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The trapezoid shown in the figure above represents a cross

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The trapezoid shown in the figure above represents a cross [#permalink]

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New post 19 Dec 2012, 05:18
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Trapezoid.png
Trapezoid.png [ 6.67 KiB | Viewed 22603 times ]
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 19 Dec 2012, 05:23
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Image
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5

Attachment:
Trapezoid2.png
Trapezoid2.png [ 9.67 KiB | Viewed 19366 times ]
The formula for calculating the area of a trapezoid is \(Area=\frac{1}{2}(base_1+base_2)(height)=\frac{1}{2}(2+5)(height)\).

So, we need to find the height AC: \(AC=\sqrt{AB^2-BC^2}=\sqrt{13^2-5^2}=12\).

Therefore, \(Area=\frac{1}{2}(2+5)*12=42\).

Answer: C.
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 29 Aug 2013, 00:24
why AC is height here??? couldnt get it
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 29 Aug 2013, 01:51
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nahianegmat wrote:
why AC is height here??? couldnt get it


Rotate the figure:
Attachment:
Untitled.png
Untitled.png [ 6.29 KiB | Viewed 18343 times ]
Hope it's clear.
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 29 Aug 2013, 22:57
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Bunuel wrote:
Image
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5

Attachment:
Trapezoid2.png
The formula for calculating the area of a trapezoid is \(Area=\frac{1}{2}(base_1+base_2)(height)=\frac{1}{2}(2+5)(height)\).

So, we need to find the height AC: \(AC=\sqrt{AB^2-BC^2}=\sqrt{13^2-5^2}=12\).

Therefore, \(Area=\frac{1}{2}(2+5)*12=42\).

Answer: C.


You could save yourself some calculation here if you knew the pythagorean triple 5-12-13. That would shave a few seconds off of the problem.
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 30 Aug 2013, 00:02
Well I don't know the formula for calculating the area of trapezoid. Also, I like to solve sums without using paper pen and with few basic formulae.

I first found out the base (the bottomline in the question figure) using pythogoras triplet (5-12-13).
Then I imagined the figure as right-angled triangle sitting on top of a rectangle.
A(Triangle) = 1/2*3*12 +
A(Rectangle) = 2*12
Total = 42.
I don't know how to insert figures here so can't explain properly
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 10 Sep 2013, 11:25
Walkabout wrote:
Attachment:
Trapezoid.png
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5


If the trapezoid represents a cross section of the rudder of a ship, then the area of the cross section of the rudder in square feet should have been the area of both the triangles represented in the figure. I know I am obviously wrong here but the way I mentioned is stuck in my head. :|
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 11 Sep 2013, 02:54
aakrity wrote:
Walkabout wrote:
Attachment:
Trapezoid.png
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5


If the trapezoid represents a cross section of the rudder of a ship, then the area of the cross section of the rudder in square feet should have been the area of both the triangles represented in the figure. I know I am obviously wrong here but the way I mentioned is stuck in my head. :|


That's not wrong at all. But the area of a trapezoid can also be found with the direct formula as shown in posts above. If you expand that formula you'll see that you basically get the sum of the areas of the two triangles.

Does this make sense?
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 11 Sep 2013, 04:57
Bunuel wrote:
aakrity wrote:
Walkabout wrote:
Attachment:
Trapezoid.png
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5


If the trapezoid represents a cross section of the rudder of a ship, then the area of the cross section of the rudder in square feet should have been the area of both the triangles represented in the figure. I know I am obviously wrong here but the way I mentioned is stuck in my head. :|


That's not wrong at all. But the area of a trapezoid can also be found with the direct formulas as shown in posts above. If you expand that formula you'll see that you basically get the sum of the areas of the two triangles.

Does this make sense?


Oh right. Thank you. That is some relief definitely ! :-D
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 10 Jul 2014, 10:05
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See Image for half triangle + rectangle solution.
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 23 Jun 2016, 09:28
1
Walkabout wrote:
Attachment:
Trapezoid.png
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5


Although this problem may seem confusing based on the description of the shape, we must keep in mind that the diagram provided is simply a trapezoid. So when we are asked to determine the area of the “cross section of the rudder,” this really means "What is the area of the trapezoid?"

The formula for the area of a trapezoid is:

area = (base 1 + base 2) x height/2

The two bases are given as 2 feet and 5 feet. Because the height is always perpendicular to the base, the height of this particular trapezoid is the side that begins at A and ends at the right angle located at the bottom right of the figure. We don't know this value, so we must calculate it. We could use the Pythagorean Theorem to determine the height, but instead we note that we have a 5-12-13 right triangle, where the unknown side (or the height) is 12.

We can now substitute the values for the two bases (2 and 5) and the height (12) into our area equation.

area = (base 1 + base 2) x height/2

area = (2 + 5) x 12/2

area = 7 x 6

area = 42

Answer is C.
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The trapezoid shown in the figure above represents a cross [#permalink]

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New post Updated on: 16 Apr 2018, 12:13
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Walkabout wrote:
Attachment:
Trapezoid.png
The trapezoid shown in the figure above represents a cross section of the rudder of a ship. If the distance from A to B is 13 feet, what is the area of the cross section of the rudder in square feet?

(A) 39
(B) 40
(C) 42
(D) 45
(E) 46.5


Let's add a few things to the diagram to make the solution easier to follow.
Image

Okay, first we should recognize that the two 90-degree angles (at vertices A and D) mean that sides AC and BD are PARALLEL, which means the rudder (quadrilateral ACBD) is a TRAPEZOID.

Area of trapezoid = (sum of the parallel sides)(distance between parallel sides)/2
So, all we need to do now is determine the length of side AD.

Now recognize that ∆ABD is a RIGHT TRIANGLE.
So, we can use the Pythagorean Theorem to get the equation: 5² + (AD)² = 13²
Evaluate: 25 + (AD)² = 169
Simplify: (AD)² = 144
Solve: AD = 12

Aside: We could have saved time be recognizing that this is a 5-12-13 (since we already knew the 5 and 13 lengths)

Okay, now that we know the length of side AD, we can find the area of this trapezoid.

Area = (2 + 5)(12)/2
= 42

Answer: C

Cheers,
Brent
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Originally posted by GMATPrepNow on 13 Mar 2018, 14:41.
Last edited by GMATPrepNow on 16 Apr 2018, 12:13, edited 1 time in total.
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Re: The trapezoid shown in the figure above represents a cross [#permalink]

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New post 14 Apr 2018, 18:58
samsonfred76 wrote:
See Image for half triangle + rectangle solution.



Wow ..Neat design :)
Re: The trapezoid shown in the figure above represents a cross   [#permalink] 14 Apr 2018, 18:58
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