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In a right triangle PQR, X and Y are mid points of PQ and PR

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vinod1986
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   13 Oct 2021, 03:11
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Carcass wrote:
Updated the OA

Thank you guys


Carcass,

Is the answer A or C because I got answer A.
T cannot be midpoint unless it is specified in the QC question stem.
If answer is C can you provide the explanation.
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In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   13 Oct 2021, 03:42
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The answer is C

here in the discussion a well done explanation https://gre.myprepclub.com/forum/in-a-righ ... tml#p56376

Ask for further assistance. maybe GreenlightTestPrep could delve into this also
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   20 Feb 2022, 02:13
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Since X and Y midpoints,
XY||QR and XY=QR/2=4
for any position of T on QR, Area of triangle XYT will be same and that for midpoint, 1/2*XY*YT=1/2*(8/2)*(6/2)=6
Area of triangle PXY=1/2*PX*XY=1/2*3*4=6
so area of PXTY=area of PXY+area of XYT=6+6=12
so OPTION C is the answer.
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   19 Aug 2022, 03:26
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Area of PXTY is not 15 as multiplying adjacent sides to find area only works for rectangles and squares and PXTY is neither.

heres how I did it. pls correct me if I'm wrong.

we have PX = XQ = 3, PY= YR = 5 ( not explaining this since other users have explained already)

now area of PQR = 1/2 * BASE * HEIGHT = 6*8/2 = 24.
and area of PXTY = area of ▲PQR - (Area of ▲XQT + Area of ▲YTR)

consider QT = x , then TR = 8-x.

then area of ▲XQT = 1/2 * 3 * x = 3x/2 (since this is a rt. triangle length of perpendicular = height)

however ▲YTR is not right angled, so we draw a line from Y that intersects QR perpendicularly at point Z.
Now midpoint theorem states that XY || QR. since YZ is perpendicular to QR, it must also perpendicular to XY. So all angles of quadrilateral XYZQ are 90° thus XYZQ is a square. hence XQ = YZ = 3.
So area of ▲YTR = 1/2 * TR * YZ = 1/2 * (8-x) * 3 = 12 - (3x/2).

Now add area of ▲XQT + ▲YTR = [3x/2] + [12 - (3x/2)] = 12.

Area of PXTY = Area of ▲PQR - (Area of ▲XQT + ▲YTR)
= 24 - 12
= 12.

Option C is the correct answer.
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In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   28 Jan 2023, 20:19
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It is a very easy question if you know the property that the median of a triangle divides the triangle into 2 equal halves.
As Line TX is the median, triangle PTQ have two equal area triangles: PXT and QXT.
Similarly, triangle PRT is divided into two equal are triangles: PTY and TRY.

In this way we get area of full triangle, PQR = 1/2 * PXTY

We know area of PQR = 1/2*base*height = 1/2*6*8= 24.
so PXTY = 1/2*24=12

Attachment:
GRE triangle (10).jpg
GRE triangle (10).jpg [ 77.49 KiB | Viewed 812 times ]


So the correct answer is C.
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   20 Jun 2023, 04:39
Carcass how do we tell the quadrilateral is a rectangle?
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   20 Jun 2023, 05:20
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Sorry, where you saw we are dealing with a rectangle ?
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   21 Jun 2023, 20:02
Carcass wrote:
PQ=6 and QR = 8
AND

PR = 10

PX is = 3 and PY=5

You know these two sides. \(3 \times 5 = 15\) regardless you do not know where is T. Actually is it a rectangle.

Regards


Carcass how do we tell PXTY is a rectangle?
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   21 Jun 2023, 22:54
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Attachment:
GRE triangle.jpg
GRE triangle.jpg [ 2.31 MiB | Viewed 732 times ]
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In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   24 Jun 2023, 04:13
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Another solution using Ratios

Area of PXTY = Area of PQR - (Area of XQT + Area of YTR)
= 24 - (24 * 3/6 * QT/QR + 24 * 5/10 * TR/QR)
= 24 - 12( QT + TR )/QR
= 12

Option C
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink] New post   30 Sep 2024, 04:11
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Re: In a right triangle PQR, X and Y are mid points of PQ and PR [#permalink]
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