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In the rectangular coordinate system above, the area of triangular

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Posts: 230
Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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27 Mar 2018, 01:02
nss123 wrote:
Attachment:
IMAGE PT1.jpg
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

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.
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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19 Apr 2018, 14:30
Top Contributor
nss123 wrote:

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

Attachment:
IMAGE PT1.jpg

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:

So, the area of PQR = 28 - (3.5 + 6 + 6) = 12.5

Cheers,
Brent
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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03 May 2018, 07:48
I was struggling with this question- but I've found another method to solving this question, as I was unsure if PQR is a right triangle or not. It is to do with lengths and midpoint calculation.

first lets solve for the length of the base which is P (0,3) and R(7,4) = (√(0-7)²+(3-4)²=> √49+1=> √50=> 5√2

For the height- fist establish the mid point on the base PR= (0+7)/2, (3+4)/2=> 3.5,3.5
height is therefore length from Q(4,0) to midpoint (3.5,3.5)= √(4-3.5)²+(0-3.5)² => √(.5)²+(-.35)² => √.25+12.25=> √12.50
This can be further calculated as √5x5x5x5x2 x 10⁻²=> hence Height= 10⁻¹x 25√2

Put all together in triangle formula= 1/2 x 5√2 x 25√2 x 10⁻¹=> 125 x 10⁻¹= 12.50
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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13 Aug 2018, 08:06
Top Contributor
nss123 wrote:

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

Attachment:
IMAGE PT1.jpg

We're trying to find the area of triangle PQR.

Let's draw a rectangle around the triangle (as shown above) 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:

So, the area of PQR = 28 - (3.5 + 6 + 6) = 12.5

Cheers,
Brent
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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21 Jan 2019, 00:02
nss123 wrote:

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

So area of Trapezium = 1/2 * sum of parallel sides * height
= 1/2 * (OQ + O'R) * 7
= 49/2
=24.5

Now from this we need to subtract the two tingles to give area of triangle QPR
= 24.5 - ( 6*2)
= 12.5

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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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19 Jul 2019, 19:20
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).
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In the rectangular coordinate system above, the area of triangular  [#permalink]

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16 Oct 2019, 08:12
Pathfinder wrote:
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.

How do we know that PGR is the right triangle? It is also can be isosceles?
In the rectangular coordinate system above, the area of triangular   [#permalink] 16 Oct 2019, 08:12

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