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Area of ship rudder

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Area of ship rudder

by GmatGreen » Fri Feb 21, 2014 8:22 am
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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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by Brent@GMATPrepNow » Fri Feb 21, 2014 8:37 am
GmatGreen 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
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 = C

Cheers,
Brent



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by Patrick_GMATFix » Fri Feb 21, 2014 8:39 am
Be familiar with the common right triangles 3-4-5 and 5-12-13 (tested here) because they come up over and over on the test.

Area of a trapezoid can be found in at least two ways:
1 - average of parallel sides * perpendicular side
2 - break trapezoid into a triangle sitting on top of a rectangle (or on the side depending on orientation) and find the areas of the individual figures.

The full solution below is taken from the GMATFix App.

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by Abhishek009 » Fri Feb 21, 2014 9:23 am
There is a very cool concept known as Pythegorean Triplets -

https://www.mathsisfun.com/pythagorean_triples.html

Applying it here

AB = 13

BC = 5

So , AC will be 12

Now we have a Triangle ( BDF )and a Rectangle ( ACDF )

Calculate Areas Individually -

AC = FD = 12

BD = BC - CD ( Or AF )

BD = 3

We get Height and Base of a right Angled Triangle , so Area of this Triangle will be 1/2*3*12 => 18

Rectangle ( ACDF )

Length = AC = DF = 12

Breadth = AF = CD = 2

Area of the rectangle is = AF * FD => 2*12 = 24

Total Area of the figure ACBF = Area of Triangle FDB + Area of the Rectangle ACDF

Total Area = 18 + 24 = 42

Hence Answer is (C) 42
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by [email protected] » Fri Feb 21, 2014 2:45 pm
Hi GmatGreen,

Each of the explanations in this string properly shows how to do the "math" behind this question, so I won't rehash that here. Instead I'll offer this takeaway about GMAT Geometry questions: they're almost always based on standard shapes and formulas. Any time you encounter a GMAT question with a non-standard shape, you should think about how you can "break" that shape into pieces that ARE standard shapes. In this question, the trapezoid can be easily broken down into a rectangle and a right triangle. Dealing with those two pieces individually makes this question fairly easy to handle.

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by mcdesty » Thu Jul 10, 2014 9:05 am
See Image: your scratch paper should look like this.
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by Scott@TargetTestPrep » Tue Jul 07, 2015 3:32 am
GmatGreen 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
Solution:

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 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: C

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by seandcarey » Tue Jul 07, 2015 6:13 pm
They had it right, either use the formula or think of it like a square and a triangle.

To solve for the bottom in this instance
a^2 + b^2 = c^2

5^2 + b^2 = 13^2
25 + b^2 = 169
b^2 = 144
b = sqrt(144)
b = 12

The area of the square made out on the bottom is just 12*2 = 24

Now just think a triangle is a square folded in half. The top part of the trapezoid is your imaginary triangle. We know the bottom length is 12 feet from above and the height is 3 feet.
12 * 3 / 2 (divide by two, think again its a square cut in half) 36/2 = 18

The area of the trapezoid is 24 + 18 (area of the square plus the area of the triangle) and you get 42 or answer C.
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