In the rectangular coordinate system above, the area of triangular

I came to the same result but my way of solving was a little bit different.

Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.

Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\).

We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5.

If the product of slopes of two lines is -1, those two lines are perpendicular.

Here; PR is perpendicular to PQ.

Point \(P=(x,y)=(4,0)\)
Point \(Q=(x,y)=(0,3)\)
Point \(R=(x,y)=(7,4)\)

Slope of PR\(=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{7-4}=\frac{4}{3}\)
Slope of PQ\(=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{0-4}=\frac{3}{-4}==\frac{-3}{4}\)

Product of slopes \(= \frac{4}{3}*\frac{-3}{4}=-1\)

Hence, \(PQ \perp PR\) and PQR is a right angled triangle.

Formula of distance between two points

Distance between two points \((x_1,y_1) & (x_2,y_2) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Distance between point P and point Q \(=\sqrt{(3-0)^2+(0-4)^2} = \sqrt{9+16} = \sqrt{25} = 5\)
Likewise,
Distance between point P and point R \(=\sqrt{(4-0)^2+(7-4)^2} = \sqrt{16+9} = \sqrt{25} = 5\)

Area of a triangle = 1/2*(PQ)*(PR) = 1/2*5*5=12.5

Ans: "A"

Please visit the following link for more on coordinate geometry:
https://gmatclub.com/forum/math-coordinate-geometry-87652.html

PQR just happened to be right triangle, so if we noticed this fact we could use properties of a right triangle to solve the problem (for example the way Pathfinder did). On the other hand solution provided by nookway works no matter whether PQR is right or not, also it requires much less calculations.

what formula did you apply.

If you calculate -
RP = \(\sqrt{(7-4)^2 + (4-0) ^2}\) = \(\sqrt{(3)^2 + (4) ^2}\) = 5
similarly,
PQ = \(\sqrt{(4-0)^2 + (0-3) ^2}\) = \(\sqrt{(4)^2 + (4) ^2}\) = 5

and,
QR = \(\sqrt{(0-7)^2 + (3-4) ^2}\) = \(\sqrt{(7)^2 + (1) ^2}\) = \(\sqrt{50}\) = 5\(\sqrt{2}\)

So, you can see from the length of the sides, that the triangle is right angled at P.

So, the area will be -
\(\frac{1}{2} (base * height)\) = \(\frac{1}{2}(5 * 5)\) = 12.5

by calculating area of different different region is a good approach but not effective when you just have 2 sec. left.
I have generated one effective way to solve these type of problem.
here we know that Q has coordinates (0,3) so measure it roughly through your pencil or pen, now we have one task left that is to measure height of a triangle.
So from the measured distance we can predict the height of a triangle which comes to be 3.60 approx. in this question thus Area will be 12.6(approx.) hence option A is 99.99% correct.
Hopefully you will enjoy to use this approach!

In the rectangular coordinate system above, the area of triangular region PQR is

(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25

We begin by drawing a rectangle that circumscribes the given triangle, creating 3 right triangles, which we label as A, B, and C. Notice that each of these 3 triangles is a right triangle. To determine the area of triangular region PQR, we can subtract the combined areas of triangles A, B, and C from the area of the rectangle.



Let’s determine the area of each right triangle.

Triangle A:

Area = base x height x 1/2

A = 7 x 1 x ½ = 3.5

Triangle B:

A = 4 x 3 x ½ = 6

Triangle C:

A = 3 x 4 x ½ = 6

The sum of the areas of triangles A, B, and C is 3.5 + 6 + 6 = 15.5

Finally we need the area of the rectangle:

Area = length x width

Area = 7 x 4 = 28.

So the area of triangle PQR is 28 – 15.5 = 12.5.

Answer: A

This answer is (A) 12.5. We can easily prove PQR is a right triangle with the length of PQ and PR are 5, and QR is 5\sqrt{2}. And the area of triangle PQR = square of PQ*square of PR/2=12.5.

And Nookway gave a correct result in very good ways.

Nookway...thanks for the colored graph. It's a great visual and helpful reminder of how to be smarter/more efficient.

Pathfinder...is it necessary to determine whether PQR is a right triangle? I'm unsure about the relevance.

Method - distance formula for PQ & PR (because they're perpendicular), area formula (1/2B*H) = 1/2(PQ)(PR)=1/2(5)(5)

Do not overwhelm yourself with unnecessary and time-consuming fomulas. There are two common types of triangles on the GMAT, besides 30-60-90 and 45-45-90. The 3:4:5 and 5:12:13 and the apply with multiples.

In this case, the base of the triangle on the right would be 3 (7 - 4) and the height 4, therefore the hypotenuse must be 5. So now we have the base for our formula. For the height, focus on the triangle on the left, which base would be 4 and height 3, therefore its hypotenuse must be 5. Now we can apply the formula. Area= 1/2 * 5 * 5 = 12,5.

Option A

Hope it helps!

For me here estimation works as following:-
we know base is 7 and by looking at figure we can say maximum height is 4 and minimum is 3 and our triangle height will like somewhere in between so
D and E out as maximum is \(\frac{1}{2}* 7*4= 14\) and minimum is\(\frac{1}{2}*7*3= 10.5\)
now we can also eliminate C AND B as from the figure itself we know its some where around between so height can be 3.4,3.5 or 3.6 and we can also say\(\sqrt{2}\) is also not possible then only option A left.

i would use the formula;
1/2 * mod [ x1(y2-y3)+ x2(y3-y1)+x3(y1-y2)]
1/2mod(-25)
12.5

It is a good approach, the way I solved it was:

the base is 7 (fact). the height is no more than 4 and no less than 3 (fact) so the area:

\(21*0.5 < A < 28*0.5\)

So answer MUST be between 10.5 and 14. only one answer fits

Multiple methods exist to solving this problem.

The easiest, at least for me, was pythag triples. See my diagram

We can actually deduce the distance of each point using the x-y axis, giving us 5,5 (legs of the triangle).

That is relevant so that you can conclude that PQ can act as a height to the base PR in the area formula, i.e, is it the perpendicular from the opposite vertex on to this side.

This is a great way to think about this. Kudos to you thanks so much! I hit this problem within the last minute of my practice exam and had to guess. Will think about this if something like it pops up again!

I have applied the formulas, but still not clear on this problem. please help

Please be more specific. Thank you.

Check this: Asking Excellent Questions.