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Re: Area of Trapezoid ABCD? [#permalink]
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? _________________

Re: Area of Trapezoid ABCD? [#permalink]
07 Feb 2012, 16:22

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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 6301 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 \(area=AE*\frac{AB+CD}{2}=12*\frac{(9-x)+(9+x)}{2}=12*9=108\).

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

The height of isosceles trapezoid ABDC is 12 units. The [#permalink]
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 \(area=AE*\frac{AB+CD}{2}=12*\frac{(9-x)+(9+x)}{2}=12*9=108\).

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.

Please clarify

Thanks.

gmatclubot

The height of isosceles trapezoid ABDC is 12 units. The
[#permalink]
01 Aug 2014, 06:39

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