In the rectangular coordinate system above, the area of triangular

Question

https://gmatclub.com/forum/download/file.php?id=8848 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 IMAGE PT1.jpg

nookway · Accepted Answer

The answer is 12.5

If you look carefully, the triangle is enclosed within a rectangle with dimensions 7 x 4. See attached figure.

Area of the  	riangle PQR

= Area of rectangle - Area of yellow triangle - Area of blue triangle - Area of red triangle

= (7 	imes 4) - (\frac{1}{2} 	imes 3 	imes 4) - (\frac{1}{2} 	imes 4 	imes 3) - (\frac{1}{2} 	imes 1 	imes 7)

= 28 - 6 - 6 - 3.5

= 12.5

oldstudent · Answer

Direct formula for finding area of a triangle in coordinate system

1/2 { (x1-x2).(y2-y3) - (y1-y2).(x2-x3) }

When we have all the coordinate points of vertices, we can directly substitute and get the area
=>
substituting above, we get:

1/2 {(-3)(1) - (-4).(7)} = 12.5 - (A)

Pathfinder · Answer

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.

fluke · Answer

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

gottabwise · Answer

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)

Bunuel · Answer

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.

MrMicrostrip · Answer

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!

akriti1 · Answer

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

vivekgautam1 · Answer

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

JeffTargetTestPrep · Answer

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.

https://i.imgur.com/9ZDNaaH.png

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

BrentGMATPrepNow · Answer

Let's  draw a  rectangle  around the triangle  (as shown below) and then subtract from the rectangle's area (28) the areas of the 3 right triangles that surround the triangle in question.

We get the following:
 https://i.imgur.com/eFvMeIF.png

So, the area of PQR =  Area of rectangle  -  (area of 3 right triangles) 
=  28  -  (3.5 + 6 + 6)  
= 12.5

Answer: A
Cheers,
Brent

dcummins · Answer

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).

ManifestDreamMBA · Answer

Used area of trapezium
= (3+4) *7/2 = 49/2 = 24.5
Area of triangle OQP = 3*4/2 = 6
Area of triangle RPx = (7-4)*4/2 = 6 (assume a perpendicular point from R to the x-axis)
Area of PQR = 24.5 - 6 - 6 = 12.5

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In the rectangular coordinate system above, the area of triangular

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In the rectangular coordinate system above, the area of triangular

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