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05 Aug 2020, 00:38
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Given an integer a, a rectangle's adjacent sides measure (a + 2) inches and (a + 5) inches, respectively. Is "a" negative?
(1) The rectangle's area in square inches is less than the rectangle's perimeter in inches.
(2) The rectangle's diagonal measures √29 inches.
(1) The rectangle's area in square inches is less than the rectangle's perimeter in inches.
(2) The rectangle's diagonal measures √29 inches.
05 Aug 2020, 00:59
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buan15
(1) Area<Perimeter
(a+2) (a+5)< 2(2a+7)
We cannot find a definitive answer.
Not sufficient.
(2) (a+2)^2+(a+5)^2=√29 (Along with the diagonal the rectangle will form two right angle triangles, we will solve for one triangle)
Only a=-7 satisfies the equation.
Therefore, we can conclude that a is -ve.
B
05 Aug 2020, 01:32
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Official Solution:
Analyze the question stem:
The stem states that a is an integer and expresses the rectangle's adjacent side lengths in terms of a: (a + 2) and (a + 5). It asks, yes or no, whether a is negative. Since a polygon's side lengths are always positive, a + 2 > 0 and a > −2. The question then is whether −2 < a < 0 or whether a ≥ 0. Information that allows unambiguous determination of either range of values for a would allow you to answer the question with a clear yes or no and would be sufficient.
Evaluate the statements
Regarding Statement (1), the rectangle's area is the product of adjacent sides: (a + 2)(a + 5). The perimeter is twice the sum of adjacent sides: 2[(a + 2) + (a + 5)]. Since, according to this statement, the area is less than the perimeter, (a + 2)(a + 5) < 2[(a + 2) + (a + 5)].
Now, solve this inequality to determine a's range of values.
a^2 + 7a + 10 < 2(2a + 7)
a^2 + 7a + 10 < 4a + 14
a^2 + 3a– 4 < 0
Factoring the left side of this inequality results in (a + 4)(a – 1) < 0. Since you want to determine whether a is always negative given this statement, test a few permissible and manageable values of a in your inequality above, keeping in mind that the multiplication on the left side must result in a negative value.
For example, if a = –1, then (–1 + 4)(–1 – 1) < 0, so a could equal −1. This value answers the question with a yes. If a = 0, then (0 + 4)(0 – 1) < 0, so zero is permissible and a could equal a non-negative value.
This value answers the question with a no. Thus, a could but does not have to be a negative number, meaning that Statement (1) is insufficient. Eliminate (A) and (D).Regarding
Statement (2), the rectangle's diagonal is the hypotenuse of the two right triangles created by the diagonal. Per the Pythagorean theorem, (one leg)^2 + (the other leg)^2 = (hypotenuse)^2, so set up and solve this equation for a.
(a + 2)^2 + (a + 5)^2 = 29
(a^2 + 4a + 4) + (a^2 + 10a + 25) – 29 = 0
2a^2 + 14a = 0
2a(a + 7) = 0
a = 0 or –7
Now, a cannot equal –7, because both of the rectangle's sides would have nonpositive side lengths. Thus, a can equal only 0, providing an "always no" response to the question. This statement is sufficient, so the correct answer is (B).
TAKEAWAY: Quadratics can show up in geometry questions. When solving, remember that dimensions of geometry figures are always positive.
If you find the solution helpful please encourage with Kudos
Analyze the question stem:
The stem states that a is an integer and expresses the rectangle's adjacent side lengths in terms of a: (a + 2) and (a + 5). It asks, yes or no, whether a is negative. Since a polygon's side lengths are always positive, a + 2 > 0 and a > −2. The question then is whether −2 < a < 0 or whether a ≥ 0. Information that allows unambiguous determination of either range of values for a would allow you to answer the question with a clear yes or no and would be sufficient.
Evaluate the statements
Regarding Statement (1), the rectangle's area is the product of adjacent sides: (a + 2)(a + 5). The perimeter is twice the sum of adjacent sides: 2[(a + 2) + (a + 5)]. Since, according to this statement, the area is less than the perimeter, (a + 2)(a + 5) < 2[(a + 2) + (a + 5)].
Now, solve this inequality to determine a's range of values.
a^2 + 7a + 10 < 2(2a + 7)
a^2 + 7a + 10 < 4a + 14
a^2 + 3a– 4 < 0
Factoring the left side of this inequality results in (a + 4)(a – 1) < 0. Since you want to determine whether a is always negative given this statement, test a few permissible and manageable values of a in your inequality above, keeping in mind that the multiplication on the left side must result in a negative value.
For example, if a = –1, then (–1 + 4)(–1 – 1) < 0, so a could equal −1. This value answers the question with a yes. If a = 0, then (0 + 4)(0 – 1) < 0, so zero is permissible and a could equal a non-negative value.
This value answers the question with a no. Thus, a could but does not have to be a negative number, meaning that Statement (1) is insufficient. Eliminate (A) and (D).Regarding
Statement (2), the rectangle's diagonal is the hypotenuse of the two right triangles created by the diagonal. Per the Pythagorean theorem, (one leg)^2 + (the other leg)^2 = (hypotenuse)^2, so set up and solve this equation for a.
(a + 2)^2 + (a + 5)^2 = 29
(a^2 + 4a + 4) + (a^2 + 10a + 25) – 29 = 0
2a^2 + 14a = 0
2a(a + 7) = 0
a = 0 or –7
Now, a cannot equal –7, because both of the rectangle's sides would have nonpositive side lengths. Thus, a can equal only 0, providing an "always no" response to the question. This statement is sufficient, so the correct answer is (B).
TAKEAWAY: Quadratics can show up in geometry questions. When solving, remember that dimensions of geometry figures are always positive.
If you find the solution helpful please encourage with Kudos
