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GMATPrepNow


What is the area of the shaded region in the figure shown?

(1) The area of the rectangle ABCD is 54.
(2) AE = 2ED

Attachment:
IMG_3202.jpg

OA : A
Statement 1 : The area(\(b*h\)) of the rectangle ABCD is 54.
The area(\(\frac{1}{2}*b*h\)) of the unshaded triangle= \(\frac{1}{2}* 54=27\)
Area of shaded region = rectangle area- unshaded triangle area\(= 54-27 = 27\)
Sufficient
Statement 2 : Shape of ABCD is not defined , area of shaded area will vary according to shape of ABCD
Not sufficient
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GMATPrepNow


What is the area of the shaded region in the figure shown?

(1) The area of the rectangle ABCD is 54.
(2) AE = 2ED

Attachment:
IMG_3202.jpg

Target question: What is the area of the shaded region

Statement 1: The area of the rectangle ABCD is 54
Let's label the two shaded triangles as ∆1 and ∆2.
Also, let j be the length of the base of ∆1 and k be the length of the base of ∆2.
Let h = the height of both triangles (since ABCD is a RECTANGLE, the height is consistent for both ∆s)


Area of ∆ = (1/2)(base)(height)
So, (area of ∆1) + (area of ∆2) = [(1/2)(j)(h)] + [(1/2)(k)(h)]
= (1/2)(h)[j + k] [I factored out the (1/2)(h)]

IMPORTANT: j + k = the length of the BASE of rectangle ABCD.
In other words, (area of ∆1) + (area of ∆2) = (1/2)(h)[BASE of rectangle ABCD]
Since (h)(the BASE of rectangle ABCD) = the area rectangle ABCD, we can say that.....
(area of ∆1) + (area of ∆2) = HALF the area of rectangle ABCD

Since statement 1 tells us that the area of the rectangle ABCD is 54, we can conclude that (area of ∆1) + (area of ∆2) = (1/2)(54) = 27
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: AE = 2ED
IMPORTANT: For geometry Data Sufficiency questions, we are typically checking to see whether the statements "lock" a particular angle, length, or shape into having just one possible measurement. This concept is discussed in much greater detail in the video below

This technique can save a lot of time.

Here, we are told that line segment AE is TWICE the length of line segment ED.
Is this enough information to LOCK IN the combined areas of the two shaded triangles? NO.
Notice that statement 2 does not prohibit us from making rectangle ABCD are TALL or as SHORT as we want.


By making rectangle ABCD are TALL or as SHORT as we want, we can make the shaded area as large or as small as we wish.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO
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Easy 30 second question if you remember that by drawing a triangle within a rectangle or square, where:
1) the base is the entire line segment of the figure
2) the height touches the opposite line segment

You divide the area of that figure by half
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BrentGMATPrepNow


What is the area of the shaded region in the figure shown?

(1) The area of the rectangle ABCD is 54.
(2) AE = 2ED

Statement 1:
When a triangle inscribed in a rectangle shares a side with the rectangle, the area of the triangle is equal to half the area of the rectangle.
Here, inscribed triangle BCE and rectangle ABCD share side BC.
Thus:
\(BCE = \frac{1}{2}ABCD = \frac{1}{2}*54 = 27\), with the result that the shaded region = rectangle ABCD - triangle BCE = 54-27 = 27.
SUFFICIENT.

Statement 2:
Since ED can be any nonnegative value, the area of the shaded region cannot be determined.
INSUFFICIENT.

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The same concept was tested on the SweSat in 2016.

Here is a copy of that question.

ABCD is a rectangle. E is a point on the side BC. If BC is 4 cm, what is the area of the triangle AED?

(1) CD is 19/8 of AD

(2) BE = EC
Posted from my mobile device
Attachments

File comment: Question from SweSat.
Screenshot_20210617-150226_Samsung Internet.jpg
Screenshot_20210617-150226_Samsung Internet.jpg [ 105.37 KiB | Viewed 18439 times ]

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BrentGMATPrepNow
GMATPrepNow


What is the area of the shaded region in the figure shown?

(1) The area of the rectangle ABCD is 54.
(2) AE = 2ED

Attachment:
IMG_3202.jpg

Target question: What is the area of the shaded region

Statement 1: The area of the rectangle ABCD is 54
Let's label the two shaded triangles as ∆1 and ∆2.
Also, let j be the length of the base of ∆1 and k be the length of the base of ∆2.
Let h = the height of both triangles (since ABCD is a RECTANGLE, the height is consistent for both ∆s)


Area of ∆ = (1/2)(base)(height)
So, (area of ∆1) + (area of ∆2) = [(1/2)(j)(h)] + [(1/2)(k)(h)]
= (1/2)(h)[j + k] [I factored out the (1/2)(h)]

IMPORTANT: j + k = the length of the BASE of rectangle ABCD.
In other words, (area of ∆1) + (area of ∆2) = (1/2)(h)[BASE of rectangle ABCD]
Since (h)(the BASE of rectangle ABCD) = the area rectangle ABCD, we can say that.....
(area of ∆1) + (area of ∆2) = HALF the area of rectangle ABCD

Since statement 1 tells us that the area of the rectangle ABCD is 54, we can conclude that (area of ∆1) + (area of ∆2) = (1/2)(54) = 27
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: AE = 2ED
IMPORTANT: For geometry Data Sufficiency questions, we are typically checking to see whether the statements "lock" a particular angle, length, or shape into having just one possible measurement. This concept is discussed in much greater detail in the video below

This technique can save a lot of time.

Here, we are told that line segment AE is TWICE the length of line segment ED.
Is this enough information to LOCK IN the combined areas of the two shaded triangles? NO.
Notice that statement 2 does not prohibit us from making rectangle ABCD are TALL or as SHORT as we want.


By making rectangle ABCD are TALL or as SHORT as we want, we can make the shaded area as large or as small as we wish.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO


Dear Sir,

I had a general DS question. May we assume that the triangle (in this question) touches the side of the rectangle? Could there be a possibility that the tip of the triangle could leave the tiniest possible gap?
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BrentGMATPrepNow
GMATPrepNow


What is the area of the shaded region in the figure shown?

(1) The area of the rectangle ABCD is 54.
(2) AE = 2ED

Attachment:
IMG_3202.jpg

Target question: What is the area of the shaded region

Statement 1: The area of the rectangle ABCD is 54
Let's label the two shaded triangles as ∆1 and ∆2.
Also, let j be the length of the base of ∆1 and k be the length of the base of ∆2.
Let h = the height of both triangles (since ABCD is a RECTANGLE, the height is consistent for both ∆s)


Area of ∆ = (1/2)(base)(height)
So, (area of ∆1) + (area of ∆2) = [(1/2)(j)(h)] + [(1/2)(k)(h)]
= (1/2)(h)[j + k] [I factored out the (1/2)(h)]

IMPORTANT: j + k = the length of the BASE of rectangle ABCD.
In other words, (area of ∆1) + (area of ∆2) = (1/2)(h)[BASE of rectangle ABCD]
Since (h)(the BASE of rectangle ABCD) = the area rectangle ABCD, we can say that.....
(area of ∆1) + (area of ∆2) = HALF the area of rectangle ABCD

Since statement 1 tells us that the area of the rectangle ABCD is 54, we can conclude that (area of ∆1) + (area of ∆2) = (1/2)(54) = 27
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: AE = 2ED
IMPORTANT: For geometry Data Sufficiency questions, we are typically checking to see whether the statements "lock" a particular angle, length, or shape into having just one possible measurement. This concept is discussed in much greater detail in the video below

This technique can save a lot of time.

Here, we are told that line segment AE is TWICE the length of line segment ED.
Is this enough information to LOCK IN the combined areas of the two shaded triangles? NO.
Notice that statement 2 does not prohibit us from making rectangle ABCD are TALL or as SHORT as we want.


By making rectangle ABCD are TALL or as SHORT as we want, we can make the shaded area as large or as small as we wish.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO


Hi BrentGMATPrepNow, can we use "Lock" concept for St.1 also? Which mean ABCD area vlaue lock in the shaded area value? Thanks
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Statement A is sufficient
1/2*AE*h + 1/2*ED*h
1/2*h*[AE+ED]
1/2*h*b= 1/2*54

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