Re: In a right triangle PQR, X and Y are mid points of PQ and PR
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19 Aug 2022, 03:26
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.