Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44288

ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 04:59
1
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
Question Stats:
62% (01:13) correct 38% (01:33) wrong based on 160 sessions
HideShow timer Statistics



Intern
Joined: 16 May 2015
Posts: 14

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 05:25
1
This post received KUDOS
Bunuel wrote: ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?
(1) The height from point B to the hypotenuse is 120. (2) The perimeter of the triangle is 680. Answer is D using hypotenuse = X(89)^1/2 we get 1/2B*H St 1 that is 1/2*8X*5X = 1/2*X(89)^1/2*120 calculate X and get the area from there. St 2 8X + 5X + X(89)^1/2 = 680 calculate X and obtain area. hence both statements individually Suff.



Manager
Joined: 13 Nov 2014
Posts: 116
Location: United States (GA)
GPA: 3.11
WE: Supply Chain Management (Consumer Products)

ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 06:12
1
This post received KUDOS
shallow9323 wrote: Bunuel wrote: ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?
(1) The height from point B to the hypotenuse is 120. (2) The perimeter of the triangle is 680. Answer is D using hypotenuse = X(89)^1/2 we get 1/2B*H St 1 that is 1/2*8X*5X = 1/2*X(89)^1/2*120 calculate X and get the area from there. St 2 8X + 5X + X(89)^1/2 = 680 calculate X and obtain area. hence both statements individually Suff. I too believe the answer is D, but just wanted to correct a typo. The question stem says the ratio is 8:15 and your solution shows 8:5. So for statement 2 we would have 8x+15x+x*(289)^1/2 =680. Answer is D.
_________________
Gmat prep 1 600 Veritas 1 650 Veritas 2 680 Gmat prep 2 690 (48Q 37V) Gmat prep 5 730 (47Q 42V) Gmat prep 6 720 (48Q 41V)



Intern
Joined: 16 May 2015
Posts: 14

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 06:19
Oh yea,! thanks. good luck



Intern
Status: Online
Joined: 07 Feb 2015
Posts: 28
Location: India
Rudey: RD
Concentration: Marketing, General Management
GMAT 1: 620 Q45 V31 GMAT 2: 640 Q46 V31
GPA: 3.29
WE: Sales (Hospitality and Tourism)

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 10:52
Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.
Warm Regards,
Rudraksh.



Intern
Joined: 16 May 2015
Posts: 14

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 11:20
using any side as the base, the area should be the same, but the corresponding height from that base needs to be used. half base into height.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11240
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 11:31
Hi All, When a Geometry question does not include a picture, you have to be careful about the assumptions you make about the shape(s) involved. Here, do you KNOW which side of the triangle is the height: AB or BC.....? If you drew 2 different pictures, and did the necessary work, would the answer to the question change? GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************



Intern
Joined: 16 May 2015
Posts: 14

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
21 May 2015, 11:49
ABC is right angle at B hence, any line touching B cant be the hypotenuse.



eGMAT Representative
Joined: 04 Jan 2015
Posts: 865

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
22 May 2015, 00:06
2
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
RudeyboyZ wrote: Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.
Warm Regards,
Rudraksh. Dear RudeyboyZPlease refer to the image below. It represents the information that we were given in the Question Statement and in St. 1.Also, if AB = 8x and BC = 15x, can you calculate AC? Sure, using Pythagoras Theorem, you know that it's 17x. Now, area of triangle ABC =\(\frac{1}{2}*AB*BC\) = \(\frac{1}{2}*8x*15x\) But, suppose you take the base as AC. Then, what is the height? BD. And St. 1 tells us that BD = 120 So, the area of triangle ABC can also be written as = \(\frac{1}{2}*BD*AC\) = \(\frac{1}{2}*120*17x\) By equating the two expressions for area of triangle ABC, you can find the value of x and therefore, you can find a unique value of the area of triangle ABC. So, St. 1 is sufficient. Note that I could analyze Statement 1 till its logical conclusion because: 1. I drew a neat diagram that represented all the pieces of information that were given in Question statement + St. 1 and also the information that I had inferred (AC = 17x, using Pythagoras Theorem) 2. I was able to mentally rotate the triangle ABC such that AC became its base. By doing so, I could see that the area of triangle ABC can also be written as \(\frac{1}{2}*BD*AC\). This ability to mentally rotate a given image comes in handy in many questions. Here's an official question that involves rotation of a triangle: http://gmatclub.com/forum/inthefigureabovetriangleabcisequilateralandpoint143496.htmlHope this helped! Japinder
_________________
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



Intern
Status: Online
Joined: 07 Feb 2015
Posts: 28
Location: India
Rudey: RD
Concentration: Marketing, General Management
GMAT 1: 620 Q45 V31 GMAT 2: 640 Q46 V31
GPA: 3.29
WE: Sales (Hospitality and Tourism)

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
22 May 2015, 02:42
EgmatQuantExpert wrote: RudeyboyZ wrote: Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.
Warm Regards,
Rudraksh. Dear RudeyboyZPlease refer to the image below. It represents the information that we were given in the Question Statement and in St. 1.Also, if AB = 8x and BC = 15x, can you calculate AC? Sure, using Pythagoras Theorem, you know that it's 17x. Now, area of triangle ABC =\(\frac{1}{2}*AB*BC\) = \(\frac{1}{2}*8x*15x\) But, suppose you take the base as AC. Then, what is the height? BD. And St. 1 tells us that BD = 120 So, the area of triangle ABC can also be written as = \(\frac{1}{2}*BD*AC\) = \(\frac{1}{2}*120*17x\) By equating the two expressions for area of triangle ABC, you can find the value of x and therefore, you can find a unique value of the area of triangle ABC. So, St. 1 is sufficient. Note that I could analyze Statement 1 till its logical conclusion because: 1. I drew a neat diagram that represented all the pieces of information that were given in Question statement + St. 1 and also the information that I had inferred (AC = 17x, using Pythagoras Theorem) 2. I was able to mentally rotate the triangle ABC such that AC became its base. By doing so, I could see that the area of triangle ABC can also be written as \(\frac{1}{2}*BD*AC\). This ability to mentally rotate a given image comes in handy in many questions. Here's an official question that involves rotation of a triangle: http://gmatclub.com/forum/inthefigureabovetriangleabcisequilateralandpoint143496.htmlHope this helped! Japinder Dear Japinder, The detailed explanation is really helpful. Thank you! Warm Regards, Rudey.



Intern
Status: Online
Joined: 07 Feb 2015
Posts: 28
Location: India
Rudey: RD
Concentration: Marketing, General Management
GMAT 1: 620 Q45 V31 GMAT 2: 640 Q46 V31
GPA: 3.29
WE: Sales (Hospitality and Tourism)

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
22 May 2015, 02:43
EgmatQuantExpert wrote: RudeyboyZ wrote: Could you guys explain it a little in detail as to why is statement 1 sufficient. I did not get it.
Warm Regards,
Rudraksh. Dear RudeyboyZPlease refer to the image below. It represents the information that we were given in the Question Statement and in St. 1.Also, if AB = 8x and BC = 15x, can you calculate AC? Sure, using Pythagoras Theorem, you know that it's 17x. Now, area of triangle ABC =\(\frac{1}{2}*AB*BC\) = \(\frac{1}{2}*8x*15x\) But, suppose you take the base as AC. Then, what is the height? BD. And St. 1 tells us that BD = 120 So, the area of triangle ABC can also be written as = \(\frac{1}{2}*BD*AC\) = \(\frac{1}{2}*120*17x\) By equating the two expressions for area of triangle ABC, you can find the value of x and therefore, you can find a unique value of the area of triangle ABC. So, St. 1 is sufficient. Note that I could analyze Statement 1 till its logical conclusion because: 1. I drew a neat diagram that represented all the pieces of information that were given in Question statement + St. 1 and also the information that I had inferred (AC = 17x, using Pythagoras Theorem) 2. I was able to mentally rotate the triangle ABC such that AC became its base. By doing so, I could see that the area of triangle ABC can also be written as \(\frac{1}{2}*BD*AC\). This ability to mentally rotate a given image comes in handy in many questions. Here's an official question that involves rotation of a triangle: http://gmatclub.com/forum/inthefigureabovetriangleabcisequilateralandpoint143496.htmlHope this helped! Japinder Dear Japinder, The detailed explanation is really helpful. Thank you! Warm Regards, Rudraksh.



Math Expert
Joined: 02 Sep 2009
Posts: 44288

ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
25 May 2015, 02:50
Bunuel wrote: ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?
(1) The height from point B to the hypotenuse is 120. (2) The perimeter of the triangle is 680. OFFICIAL SOLUTION:ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?First of all since the ration of AB to BC is 8x:15x, then the hypotenuse is \(\sqrt{(8x)^2+(15x)^2}=17x\). (1) The height from point B to the hypotenuse is 120. Consider the image below: The area of a right triangle equals to \(\frac{1}{2}*leg_1*leg_2=\frac{1}{2}*AB*BC\). But the area of a right triangle can also be found by \(\frac{1}{2}*(altitude \ from \ right \ angle)*(hypotenuse)=\frac{1}{2}*BD*AC\). Equate these expressions: \(\frac{1}{2}*AB*BC=\frac{1}{2}*BD*AC\); \(\frac{1}{2}*8x*15x=\frac{1}{2}*120*17x\); \(120x=120*17\); \(x=17\). The area = \(\frac{1}{2}*AB*BC=\frac{1}{2}*8x*15x=60x=60*17\). Sufficient. (2) The perimeter of the triangle is 680. The perimeter = \(8x+15x+17x=680\). We can find x, thus we can find the area. Sufficient. Answer: D. P.S. Thank you Japinder for the drawing. Attachment:
image.png [ 26.82 KiB  Viewed 4182 times ]
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 01 Jan 2015
Posts: 56

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
14 Aug 2015, 12:27
Bunuel wrote: Bunuel wrote: ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?
(1) The height from point B to the hypotenuse is 120. (2) The perimeter of the triangle is 680. OFFICIAL SOLUTION:ABC is a right triangle with right angle at point B. If the ratio of AB to BC is 8 to 15, what is the area of the triangle?First of all since the ration of AB to BC is 8x:15x, then the hypotenuse is \(\sqrt{(8x)^2+(15x)^2}=17x\). (1) The height from point B to the hypotenuse is 120. Consider the image below: The area of a right triangle equals to \(\frac{1}{2}*leg_1*leg_2=\frac{1}{2}*AB*BC\). But the area of a right triangle can also be found by \(\frac{1}{2}*(altitude \ from \ right \ angle)*(hypotenuse)=\frac{1}{2}*BD*AC\). Equate these expressions: \(\frac{1}{2}*AB*BC=\frac{1}{2}*BD*AC\); \(\frac{1}{2}*8x*15x=\frac{1}{2}*120*17x\); \(120x=120*17\); \(x=17\). The area = \(\frac{1}{2}*AB*BC=\frac{1}{2}*8x*15x=60x=60*17\). Sufficient. (2) The perimeter of the triangle is 680. The perimeter = \(8x+15x+17x=680\). We can find x, thus we can find the area. Sufficient. Answer: D. P.S. Thank you Japinder for the drawing. Hi Bunuel, Can you please help me with this question, i have a specific doubt. St 1 says, The height from point B to the hypotenuse is 120; We know that Perpendicular to the hypotenuse(height from point B to the hypotenuse) will always divide the triangle into two triangles with the same properties as the original triangle and it'll be half the length of the hypotenuse? Which only goes on to prove hypotenuse = 2x120=240 St 2 says, perimeter of the triangle is 680; 8x+15x+17x=680 x=17 hypotenuse = 17x17 = 289 How can we have 2 different hypotenuse measures for the same triangle?



Current Student
Joined: 20 Mar 2014
Posts: 2685
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
14 Aug 2015, 12:41
2
This post received KUDOS
Ted21 wrote: Hi Bunuel, Can you please help me with this question, i have a specific doubt. St 1 says, The height from point B to the hypotenuse is 120; We know that Perpendicular to the hypotenuse(height from point B to the hypotenuse) will always divide the triangle into two triangles with the same properties as the original triangle and it'll be half the length of the hypotenuse? Which only goes on to prove hypotenuse = 2x120=240 St 2 says, perimeter of the triangle is 680; 8x+15x+17x=680 x=17 hypotenuse = 17x17 = 289 How can we have 2 different hypotenuse measures for the same triangle? you are quoting statement 2 correctly but your interpretation of statement 1 is not correct. The property that you have mentioned in the red text above is ONLY possible when the triangle is a 454590 triangle and in this triangle AB = BC which is not the case provided to us. Triangle ABC is NOT 454590 triangle and hence you can not say that the hypotenuse will be divided into 2 by the height from B. The way you will be able to calculate the impact of height is as follows: Let the 2 sides of the triangle be 8k and 15k (as the ratio is 8:15) Thus the hypotenuse is = ((8k)^2+(15k)^2)^0.5 = 17k. Thus the area of the triangle = 0.5*AB*BC = 0.5*Height*Hypotenuse > 0.5*8k*15k = 0.5*120*17k > k = 17 and thus hypotenuse = 17*17= 289 same as Statement 2. Hope this helps.



NonHuman User
Joined: 09 Sep 2013
Posts: 6530

Re: ABC is a right triangle with right angle at point B. If the ratio of [#permalink]
Show Tags
14 Apr 2017, 21:31
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: ABC is a right triangle with right angle at point B. If the ratio of
[#permalink]
14 Apr 2017, 21:31






