In the rectangular coordinate system above, if OP < PQ, is the area of

From the question we have OP < OQ .....(a)
also from Pythagoras theorem we have \(OP^2\) = \(PS^2\) +\(OS^2\) and \(OQ^2\) = \(PS^2\) +\(SQ^2\)
Putting these value in equation a we have
\(PS^2\) +\(OS^2\) < \(PS^2\) +\(SQ^2\)
=> OS<SQ

Hope its clear now

More simplified version of what Bunuel explained.

We MUST test OP = PQ because under this constraint we can see that the base equals 12 and thus the area =1/2 *12*8=48, but since OP is actually less than PQ, the base must be larger (can deduce through inequality theorem).

Hi Bunuel VeritasKarishma - can i make the following general take-away ?

In any isosceles triangle
- Right angle isosceles like 45-45-90 or
- Non right angle isosceles triangle like 20-20-140

The perpendicular from the vertex to the base line (the height) in both types of isosceles triangles divides the base into two equal lengths ?

Yes, but only if the base is the non-equal side.

This is clever, but I am wondering why my approach was wrong?

Given P(6,8), we can calculate the distance OP --> √(6-0)^2 + (8-0)^2 --> OP = 10.

Since PQ > OP, we can do some test values

PQ = 11 --> Area = 44
PQ = 13 --> Area = 104/2 = 52

Insufficient.

Just wondering why that's not an equally valid approach?

Not sure how you calculated the areas but in both cases the area is different and greater than 48.

So given a height of 8 ---> 8 x 11 /2 = 88 /2 = 44 <48 and 8 x 13 /2 = 104 / 2 = 52 > 48...two different solutions?

The area of a triangle 1/2*height*base = 1/2*PS*OQ.

Height = PS = 8;
Base = OQ. (If PQ = 11, then \(OQ = 6 + \sqrt{57}\) and if PQ = 13, then \(OQ = 6 + \sqrt{105}\)).

You are calculating the area as 1/2*PS*PQ, which is wrong.

Isn't assuming that OQ is a base an improper assumption? brunel

Good Question +1

This question reminds me that GMAT is not only about math, Its MATH + LOGIC skills.

Let's try to analyze this question.

In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48?

(1) The coordinates of point P are (6,8)



If you drop a perpendicular from point P to the base OQ and meet at a point S and this line will divide the △ POQ into 2 triangles.

△1 would be a right-angled triangle with base OS and height PS.

By using the coordinates of point P (6,8 ), we can say that side OS = 6 and PS = 8.

Hence, Area of △1 = 1/2 * base * height = 1/2 * 6 * 8 = 24.

Now let us consider △2. It will also be a right-angled triangle with base SQ and height PS.

Since it is given in the question that the hypotenuse PQ > OP,

\(OS^2 + PS^2 = OP^2 \) in △1

\(SQ^2 + PS^2 = PQ^2 \) in △ 2

Comparing both equations, since PQ > OP, we can conclude that the base SQ > OS.

Since the height of 2 triangles, PS is the same. The base of △ 2 is greater than △ 1.

Area of △2 = 1/2 * base SQ * height PS

Area of △1 = 1/2 * base OS * height PS = 24.

Therefore, we conclude that the area of △2 is greater than △ 1 i.e 24.

So the combined area of 2 triangles should be greater than 24+24 i.e 48

This will answer the question stem. Hence, Statement 1 alone is sufficient.

(2) The coordinates of point Q are (13,0)

From the coordinates of Q, we can say that the base OQ=13 but we don't have any idea about the coordinates of P.
Coordinates of P will determine the height of the triangle. Since the height is unknown, we will not be able to figure out if the area of △ OPQ is greater than 48 or not.

Statement 2 alone is not sufficient.

Option A is the correct answer.

Thanks,
Clifin J Francis,
GMAT QUANT SME

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