For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. if point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?
(1) Triangle ABC is isosceles.
(2) Line segment BD is perpendicular to side AC.
[spoiler]OA C
why is answer not A. I always get confused with a line drawn from the right angle to the hypotenuse of that Right triangle.
Isn't this alone sufficient that a line drawn from the right angle of an isosceles right triangle to its hypotenuse will divide the hypotenuse into two equal pieces?[/spoiler]
Geometry-triangles
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Statement 1 doesn't tell us the position of "D".. it can mean any possible point on AC.buoyant wrote: why is answer not A. I always get confused with a line drawn from the right angle to the hypotenuse of that Right triangle.
Isn't this alone sufficient that a line drawn from the right angle of an isosceles right triangle to its hypotenuse will divide the hypotenuse into two equal pieces?[/spoiler]
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If you draw the triangle it helps.
BD bisect angle ABC because it is symmetrical (isosceles)
So looking at a half triangle
BD = 7.5/tan45
tan45 = 1
so BD = 7.5
BD bisect angle ABC because it is symmetrical (isosceles)
So looking at a half triangle
BD = 7.5/tan45
tan45 = 1
so BD = 7.5
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IMPORTANT: For geometry DS questions, we are typically checking to see whether the statements "lock" a particular angle or length into having just one value. This concept is discussed in much greater detail in our free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1103buoyant wrote:For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. if point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?
(1) Triangle ABC is isosceles.
(2) Line segment BD is perpendicular to side AC.
This technique can save a lot of time.
Target question: What is the length of line segment BD?
Given: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15.
So, we have a shape that looks something like this . . .
. . . where the legs of the triangle can vary AND the location of point D can vary.
Statement 1: Triangle ABC is isosceles.
Since there is ONLY ONE isosceles right triangle with hypotenuse 15, this statement LOCKS triangle ABC into having one and only one shape.
However, statement 1 does NOT lock in the location of point D.
Since this statement does not lock in the location of point D, the length of BD is NOT LOCKED IN.
Consider these two examples.
Notice the different lengths of line segment BD
Since statement 1 does not lock in the length of line segment BD, it is NOT SUFFICIENT
Statement 2: Line segment BD is perpendicular to side AC.
This statement locks in the location of point D (in relation to the triangle's hypotenuse), but it does NOT lock in the shape of the triangle.
Consider these two examples:
Notice the different lengths of line segment BD
Since statement 2 does not lock in the length of line segment BD, it is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 locks in the shape of triangle ABC.
Statement 2 then locks in the location of point D as follows:
Since there's only one diagram that can be drawn with the given information, there can be ONLY ONE length of line segment BD
Are we required to find this length? No. We need only recognize that there can be only one length.
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Apr 16, 2018 12:46 pm, edited 1 time in total.
Hi theCodeToGMAT,theCodeToGMAT wrote:Answer [spoiler]{C}[/spoiler]
So, is it a rule that the perpendicular bisector drawn from the right angled vertex of any isosceles Right Triangle will always divide its hypotenuse into two equal parts?
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In an isosceles right triangle, a line drawn from the right angle to the hypotenuse forms two congruent triangles.buoyant wrote:Hi theCodeToGMAT,theCodeToGMAT wrote:Answer [spoiler]{C}[/spoiler]
So, is it a rule that the perpendicular bisector drawn from the right angled vertex of any isosceles Right Triangle will always divide its hypotenuse into two equal parts?
If you look at Brent's drawing when the statements are combined, ∆ABD and ∆BCD are both 45-45-90 triangles.
In ∆ABD, AB=BD, since the opposite angles are equal.
In ∆BCD, CD=BD, since the opposite angles are equal.
Thus, AB=CD=BD, implying that BD bisects side AC.
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Well, if you say "perpendicular bisector" then you yourself answered what you askedbuoyant wrote:Hi theCodeToGMAT,theCodeToGMAT wrote:Answer [spoiler]{C}[/spoiler]
So, is it a rule that the perpendicular bisector drawn from the right angled vertex of any isosceles Right Triangle will always divide its hypotenuse into two equal parts?
In general, for a right angled triangle:
h = ab/c
and h^2 = mn
where,
a, b and c are the measurement of sides.
and, c = m + n
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I received a private message about this thread.
I can see the problem, but, for some reason, your comments are entirely blacked out -- I can't see what you wrote at all. (I'm using a tablet, so perhaps that's the problem.)
If you still have questions, can you post them so that I can see them?
Thanks.
I can see the problem, but, for some reason, your comments are entirely blacked out -- I can't see what you wrote at all. (I'm using a tablet, so perhaps that's the problem.)
If you still have questions, can you post them so that I can see them?
Thanks.
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THAT is an awesome explanation. If you did these for tough 700 questions I would get a 700 no problem.
Brent@GMATPrepNow wrote:IMPORTANT: For geometry DS questions, we are typically checking to see whether the statements "lock" a particular angle or length into having just one value. This concept is discussed in much greater detail in our free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1103buoyant wrote:For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15. if point D lies on side AC, and a line is drawn from point B to point D, what is the length of line segment BD?
(1) Triangle ABC is isosceles.
(2) Line segment BD is perpendicular to side AC.
This technique can save a lot of time.
Target question: What is the length of line segment BD?
Given: For triangle ABC, angle ABC = 90 degrees, and side AC has a length of 15.
So, we have a shape that looks something like this . . .
. . . where the legs of the triangle can vary AND the location of point D can vary.
Statement 1: Triangle ABC is isosceles.
Since there is ONLY ONE isosceles right triangle with hypotenuse 15, this statement LOCKS triangle ABC into having one and only one shape.
However, statement 1 does NOT lock in the location of point D.
Since this statement does not lock in the location of point D, the length of BD is NOT LOCKED IN.
Consider these two examples.
Notice the different lengths of line segment BD
Since statement 1 does not lock in the length of line segment BD, it is NOT SUFFICIENT
Statement 2: Line segment BD is perpendicular to side AC.
This statement locks in the location of point D (in relation to the triangle's hypotenuse), but it does NOT lock in the shape of the triangle.
Consider these two examples:
Notice the different lengths of line segment BD
Since statement 2 does not lock in the length of line segment BD, it is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 locks in the shape of triangle ABC.
Statement 2 then locks in the location of point D as follows:
Since there's only one diagram that can be drawn with the given information, there can be ONLY ONE length of line segment BD
Are we required to find this length? No. We need only recognize that there can be only one length.
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent
Hi Ron,lunarpower wrote:I received a private message about this thread.
I can see the problem, but, for some reason, your comments are entirely blacked out -- I can't see what you wrote at all. (I'm using a tablet, so perhaps that's the problem.)
If you still have questions, can you post them so that I can see them?
Thanks.
I had a notion that any line drawn from the right angled vertex of a right triangle to the opposite side will be a perpendicular bisector. After all these explanations by other instructors, i took a note that such is the case in an isosceles right angle triangle, because we have two 45-45-90 triangles, which make the two bases equal.
I usually get confused in similar triangles and variations in right triangles.
Please add if there is anything that can be relevant to this discussion.
Thanks!