks. I think there is a typo in our explanation. Do you mean CE = FD = x?

Also, how come they will be equal? Even if this is an isosceles trapezoid then also AC = BD, or am I not getting it right.
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
Joined: 02 Sep 2009
Posts: 37153
Followers: 7275

Kudos [?]: 96850 [1] , given: 10808

The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

07 Feb 2012, 16:31
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
enigma123 wrote:
Bunuel - thanks. I think there is a typo in our explanation. Do you mean CE = FD = x?

Also, how come they will be equal? Even if this is an isosceles trapezoid then also AC = BD, or am I not getting it right.

Triangles CAE and DBF are congruent: AC=BD, AE=BF=altitude, <ACE=<BDF, <AEC=<BFD=90, ... --> CE = FD = x.
_________________
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 75

Kudos [?]: 3077 [1] , given: 217

Re: Area of Trapezoid ABCD? [#permalink]

### Show Tags

07 Feb 2012, 16:35
1
KUDOS
Many thanks Bunuel. All makes sense now to me.
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
Joined: 02 Sep 2009
Posts: 37153
Followers: 7275

Kudos [?]: 96850 [0], given: 10808

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

25 May 2013, 03:57
Bumping for review and further discussion.
_________________
Manager
Joined: 27 Feb 2012
Posts: 137
Followers: 1

Kudos [?]: 50 [3] , given: 22

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

25 May 2013, 13:14
3
KUDOS
enigma123 wrote:
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?

(A) 72
(B) 90
(C) 96
(D) 108
(E) 180

Another approach....Imagine this as the one attached below and then find the area of rectangle.
12*9 = 108

Attachment:

Symmetry.jpg [ 23.87 KiB | Viewed 10642 times ]

_________________

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Please +1 KUDO if my post helps. Thank you.

Math Expert
Joined: 02 Sep 2009
Posts: 37153
Followers: 7275

Kudos [?]: 96850 [0], given: 10808

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

25 May 2013, 13:17
BangOn wrote:
enigma123 wrote:
Attachment:
Trapezoid ABCD.PNG
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?

(A) 72
(B) 90
(C) 96
(D) 108
(E) 180

Another approach....Imagine this as the one attached below and then find the area of rectangle.
12*9 = 108
Attachment:
Symmetry.jpg

Attached the image.
_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 65

Kudos [?]: 605 [0], given: 355

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

05 Feb 2014, 14:29
enigma123 wrote:
Attachment:
Trapezoid ABCD.PNG
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?

(A) 72
(B) 90
(C) 96
(D) 108
(E) 180

Tricky problem +1

It's going to be a bit hard to explain without an image but I'll give my best shot

Isosceles trapezoid is key

So the area is the average of the bases * height

Height is 12

So we have that the triangle with hypotenuse 15 and height 12 have a base of 9. Likewise the other triangle will have the same base of 9 since it is a mirror image given that trapezoid is isosceles

Now we don't know what the smaller base is but check this out:

Let's give X to the small base and y to the other two measurements that complete the larger base

So small base : x
Large base: 2y + x

Now, we also know that x + y = 9

So the average of both bases will be : 2x + 2y = 18 / 2 = 9

So area is 9 * 12 = 108

Hope it clarifies
Cheers
J
Intern
Joined: 15 Jul 2012
Posts: 38
Followers: 0

Kudos [?]: 4 [0], given: 245

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

24 Jun 2014, 06:59
Bunuel wrote:
enigma123 wrote:
Bunuel - thanks. I think there is a typo in our explanation. Do you mean CE = FD = x?

Also, how come they will be equal? Even if this is an isosceles trapezoid then also AC = BD, or am I not getting it right.

Triangles CAE and DBF are congruent: AC=BD, AE=BF=altitude, <ACE=<BDF, <AEC=<BFD=90, ... --> CE = FD = x.

can you please explain the colored part? how are these 2 angles equal
Math Expert
Joined: 02 Sep 2009
Posts: 37153
Followers: 7275

Kudos [?]: 96850 [0], given: 10808

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

24 Jun 2014, 07:15
saggii27 wrote:
Bunuel wrote:
enigma123 wrote:
Bunuel - thanks. I think there is a typo in our explanation. Do you mean CE = FD = x?

Also, how come they will be equal? Even if this is an isosceles trapezoid then also AC = BD, or am I not getting it right.

Triangles CAE and DBF are congruent: AC=BD, AE=BF=altitude, <ACE=<BDF, <AEC=<BFD=90, ... --> CE = FD = x.

can you please explain the colored part? how are these 2 angles equal

Because triangles CAE and DBF are congruent, the angles there are also congruent.

Generally, in isosceles trapezoid the base angles have the same measure.
_________________
Senior Manager
Joined: 10 Mar 2014
Posts: 250
Followers: 2

Kudos [?]: 85 [0], given: 13

The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

01 Aug 2014, 06:39
Bunuel wrote:
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?
(A) 72
(B) 90
(C) 96
(D) 108
(E) 180
Attachment:
Trapezoid-area.PNG
ED^2+AE^2=AD^2 -->ED^2+12^2=15^2 --> ED=9. Now, as the trapezoid isosceles then CE=FD=x --> AB=9-x and CD=9+x.

Area of trapezoid $$are=a*\frac{b_1+b_2}{2}$$, where b1, b2 are the lengths of the two bases a is the altitude of the trapezoid. Hence, the are of trapezoid ABCD is $$area=AE*\frac{AB+CD}{2}=12*\frac{(9-x)+(9+x)}{2}=12*9=108$$.

Hi Bunuel,

I have one confusion here.

We say Trapezoid is having one pair of sides parallel and it is known as base of trapezoid so they should have same angle as both are parallel. Now in case of isoceles triangle it is given that base angles are same. So what is difference here for base angles in Trapezoid and isoceles trapozoid.

doubt from question explanation by you
As we say we can cut a trapezoid in one rectangle and two right triangle. so if this is not isoceles trapezoid still CE= FD=x. as both triangle are similar.

Thanks.
Intern
Joined: 16 Mar 2014
Posts: 16
GMAT Date: 08-18-2015
Followers: 0

Kudos [?]: 7 [0], given: 123

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

06 Oct 2015, 10:03
enigma123 wrote:
Attachment:
The attachment Trapezoid ABCD.PNG is no longer available
The height of isosceles trapezoid ABDC is 12 units. The length of diagonal AD is 15 units. What is the area of trapezoid ABDC?

(A) 72
(B) 90
(C) 96
(D) 108
(E) 180

Hi all,
Here is another approach. Hope it works.
Please see attached image.
BC = AD = 15, EH = BK = 12. In the right triangle AHD, AH^2 + HD^2 = AD^2 => HD = 9.
The area of the right triangle BHD = 0.5 x BK x HD = 0.5 x 12 x 9 = 54.
Similar for the right triangle AKC, S triangle AKC = 54.
We can observe that Area of BHD + Area of AKC = Area of ABDC (the overlapping area of the two triangles is OHK = The area of AOB- the one supplement BHD and AKC to make ABDC) = 54 + 54 = 108.

Hope it clear.
Attachments

geometry.png [ 6.32 KiB | Viewed 5307 times ]

Senior Manager
Joined: 26 Oct 2016
Posts: 280
Location: United States
Concentration: Marketing, International Business
Schools: HBS '19
GPA: 4
WE: Education (Education)
Followers: 13

Kudos [?]: 25 [0], given: 714

Re: The height of isosceles trapezoid ABDC is 12 units. The [#permalink]

### Show Tags

30 Dec 2016, 04:10
By sketching a drawing of trapezoid ABDC with the height and diagonal drawn in, we can use the Pythagorean theorem to see the ED = 9. We also know that ABDC is an isosceles trapezoid, meaning that AC = BD; from this we can deduce that CE = FD, a value we will call x. The area of a trapezoid is equal to the average of the two bases multiplied by the height.
The bottom base, CD, is the same as CE + ED, or x + 9. The top base, AB, is the same as ED – FD, or 9 – x.
Thus the average of the two bases is . {(9+x) + (9-x)}/2 = 9
Multiplying this average by the height yields the area of the trapezoid: 9*12 = 108.
Attachments

Area_problem.PNG [ 12.84 KiB | Viewed 819 times ]

_________________

Thanks & Regards,
Anaira Mitch

Re: The height of isosceles trapezoid ABDC is 12 units. The   [#permalink] 30 Dec 2016, 04:10
Similar topics Replies Last post
Similar
Topics:
What is the unit digit of the sum 3^47 + 5^43 + 2^12? 1 29 Nov 2016, 07:26
The area of an isosceles trapezoid with sides of length 5 and bases of 4 13 Jun 2016, 03:53
6 If an isosceles triangle an area of 2x^2 + 2x + 1/2. What is its perim 4 21 Jun 2011, 02:07
2 13 disc-shaped objects (height 12/13 cm) are to be stacked 10 26 Mar 2011, 11:14
41 If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit 17 13 Nov 2009, 23:49
Display posts from previous: Sort by

# The height of isosceles trapezoid ABDC is 12 units. The

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.