The height of isosceles trapezoid ABDC is 12 units. The

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?

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.

Many thanks Bunuel. All makes sense now to me.

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

Answer is D

Hope it clarifies
Cheers
J

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.

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.

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.

I took an educated guess, the height = 12, formula is H* average base, so answer has to be multiple of 12.
The base has to be somewhere around 15 as the hypotnuse is 15. So I selected D. being the highest and "E" was too great

The height dropped from Point A Perpendicular to Side DC ——call this Point X

The height dropped from Point B Perpendicular to Side DC ——- call this Point Y

Since it is an Isosceles Trapezoid, if we draw a Diagonal AD it will be EQUAL in Length to Diagonal BC ——> both will equal = 15


Each Diagonal will create a Right Triangle. The 2 Right Triangles will be Congruent.

(1st)
Right Triangle XAD:

Leg AX = height of trapezoid = 12

Hypotenuse = Diagonal AD = 15

This is a Multiple of a 3-4-5 Right Triangle and Length of XD = 9

Right Triangle YBC:

Will have the same congruent lengths.

———————-
Length of XD = 9
and
Length of CY = 9


(2nd)
Further, since it is an Isosceles Trapezoid, the Angles at Vertex C and Vertex D are Equal.

This means the 2 Right Triangles on the LEFT and RIGHT Side of the Trapezoid will be congruent (R-H-S Property)

Triangle XAC is congruent to Triangle YBD.

Corresponding Sides CX and YD are Equal. Call this length = N

——————————
CX = YD = length of N


(3rd)
Based on the figure Described:

Length of XD = XY + YD

From above:
XD = 9

YD = N

and XY = (9 - N)


Also, Length of CY = XY + CX

CY = 9

CX = N

and XY = (9 - N)


(Lastly)

Area of Trapezoid =

(Area of Right Triangle XAC) + (Area of Rectangle ABYX) + (Area of Right Triangle YBD)

=

( (1/2) * 12 * N ) + ( (12) * (9 - N) ) + ( (1/2) * 12 * N)

= 6 * N + (108 - 12 * N) + 6 * N

= 108


Area of Trapezoid = 108

-D-

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Dear Bunuel please advise if we can apply same to all isosceles trapezium. Where we longer base can be divided in to three segments where two segments on outside will be equal

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Yes, x's in all isosceles trapezoids will be equal.