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Re: In the rectangular coordinate system above, the area of triangular
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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
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19 Apr 2018, 14:30
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 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 Answer: A Cheers, Brent
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Re: In the rectangular coordinate system above, the area of triangular
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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) = (√(07)²+(34)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)= √(43.5)²+(03.5)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
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13 Aug 2018, 08:06
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 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 Answer = A Cheers, Brent
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Re: In the rectangular coordinate system above, the area of triangular
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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 Answer A.
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Re: In the rectangular coordinate system above, the area of triangular
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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 xy axis, giving us 5,5 (legs of the triangle).
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In the rectangular coordinate system above, the area of triangular
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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
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