In the rectangular coordinate system above, the area of tria

[r08]

Greetings,
Forgive me if this seems too simple.
I searched for the question in the forums but could not find it.
Could you help explain the answer?
(I calculated 14 but I knew it was wrong)

[Guest]

There are several ways to do this and I'm going to show you the two minute solution.

The question is asking us to calculate the different sides of the triangle, identify the base and height, and calculate.

These problems can get time consuming so the important thing is to work with what they give you.




By looking at the points of the yellow triangle, we see that it contains the points (0,3) and (4,0). This means its a 3,4,5 triangle and we know one side of the mystery triangle = 5

By drawing an imaginary line to the point (7,0) from (7,4), we can see that we have another 3,4,5 triangle and the other side of the mystery triangle is 5.

(BxH)/2 = 25/2 = 12.5


You can always use the distance formula too which is

[r08]

Thank you!
Great explanation.

[RonPurewal]

the solution is beautiful, but it also relies on the fortuitous fact that the lines QP and PR are perpendicular. you didn't explicitly acknowledge this fact in your otherwise wonderfully detailed solution, so i'm not sure you're aware of it; the only reason that the formula (1/2)(b)(h) works in this case is because PQ just so happens to be perpendicular to PR.
for instance, if you move R from (7, 4) to (8, 3), then your green triangle would still be 3-4-5, but in that case the formula (1/2)(5)(5) would yield the wrong area.

here's a much more universally applicable way to find the area of triangles like these:
surround the triangle with a rectangle, and then subtract out the unwanted area.
this will always work, regardless of the shape of your triangle, because the unwanted parts of the rectangle will always be either rectangles or right triangles with sides perpendicular to the axes (both of which have areas that are extremely easy to calculate).
in this problem:
surround the triangle with a rectangle whose vertices lie at O(0, 0), A(7, 0), R, and B(0, 4).
the area of the rectangle is (7)(4) = 28.
the area of unwanted triangle QBR is (1/2)(1)(7) = 3.5.
the area of unwanted triangle RAP is (1/2)(3)(4) = 6.
the area of unwanted triangle POQ is (1/2)(4)(3) = 6.
28 - 3.5 - 6 - 6 = 12.5

[Priyanka]

Hi the area can be calculated by solving a 3 x 3 matrix

= 1/2 | 1 x1 y1 |
| 1 x2 y2 |
| 1 x3 y3 |

where A (x1,y1) , B(x2,y2) and C (x3,y3)

[RonPurewal]

Hi the area can be calculated by solving a 3 x 3 matrix

= 1/2 | 1 x1 y1 |
| 1 x2 y2 |
| 1 x3 y3 |

where A (x1,y1) , B(x2,y2) and C (x3,y3)

yes, this works too, if you have this sort of artillery in your arsenal.

[rippersid]

An arithematic approach:

Area of Triangle = 1/2*b*h
base = sqrt(( x2-x1)^2) = 7

now, imagine PQ to be a line rather than a line segment.

The equation of the line can be calculated as y=(1/7)x + 3

now the height by definition is the perpendicular drawn from the vertex, in this case (4,0)
substitute into the equation of the line and you get y = 25/7

thus area is (1/2)*7*(25/7) = 12.5 :)

Hope that works

[Misha]

Ron's solution is the best: it's easy (we can ALL do this), efficient , and most importantly provides the correct answer.

Appreciate the other interesting ways to solve this problem, but GMAT is supposed to test our basic knowledge.

[gkhan]

the solution is beautiful, but it also relies on the fortuitous fact that the lines QP and PR are perpendicular. you didn't explicitly acknowledge this fact in your otherwise wonderfully detailed solution, so i'm not sure you're aware of it; the only reason that the formula (1/2)(b)(h) works in this case is because PQ just so happens to be perpendicular to PR.
for instance, if you move R from (7, 4) to (8, 3), then your green triangle would still be 3-4-5, but in that case the formula (1/2)(5)(5) would yield the wrong area.

here's a much more universally applicable way to find the area of triangles like these:
surround the triangle with a rectangle, and then subtract out the unwanted area.
this will always work, regardless of the shape of your triangle, because the unwanted parts of the rectangle will always be either rectangles or right triangles with sides perpendicular to the axes (both of which have areas that are extremely easy to calculate).
in this problem:
surround the triangle with a rectangle whose vertices lie at O(0, 0), A(7, 0), R, and B(0, 4).
the area of the rectangle is (7)(4) = 28.
the area of unwanted triangle QBR is (1/2)(1)(7) = 3.5.
the area of unwanted triangle RAP is (1/2)(3)(4) = 6.
the area of unwanted triangle POQ is (1/2)(4)(3) = 6.
28 - 3.5 - 6 - 6 = 12.5

Great tip, thanks...

[rchitta]

I think Ron's solution could be simplified even more. Ron's solution considers a rectangle but it would be simpler to consider a trapezium.

Let's say we draw a perpendicular line to x -axis from R and that intersects x-axis at M.

Then all we need to calculate is:
A(Trapezium OQRM) - A(Right Triangle OQP) - A(Right Triangle PRM)

Area of Trapezium OQRM = 1/2* (b1 + b2) * h = 1/2* (3 + 4) * 7 = 1/2 * 49
Area of right triangle OQP = 1/2 * b * h = 1/2 * 4 * 3 = 1/2 * 12
Area of right triangle PRM = 1/2 * b * h = 1/2 * 3 * 4 = 1/2 * 12

Area of Tri'gle PQR = 1/2 (49 - 12 - 12) = 12.5 sq units

Take away I got from this problem: think outside the box a little bit, look for familiar diagrams (for which you know the formulas). It's not a bad idea to remember the formula for the area of a trapezium.

Thanks

the solution is beautiful, but it also relies on the fortuitous fact that the lines QP and PR are perpendicular. you didn't explicitly acknowledge this fact in your otherwise wonderfully detailed solution, so i'm not sure you're aware of it; the only reason that the formula (1/2)(b)(h) works in this case is because PQ just so happens to be perpendicular to PR.
for instance, if you move R from (7, 4) to (8, 3), then your green triangle would still be 3-4-5, but in that case the formula (1/2)(5)(5) would yield the wrong area.

here's a much more universally applicable way to find the area of triangles like these:
surround the triangle with a rectangle, and then subtract out the unwanted area.
this will always work, regardless of the shape of your triangle, because the unwanted parts of the rectangle will always be either rectangles or right triangles with sides perpendicular to the axes (both of which have areas that are extremely easy to calculate).
in this problem:
surround the triangle with a rectangle whose vertices lie at O(0, 0), A(7, 0), R, and B(0, 4).
the area of the rectangle is (7)(4) = 28.
the area of unwanted triangle QBR is (1/2)(1)(7) = 3.5.
the area of unwanted triangle RAP is (1/2)(3)(4) = 6.
the area of unwanted triangle POQ is (1/2)(4)(3) = 6.
28 - 3.5 - 6 - 6 = 12.5

[RonPurewal]

I think Ron's solution could be simplified even more. Ron's solution considers a rectangle but it would be simpler to consider a trapezium.

Let's say we draw a perpendicular line to x -axis from R and that intersects x-axis at M.

Then all we need to calculate is:
A(Trapezium OQRM) - A(Right Triangle OQP) - A(Right Triangle PRM)

Area of Trapezium OQRM = 1/2* (b1 + b2) * h = 1/2* (3 + 4) * 7 = 1/2 * 49
Area of right triangle OQP = 1/2 * b * h = 1/2 * 4 * 3 = 1/2 * 12
Area of right triangle PRM = 1/2 * b * h = 1/2 * 3 * 4 = 1/2 * 12

Area of Tri'gle PQR = 1/2 (49 - 12 - 12) = 12.5 sq units

you can also do that, sure.

i'm not sure whether that qualifies as "simpler", since the additional work involved in computing the area of a trapezoid (vs. a rectangle) pretty much cancels out the time savings of having one fewer triangle.

but the point is, yeah, there are lots of ways to do problems like this one. nice job.

[salman30]

I think Ron's solution could be simplified even more. Ron's solution considers a rectangle but it would be simpler to consider a trapezium.

Let's say we draw a perpendicular line to x -axis from R and that intersects x-axis at M.

Then all we need to calculate is:
A(Trapezium OQRM) - A(Right Triangle OQP) - A(Right Triangle PRM)

Area of Trapezium OQRM = 1/2* (b1 + b2) * h = 1/2* (3 + 4) * 7 = 1/2 * 49
Area of right triangle OQP = 1/2 * b * h = 1/2 * 4 * 3 = 1/2 * 12
Area of right triangle PRM = 1/2 * b * h = 1/2 * 3 * 4 = 1/2 * 12

Area of Tri'gle PQR = 1/2 (49 - 12 - 12) = 12.5 sq units

you can also do that, sure.

i'm not sure whether that qualifies as "simpler", since the additional work involved in computing the area of a trapezoid (vs. a rectangle) pretty much cancels out the time savings of having one fewer triangle.

but the point is, yeah, there are lots of ways to do problems like this one. nice job.

Very useful thread - thanks Ron.

[RonPurewal]

Very useful thread - thanks Ron.

thanks

[nilubh]

You simply rock Ron ..:)

[RonPurewal]

thanks.