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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 14380 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 \(area=AE*\frac{AB+CD}{2}=12*\frac{(9-x)+(9+x)}{2}=12*9=108\).

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

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05 Feb 2014, 15:29

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

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

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01 Aug 2014, 07:39

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 \(area=AE*\frac{AB+CD}{2}=12*\frac{(9-x)+(9+x)}{2}=12*9=108\).

Answer: D.

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.

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

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06 Oct 2015, 11:03

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.

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

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30 Dec 2016, 05:10

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 1087 times ]

_________________

Thanks & Regards, Anaira Mitch

gmatclubot

Re: The height of isosceles trapezoid ABDC is 12 units. The
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30 Dec 2016, 05:10

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