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

Hi Bunuel:

Can you please explain how are the two triangles similar??
And can you give a detailed solution to the problem.

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.

Hope this helped!

Best Regards

Japinder

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

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

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)

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