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Re: In the figure above, ABC is a right triangle with AC as its hypotenuse [#permalink]

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01 May 2015, 01:57

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#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.

Re: In the figure above, ABC is a right triangle with AC as its hypotenuse [#permalink]

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01 Jun 2015, 11:13

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.

(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.

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

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

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

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02 Jun 2015, 23:04

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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 \(\triangle APS\) ~ \(\triangle CRQ\)

Re: In the figure above, ABC is a right triangle with AC as its hypotenuse [#permalink]

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16 Jul 2016, 00:31

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Re: In the figure above, ABC is a right triangle with AC as its hypotenuse [#permalink]

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07 Sep 2017, 06:10

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Re: In the figure above, ABC is a right triangle with AC as its hypotenuse [#permalink]

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08 Sep 2017, 00:15

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
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