Geometry

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Geometry

by BTGmoderatorRO » Sat Sep 16, 2017 10:54 am
Is the perimeter of triangle T greater than the perimeter of square S ?
(1) T is an isoceles right triangle.
(2) The length of the longest side of T is equal to the length of a diagonal of S.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(C) Statements (1) and (2) TOGETHER are NOT sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

OA is b
Can anyone please provide a detailed solution to show that option E is very correct.

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by Jay@ManhattanReview » Sun Sep 17, 2017 7:20 am
Roland2rule wrote:Is the perimeter of triangle T greater than the perimeter of square S ?

(1) T is an isosceles right triangle.
(2) The length of the longest side of T is equal to the length of a diagonal of S.

OA is b
Can anyone please provide a detailed solution to show that option E is very correct.
The correct answer is C.

Statement 1: T is an isosceles right triangle.

Certainly insufficient as we do not have any information about sides.

Statement 2: The length of the longest side of T is equal to the length of a diagonal of S.

Say the side of the square S = a, thus, the diagonal = a√2
Longest side of triangle T = a√2

Case 1: If the triangle T is an equilateral triangle.

Perimeter of the triangle T = 3 x a√2 = 3a√2 = 3 x 1.414 x a = ~4.2a

Perimeter of the square S = 4a

Thus, the perimeter of the triangle T (~4.2a) > Perimeter of the square S (4a). The answer is Yes.

Case 2: If the triangle T is an isosceles triangle or a scalene triangle, whose smaller sides are significantly less than the longest side (a√2), the perimeter of the triangle T < Perimeter of the square S. The answer is No.

No unique answer.

Statement 1 & 2:

We know that the triangle T is an isosceles right triangle.

Say the equal sides are b each.

We know that the longest side = a√2

From Pythagoras theorem, we get,

(a√2)^2 = b^2 + b^2

2a^2 = 2b^2

b = a

Thus, the perimeter of the triangle T = a√2 + a + a = (2 + √2)a = 3.141a

We see that the perimeter of the triangle T (3.141a) < Perimeter of the square S (4a). The answer is No.

Sufficient.

The correct answer: C

Hope this helps!

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