Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 16 Jul 2019, 11:31

### 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

# In the figure above, ABC is a right triangle with AC as its hypotenuse

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56244
In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

01 May 2015, 02:23
19
00:00

Difficulty:

95% (hard)

Question Stats:

28% (02:18) correct 72% (02:38) wrong based on 208 sessions

### HideShow timer Statistics

In the figure above, ABC is a right triangle with AC as its hypotenuse, and PQRS is a square. What is the area of the square?

(1) AC is 70 units long.

(2) The product of the length of AS and the length of RC is 396.

Kudos for a correct solution.

Attachment:

conundrum.gif [ 1.93 KiB | Viewed 5946 times ]

_________________
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2959
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

02 Jun 2015, 08:13
5
5
Bunuel wrote:

In the figure above, ABC is a right triangle with AC as its hypotenuse, and PQRS is a square. What is the area of the square?

(1) AC is 70 units long.

(2) The product of the length of AS and the length of RC is 396.

Kudos for a correct solution.

Attachment:
conundrum.gif

I never suggest students to take help of any Trigonometry therefore the important way to learn to solve this question is identifying the similar triangles

Look at Triangle PAS and Triangle QRC

in Triangle PAS
Let Angle PAS = x
then Angle APS = (90-x)
and angle PSA = 90

Also look at triangle ABC where if Angle A = x then angle C = (90-x) because angle B=90

Now in Triangle QRC

Since Angle QCR = (90-x)
therefore, Angle RQC = x
and Angle QRC = 90

Since angles between Triangle Triangle PAS and Triangle QRC are equal hence they become Similar Triangle

i.e. PS/AS = RC/QR

i.e. AS x RC = PS x QR
But PS and QR are equal as they are sides of Square therefore PS x QR represents area of Square
hence, Area of Square = AS x RC

Statement 1: AC = 70
NOT SUFFICIENT

Statement 2: AS x RC = 396
SUFFICIENT

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Manager
Joined: 17 Mar 2015
Posts: 116
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

01 May 2015, 02:57
3
2
#1 is clearly insufficient.
#2 - since PQ and AC are parallel, we can figure out that triangles APS and QRC are similar. Going by angles we get the relation:QR/AS = RC/PS
QR*PS = AS*RC = 396. QR*PS is pretty much the area of the square, sufficient.

B
##### General Discussion
Manager
Joined: 06 Mar 2014
Posts: 89
In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

01 Jun 2015, 10:45
Hi Bunuel:

Can you please explain how are the two triangles similar??
And can you give a detailed solution to the problem.
Manager
Joined: 17 Mar 2015
Posts: 116
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

01 Jun 2015, 12:13
The only reason they are similar is coz angle B is 90.
Say angle BPQ is q, then angle BQP = 90 - q, Since PQRS is a square, PQ is parallel to AC which means that angle A = angle BPQ = q and angle C = angle BQP = 90 - q. Now if you look closely at these 2 triangles which are right triangles by the way (courtesy of PQRS being a square), you can infer that angle A = angle RQC = q and angle C = angle APS = 90 - q which makes these 2 triangles similar.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2942
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

02 Jun 2015, 07:48
2
1

(1) AC is 70 units long.
That is, AS + RS + RC = 70

We need to know the value of RS in order to find the area of the square. But the above equation also has 2 other unknowns. Even if we try to express AS and RC in terms of RS, we cannot do so without involving other dimensions of the triangles in this figure. Therefore, St. 1 is not sufficient to determine a unique value of RS.

(2) The product of the length of AS and the length of RC is 396.

That is, AS*RC = 396

In right triangle ABC,

$$tanC = \frac{AB}{BC}$$ . . . (1)

In right triangle CRQ,

$$tanC = \frac{QR}{RC}$$ . . . (2)

By equating (1) and (2), we get:

$$\frac{AB}{BC} = \frac{QR}{RC}$$

That is, $$RC = \frac{QR*BC}{AB}$$ . . . (3)

Now, in right triangle ABC,

$$tanA = \frac{BC}{AB}$$ . . . (1')

In right triangle ASP,

$$tanA = \frac{PS}{AS}$$ . . . (2')

By equating (1') and (2'), we get:

$$\frac{BC}{AB} = \frac{PS}{AS}$$

That is, $$AS = \frac{PS*AB}{BC}$$ . . . (3')

By substituting equations (3) and (3') in the red equation above, we get:

$$\frac{PS*AB}{BC}*\frac{QR*BC}{AB} = 396$$

That is, PS*QR = 396
$$(Side of square)^2 = 396$$

So, the area of square = 396 sq. units.

Thus, St. 2 is sufficient to find the area of the square.

Hope this helped!

Best Regards

Japinder
_________________
Manager
Joined: 18 Nov 2013
Posts: 78
Concentration: General Management, Technology
GMAT 1: 690 Q49 V34
In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

03 Jun 2015, 00:04
2
Attachment:
File comment: triangle.png

triangle.png [ 27.36 KiB | Viewed 4689 times ]

to get area of square we need one side of square or some relation with square sides and triangle

stmt 1: AC = 70 ; not sufficient
one value and lots of unknowns

stmt 2: AS x RC = 396 ; looks promising lets evaluate these sides

now refer to these sides and triangles consisted with these side {AS} and {RC} , they have one side of square common, one right angle common, possibility of similar triangles and (ratio of sides can lead us to $$side^{2}$$)

lets see $$\triangle APS$$ ~ $$\triangle CRQ$$

$$\angle a = \angle x$$ ; this is because a, b, c in bigger right $$\triangle$$ have eq (a + c =90) and similarly in smaller right $$\triangle$$ APS (a + y = 90) and CRQ (x + c = 90)

$$\angle y = \angle c$$

$$\angle S = \angle R$$ ; both $$90^{\circ}$$

now in similar triangles $$\triangle APS$$ ~ $$\triangle CRQ$$

$$\frac{PS}{AS}$$ $$= \frac{RC}{QR}$$

$$side^{2} =$$ AS x RC = 396 : sufficient

Ans: B
_________________
_______
- Cheers

+1 kudos if you like
Intern
Joined: 21 Mar 2017
Posts: 37
Location: Zimbabwe
Concentration: General Management, Entrepreneurship
GMAT 1: 680 Q45 V38
GMAT 2: 750 Q49 V42
GPA: 3.3
WE: Accounting (Accounting)
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

08 Sep 2017, 01:15
EgmatQuantExpert wrote:

(1) AC is 70 units long.
That is, AS + RS + RC = 70

We need to know the value of RS in order to find the area of the square. But the above equation also has 2 other unknowns. Even if we try to express AS and RC in terms of RS, we cannot do so without involving other dimensions of the triangles in this figure. Therefore, St. 1 is not sufficient to determine a unique value of RS.

(2) The product of the length of AS and the length of RC is 396.

That is, AS*RC = 396

In right triangle ABC,

$$tanC = \frac{AB}{BC}$$ . . . (1)

In right triangle CRQ,

$$tanC = \frac{QR}{RC}$$ . . . (2)

By equating (1) and (2), we get:

$$\frac{AB}{BC} = \frac{QR}{RC}$$

That is, $$RC = \frac{QR*BC}{AB}$$ . . . (3)

Now, in right triangle ABC,

$$tanA = \frac{BC}{AB}$$ . . . (1')

In right triangle ASP,

$$tanA = \frac{PS}{AS}$$ . . . (2')

By equating (1') and (2'), we get:

$$\frac{BC}{AB} = \frac{PS}{AS}$$

That is, $$AS = \frac{PS*AB}{BC}$$ . . . (3')

By substituting equations (3) and (3') in the red equation above, we get:

$$\frac{PS*AB}{BC}*\frac{QR*BC}{AB} = 396$$

That is, PS*QR = 396
$$(Side of square)^2 = 396$$

So, the area of square = 396 sq. units.

Thus, St. 2 is sufficient to find the area of the square.

Hope this helped!

Best Regards

Japinder

Thanks for this detailed explanation.

1 question though, do we need to know trigonometry to this level for the GMAT. I would like a Q51
_________________
Kudos if you like my response please
Senior Manager
Joined: 02 Apr 2014
Posts: 472
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

06 Mar 2018, 01:02
1
Statement 1 : clearly insufficient

Statement 2: AS * RC = 396

All the triangles are similiar

AS/PS = QR/RC

PS = QR = side of square

AS * RC = PS * QR = 396 = area of square => sufficient (B)
Non-Human User
Joined: 09 Sep 2013
Posts: 11658
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse  [#permalink]

### Show Tags

26 Jun 2019, 03:09
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.
_________________
Re: In the figure above, ABC is a right triangle with AC as its hypotenuse   [#permalink] 26 Jun 2019, 03:09
Display posts from previous: Sort by