GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Jul 2018, 23:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M04-12

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

16 Sep 2014, 00:22
14
00:00

Difficulty:

45% (medium)

Question Stats:

62% (01:03) correct 38% (01:19) wrong based on 214 sessions

### HideShow timer Statistics

A, B, C, and D are distinct points on a plane. If triangle ABC is right angled and BD is a height of this triangle, what is the value of AB times BC ?

(1) $$AB = 6$$

(2) The product of the non-hypotenuse sides of triangle ABC is equal to 24.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

16 Sep 2014, 00:22
Official Solution:

A, B, C, and D are distinct points on a plane. If triangle ABC is right angled and BD is a height of this triangle, what is the value of AB times BC ?

Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.

(2) The product of the non-hypotenuse sides of triangle ABC is equal to 24 $$\rightarrow$$ directly gives us the value of AB*BC. Sufficient.

_________________
Intern
Joined: 04 Nov 2014
Posts: 1

### Show Tags

30 Jan 2015, 14:55
Bunuel wrote:
Official Solution:

Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.

(2) The product of the non-hypotenuse sides is equal to 24 $$\rightarrow$$ directly gives us the value of AB*BC. Sufficient.

How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle?

Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

31 Jan 2015, 06:13
3
1
chieffarmer wrote:
Bunuel wrote:
Official Solution:

Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.

(2) The product of the non-hypotenuse sides is equal to 24 $$\rightarrow$$ directly gives us the value of AB*BC. Sufficient.

How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle?

BD is a height means that B is a right angle and AC is a hypotenuse (so BD is a height from right angle B to the hypotenuse AC).

_________________
Current Student
Joined: 21 Apr 2015
Posts: 13
Schools: Broad '18 (A)

### Show Tags

26 Apr 2015, 18:31
first option says AB =6
doesnt it imply that the sides are 6,8,10?
Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

27 Apr 2015, 01:44
ishitathukral wrote:
first option says AB =6
doesnt it imply that the sides are 6,8,10?

No.

Knowing that one side of a right triangle is 6 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if $$6^2+y^2=z^2$$ DOES NOT mean that $$y=8$$ and $$z=10$$. Certainly this is one of the possibilities but definitely not the only one. In fact $$6^2+y^2=z^2$$ has infinitely many solutions for $$y$$ and $$z$$ and only one of them is $$y=8$$ and $$z=10$$.

For example: $$y=1$$ and $$z=\sqrt{37}$$ or $$y=2$$ and $$z=\sqrt{40}$$...

For more on this trap check the following questions:
what-is-the-area-of-parallelogram-abcd-111927.html
the-circular-base-of-an-above-ground-swimming-pool-lies-in-a-167645.html
figure-abcd-is-a-rectangle-with-sides-of-length-x-centimete-48899.html
in-right-triangle-abc-bc-is-the-hypotenuse-if-bc-is-13-and-163591.html
m22-73309-20.html
if-vertices-of-a-triangle-have-coordinates-2-2-3-2-and-82159-20.html
if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-105414.html
what-is-the-perimeter-of-rectangle-r-96381.html
pythagorean-triples-131161.html
given-that-abcd-is-a-rectangle-is-the-area-of-triangle-abe-127051.html
m13-q5-69732-20.html#p1176059
m20-07-triangle-inside-a-circle-71559.html
what-is-the-perimeter-of-rectangle-r-96381.html
what-is-the-area-of-rectangular-region-r-166186.html
if-distinct-points-a-b-c-and-d-form-a-right-triangle-abc-129328.html

Hope this helps.
_________________
Intern
Joined: 29 Jun 2014
Posts: 6

### Show Tags

08 Jul 2015, 09:42
Bunuel wrote:
chieffarmer wrote:
Bunuel wrote:
Official Solution:

Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.

(2) The product of the non-hypotenuse sides is equal to 24 $$\rightarrow$$ directly gives us the value of AB*BC. Sufficient.

How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle?

BD is a height means that B is a right angle and AC is a hypotenuse (so BD is a height from right angle B to the hypotenuse AC).

Bunuel -- in the figure above isn't BD an altitude and AB the height of triangle ABC??
Math Expert
Joined: 02 Aug 2009
Posts: 6217

### Show Tags

12 Jul 2015, 00:10
efforts wrote:

Bunuel -- in the figure above isn't BD an altitude and AB the height of triangle ABC??

Hi,
there is no difference in " altitude and height"..
" altitude or height" is the shortest distance from a point to the line..
so here BD is the altitude/height, when AC is the base and AB is the alt/height when base is BC...
Hope it clears the query..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 12 Jul 2015
Posts: 5

### Show Tags

21 Jul 2015, 04:32
Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.
Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

22 Jul 2015, 01:16
CountClaud wrote:
Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.

If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.
_________________
Intern
Joined: 22 Jun 2014
Posts: 22
Concentration: General Management, Finance
GMAT 1: 700 Q50 V34
GRE 1: Q800 V600
GPA: 3.68

### Show Tags

12 Jan 2016, 07:50
1
1
buffaloboy wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I am surprised if A,B, C, and D are distinct points and ABC is right triangle, right angled at B, then how come A , and D are not the same point. AB is the height . I am damn confused.

Each triangle can have three different bases and perpendicular to each base, three different heights. In the given description of triangle ABC, angle B being right angle, if you take BA or BC as height, then D coincides with A or C. But the question also says all four points are distinct. Hence we need to take AC as base and BD as height. Thats why this problem is in 700-800 difficulty level. Hope this helps.
Intern
Joined: 12 Mar 2015
Posts: 45
Schools: Haas '20
GPA: 2.99
WE: Corporate Finance (Aerospace and Defense)

### Show Tags

27 Mar 2016, 23:04
wow I can't believe this question tricked me so hard. Thank you for this question. It really makes me read a LOT more carefully.
Current Student
Joined: 21 Apr 2016
Posts: 29
Location: United States

### Show Tags

29 Apr 2016, 13:35
2
Bunuel wrote:
CountClaud wrote:
Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.

If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.

How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...
>> !!!

You do not have the required permissions to view the files attached to this post.

Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

02 May 2016, 04:10
nk18967 wrote:
Bunuel wrote:
CountClaud wrote:
Hi Bunuel,

Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.

If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.

How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.
_________________
Current Student
Joined: 21 Apr 2016
Posts: 29
Location: United States

### Show Tags

02 May 2016, 04:26
How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...[/quote]

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.[/quote]

-----

Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'.
Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

02 May 2016, 04:30
nk18967 wrote:
How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.[/quote]

-----

Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'.
Thanks![/quote]

It's not necessary to be inside a triangle:
Attachment:
alt2.gif

>> !!!

You do not have the required permissions to view the files attached to this post.

_________________
Current Student
Joined: 21 Apr 2016
Posts: 29
Location: United States

### Show Tags

02 May 2016, 07:04
Bunuel wrote:
nk18967 wrote:
How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...

The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.

-----

Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'.
Thanks![/quote]

It's not necessary to be inside a triangle:
Attachment:
alt2.gif
[/quote]

'VERTICES' is the magic word!! Gotcha, Thanks!
Intern
Joined: 26 Apr 2016
Posts: 8

### Show Tags

24 May 2016, 18:57
I think this is a poor-quality question and I don't agree with the explanation. Solution is incorrect. Answer should be (E). There is no way to prove that AB or BC are both legs, or alternatively that one is a leg and the other a hypotenuse. The height BD can lie outside the triangle contrary to the discussion posted in the forum.
Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

25 May 2016, 08:43
DavidFox wrote:
I think this is a poor-quality question and I don't agree with the explanation. Solution is incorrect. Answer should be (E). There is no way to prove that AB or BC are both legs, or alternatively that one is a leg and the other a hypotenuse. The height BD can lie outside the triangle contrary to the discussion posted in the forum.

That's not correct. The height in a right triangle is either one of the legs or the perpendicular from right angle to the hypotenuse. Thus the height of a right triangle cannot lie outside the triangle. Moreover, since A, B, C, and D are distinct points then BD cannot be the height from non-right angle because in this case it would coincide with one of the legs.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47030

### Show Tags

15 Jul 2016, 07:25
banty1987 wrote:
I think this is a high-quality question and I don't agree with the explanation. In statement (2) it say's product of the non-hypotenuse sides. from the figure in the explanation even AB and BC are hypotenuse to triangle's ADB & BDC respectively. Please explain if I am wrong. in the question it does not say hypotenuse as AC only.

Please read the discussion on previous pages.
_________________
Re: M04-12   [#permalink] 15 Jul 2016, 07:25

Go to page    1   2    Next  [ 36 posts ]

Display posts from previous: Sort by

# M04-12

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.