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Bunuel
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can someone explain this?
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can anyone share the solution for this question?
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The answers here are amazing and noteworthy.
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Statement 2 gives no new info. Still can't work out the math from the info given in statement 1.

BrentGMATPrepNow care to weigh in ?
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Bunuel chetan2u VeritasKarishma

Could you please share the solution. Unable to understand how A is the answer.
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In many of the very difficult data sufficiency questions involving triangles and other figures, whether the statement is sufficient comes down to whether there is 1 unique figure.

What I mean is, if we know that the triangle ABC is one, specific figure, then there is only one unique value for whatever the question may ask:

perimeter
area
A particular interior angle measure

Or as in this case, the perpendicular distance from the point of intersection of the Angle Bisectors (called the In center, but you don’t have to know this)


We are given that this triangle is a Right Triangle at vertex B.

If we knew the 2 adjacent sides that include Angle <B, the triangle will be “uniquely determined” (S-A-S ——> knowing these 3 unique values fixes the triangle in place)


Statement 1: the legs are 6 and 8 —-> this is a 6-8-10 right triangle.

The triangle is therefore uniquely set in one place. In other words, there is one particular set of interior angles, 3 unique angle Bisectors that meet at point P, etc.

Since we know that the triangle is fixed in place, we could theoretically answer the question with one unique value.

S1 sufficient

S2: all we have is the perimeter. There is no integer constraint for the side lengths.

Therefore, there exists more than one right triangle that has the perimeter of 24.

S2 not sufficient.

A

Note, if you wanted to solve the problem:

The angle Bisectors of the 3 interior angles of a triangle meet at the Incenter.

The Incenter will be the center of the circle that can be inscribed within the right triangle ABC

The perpendicular distance from point P to side AB will be the Inradius for this inscribed circle.

Question is really asking what is the inradius for the circle that can be inscribed perfectly within right triangle with the sides 6-8-10


Step 1: draw the inscribed circle and draw 2 radii from center P to the point of Tangency at which the circle intersects with the Legs of the right triangle (AB and BC)

Call the radii = r

Rule: the radius of a circle drawn to the point of Tangency will be perpendicular to the tangent line.

In this case, r will form a 90 degree angle with both sides AB and BC of the right triangle (the Legs)

Step 2: this creates a Square of side length r around Vertex B

Next rule: from a single exterior point, 2 lines drawn tangent to a circle will be equal.

Therefore:

Side AB = 8

From Vertex A to the point of Tangency on side AB ——> length = 8 - r

8 - r = length of Vertex A to the point of Tangency on the hypotenuse.


Then do the same for Side BC

6 - r = length of Vertex C to the point of Tangency on the hypotenuse

The entire hypotenuse length is:

(8 - r) + (6 - r)

And we know from the fact that this is a Pythagorean Triplet that the hypotenuse us 10

(8 - r) + (6 - r) = 10

14 - 2r = 10

4 = 2r

r = 2

Posted from my mobile device
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