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Is the area of the triangular region above less than 20? [#permalink]

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26 Jun 2017, 02:27

Is the area of the triangular region above less than 20?

1) \(x^2 + y^2≠ z^2\) \(x^2 + y^2\) , not being equal to \(z^2\) could mean any of the following : \(x^2 + y^2 < z^2\) \(x^2 + y^2 > z^2\) We cannot clearly ascertain what the values of x and y are. Therefore, we cannot tell anything about the area of the triangular region. Insufficient.

2) x + y < 13 If x+y < 13 the sum could be as less as 3 and as big as 12. x = 1,y=2 will give us an area lesser than 20 x = 5,y=7 will give us an area greater than 20. Insufficient

Combining the statements, we can't clearly tell anything about the area of the triangle. Insufficient(Option E)
_________________

Target question:Is the area of the triangular region above less than 20?

I'm going to head straight to....

Statements 1 and 2 combined Consider the following two scenarios, each of which satisfies BOTH statements.

case a: The triangles is an equilateral triangles and each side has length 0.001. As you can imagine, we need not find the area of this triangle since it's clear that the area is LESS THAN 20

case b: Even though statement 1 tells us that x and y cannot be the legs of a right triangle, let's see what would happen if those sides WERE the legs of a right triangle. Also, let's say that x and y are both equal to 6.5. In this case, the base has length 6.5 and the height has length 6.5. So, the area = (base)/(height)/2 = (6.5)(6.5)/2 = 21.125 Granted this scenario breaks both of the given conditions (x and y ARE the legs of a right triangle AND x+y is NOT less than 13, HOWEVER, we need only recognize that if were were to reduce the angle between x and y from 90 degrees to 89.9999999 degrees, then the area of the triangle would be a teeeeny bit less than 21.125. This would mean that x and y are NOT the legs of a right triangle, so statement 1 is satisfied. Likewise, if we were to reduce the length of x from 6.5 to 6.49999999, then the area of the triangle would be a teeeeny bit less than 21.125. This would mean that x+y < 13, so statement 2 is satisfied. As we can see, we can make it so that case b satisfies both conditions, and have it so that the area of the triangle is GREATER THAN 20 Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Is the area of the triangular region above less than 20? [#permalink]

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26 Jun 2017, 10:22

Area of triangle= \(\frac{1}{2}*Base * height\) (1) x^2 + y^2≠ z^2 This option states that the triangle is not right angled No data on base and height NS (2) x + y < 13 Two variable one equation and an inequality Clearly NS

St 1+2 states no info on base and height

Option E
_________________

Never stop fighting until you arrive at your destined place - that is, the unique you. Have an aim in life, continuously acquire knowledge, work hard, and have the perseverance to realise the great life.A. P. J. Abdul Kalam

Area of triangle= \(\frac{1}{2}*Base * height\) (1) x^2 + y^2≠ z^2 This option states that the triangle is not right angled No data on base and height NS (2) x + y < 13 Two variable one equation and an inequality Clearly NS

St 1+2 states no info on base and height

Option E

I thought I'd point out that your rationale to conclude that the combined statement are not sufficient (i.e., "1+2 states no info on base and height") is not quite correct. If the target question had asked "Is the area less than 22?" (instead of asking "Is the area less than 20?", then statement 2 would be sufficient.

Is the area of the triangular region above less than 20? [#permalink]

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26 Jun 2017, 14:34

GMATPrepNow wrote:

If the target question had asked "Is the area less than 22?" (instead of asking "Is the area less than 20?", then statement 2 would be sufficient.

Cheers, Brent

@Brent, thankyou for your observation on my approach. But I need clarification on the proposed approach.

So if the question had been \(Area<22\)

B.\(x+y<13\)

Approach 1: There are two cases

Case 1: Right angle triangle Case 2: Other triangles (Scalene, equilateral)

Case 1: Assume any values for x,y such that the sum is less than 13, And the area can be found and it is less than 22.

Case 2: How can I find the area of other triangles.

Should I assume a Value for x and y and get the range of values for z by formula \(x-y<z<x+y\) and then I can find height and consider z as base

Approach 2: Orelse it's safer to find the case for equilateral triangle area, as it has the greatest area of all the triangles
_________________

Never stop fighting until you arrive at your destined place - that is, the unique you. Have an aim in life, continuously acquire knowledge, work hard, and have the perseverance to realise the great life.A. P. J. Abdul Kalam

@Brent, thankyou for your observation on my approach. But I need clarification on the proposed approach.

So if the question had been \(Area<22\)

B.\(x+y<13\)

Approach 1: There are two cases

Case 1: Right angle triangle Case 2: Other triangles (Scalene, equilateral)

Case 1: Assume any values for x,y such that the sum is less than 13, And the area can be found and it is less than 22.

Case 2: How can I find the area of other triangles.

Should I assume a Value for x and y and get the range of values for z by formula \(x-y<z<x+y\) and then I can find height and consider z as base

Approach 2: Orelse it's safer to find the case for equilateral triangle area, as it has the greatest area of all the triangles

If the question were "Is the area < 22?" , we don't need to do as much work.

(2) x+y<13 Let's just consider a triangle where two of the sides have lengths 6.5 and 6.5 What's the greatest area possible? Let's make the base have length 6.5 Since area = (base)(height)/2, we will MAXIMIZE the area if we can MAXIMIZE the height. If the other side has length 6.5, we can maximize the height by making that side PERPENDICULAR to the base, in which case the area = (6.5)(6.5)/2 = 21.125 If x+y = 13, then the MAXIMUM area = 21.25 If x+y < 13, then the MAXIMUM area is less than 21.25 If the maximum area is less than 21.25, then the answer to the target question is "No! The area is definitely less than 22" So, statement 2 is sufficient.

Re: Is the area of the triangular region above less than 20? [#permalink]

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07 Jul 2017, 07:03

GMATPrepNow wrote:

RaguramanS wrote:

Area of triangle= \(\frac{1}{2}*Base * height\) (1) x^2 + y^2≠ z^2 This option states that the triangle is not right angled No data on base and height NS (2) x + y < 13 Two variable one equation and an inequality Clearly NS

St 1+2 states no info on base and height

Option E

I thought I'd point out that your rationale to conclude that the combined statement are not sufficient (i.e., "1+2 states no info on base and height") is not quite correct. If the target question had asked "Is the area less than 22?" (instead of asking "Is the area less than 20?", then statement 2 would be sufficient.

Cheers, Brent

Hello Brent

Just by looking at the figure how can we determine which one is the height?

Re: Is the area of the triangular region above less than 20? [#permalink]

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10 Jul 2017, 13:57

GMATPrepNow wrote:

RaguramanS wrote:

@Brent, thankyou for your observation on my approach. But I need clarification on the proposed approach.

So if the question had been \(Area<22\)

B.\(x+y<13\)

Approach 1: There are two cases

Case 1: Right angle triangle Case 2: Other triangles (Scalene, equilateral)

Case 1: Assume any values for x,y such that the sum is less than 13, And the area can be found and it is less than 22.

Case 2: How can I find the area of other triangles.

Should I assume a Value for x and y and get the range of values for z by formula \(x-y<z<x+y\) and then I can find height and consider z as base

Approach 2: Orelse it's safer to find the case for equilateral triangle area, as it has the greatest area of all the triangles

If the question were "Is the area < 22?" , we don't need to do as much work.

(2) x+y<13 Let's just consider a triangle where two of the sides have lengths 6.5 and 6.5 What's the greatest area possible? Let's make the base have length 6.5 Since area = (base)(height)/2, we will MAXIMIZE the area if we can MAXIMIZE the height. If the other side has length 6.5, we can maximize the height by making that side PERPENDICULAR to the base, in which case the area = (6.5)(6.5)/2 = 21.125 If x+y = 13, then the MAXIMUM area = 21.25 (21.125) If x+y < 13, then the MAXIMUM area is less than 21.25 If the maximum area is less than 21.25, then the answer to the target question is "No! The area is definitely less than 22" So, statement 2 is sufficient.

Cheers, Brent

Hey Brent,

I think the maximum area, if 2 perpendicular sides totalled 13, would be 21.125

Re: Is the area of the triangular region above less than 20? [#permalink]

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02 Aug 2017, 05:31

pushpitkc wrote:

Is the area of the triangular region above less than 20?

1) \(x^2 + y^2≠ z^2\) \(x^2 + y^2\) , not being equal to \(z^2\) could mean any of the following : \(x^2 + y^2 < z^2\) \(x^2 + y^2 > z^2\) We cannot clearly ascertain what the values of x and y are. Therefore, we cannot tell anything about the area of the triangular region. Insufficient.

2) x + y < 13 If x+y < 13 the sum could be as less as 3 and as big as 12. x = 1,y=2 will give us an area lesser than 20 x = 5,y=7 will give us an area greater than 20. Insufficient

Combining the statements, we can't clearly tell anything about the area of the triangle. Insufficient(Option E)

This is not the best explanation: "x = 5,y=7 will give us an area greater than 20. Insufficient"

For X=5;Y=7, Area will be less than 20 (to be precise - less than or equal to 17.5).

Only for X >= 5.72 and Y = 7, area will be equal to or more than 20

Area of triangular region is the greatest for right triangle: Formula is: 1/2* X * Y = 1/2*5*7 = 17.5 (less than 20) For all triangles other than right triangle, area will be even smaller for given X and Y.

Re: Is the area of the triangular region above less than 20? [#permalink]

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28 Sep 2017, 05:15

GMATPrepNow wrote:

RaguramanS wrote:

@Brent, thankyou for your observation on my approach. But I need clarification on the proposed approach.

So if the question had been \(Area<22\)

B.\(x+y<13\)

Approach 1: There are two cases

Case 1: Right angle triangle Case 2: Other triangles (Scalene, equilateral)

Case 1: Assume any values for x,y such that the sum is less than 13, And the area can be found and it is less than 22.

Case 2: How can I find the area of other triangles.

Should I assume a Value for x and y and get the range of values for z by formula \(x-y<z<x+y\) and then I can find height and consider z as base

Approach 2: Orelse it's safer to find the case for equilateral triangle area, as it has the greatest area of all the triangles

If the question were "Is the area < 22?" , we don't need to do as much work.

(2) x+y<13 Let's just consider a triangle where two of the sides have lengths 6.5 and 6.5 What's the greatest area possible? Let's make the base have length 6.5 Since area = (base)(height)/2, we will MAXIMIZE the area if we can MAXIMIZE the height. If the other side has length 6.5, we can maximize the height by making that side PERPENDICULAR to the base, in which case the area = (6.5)(6.5)/2 = 21.125 If x+y = 13, then the MAXIMUM area = 21.25 If x+y < 13, then the MAXIMUM area is less than 21.25 If the maximum area is less than 21.25, then the answer to the target question is "No! The area is definitely less than 22" So, statement 2 is sufficient.

Cheers, Brent

Hi Brent,

I couldn't understand this part of your explanation "If the other side has length 6.5, we can maximize the height by making that side PERPENDICULAR to the base, in which case the area = (6.5)(6.5)/2 = 21.125". Is area of a triangle highest when the triangle is a right angle triangle? Also, i presume that when two sides are equal and the given triangle is a right triangle, then both 6.5 ought to be legs as third side will be hypotenuse and hence greater than two other sides. Hence, you took 6.5 as a height.

@Brent, thankyou for your observation on my approach. But I need clarification on the proposed approach.

So if the question had been \(Area<22\)

B.\(x+y<13\)

Approach 1: There are two cases

Case 1: Right angle triangle Case 2: Other triangles (Scalene, equilateral)

Case 1: Assume any values for x,y such that the sum is less than 13, And the area can be found and it is less than 22.

Case 2: How can I find the area of other triangles.

Should I assume a Value for x and y and get the range of values for z by formula \(x-y<z<x+y\) and then I can find height and consider z as base

Approach 2: Orelse it's safer to find the case for equilateral triangle area, as it has the greatest area of all the triangles

If the question were "Is the area < 22?" , we don't need to do as much work.

(2) x+y<13 Let's just consider a triangle where two of the sides have lengths 6.5 and 6.5 What's the greatest area possible? Let's make the base have length 6.5 Since area = (base)(height)/2, we will MAXIMIZE the area if we can MAXIMIZE the height. If the other side has length 6.5, we can maximize the height by making that side PERPENDICULAR to the base, in which case the area = (6.5)(6.5)/2 = 21.125 If x+y = 13, then the MAXIMUM area = 21.25 If x+y < 13, then the MAXIMUM area is less than 21.25 If the maximum area is less than 21.25, then the answer to the target question is "No! The area is definitely less than 22" So, statement 2 is sufficient.

Cheers, Brent

Hi Brent,

I couldn't understand this part of your explanation "If the other side has length 6.5, we can maximize the height by making that side PERPENDICULAR to the base, in which case the area = (6.5)(6.5)/2 = 21.125". Is area of a triangle highest when the triangle is a right angle triangle? Also, i presume that when two sides are equal and the given triangle is a right triangle, then both 6.5 ought to be legs as third side will be hypotenuse and hence greater than two other sides. Hence, you took 6.5 as a height.

If the lengths of two sides of a triangle are x and y, then we will maximize the area of that triangle if we make those sides PERPENDICULAR to each other. Why is this? Since any side of a triangle can be considered the base, let's let the side with length x be the base. Area of triangle = (base)(height)/2 = (x)(height)/2 To maximize the area, we must now maximize the height. If we make the sides with length x and y PERPENDICULAR to each other, then the height of the triangle will EQUAL y If the sides with length x and y are NOT perpendicular, then the height of the triangle will be LESS THAN y

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