In the given figure Abcd is a rectangle with length (p2q-5) and breadth (q2+4p)findthe area of shaded triangle pab

Area of rectangle  ABCD =length l×breadth b=AB×BC=p2q-5×q2+4p    given length=p2q-5 and breadth=q2+4p=p2q-5q2+4p sq. units.Now from the figure we can observe that PBC is symmteric to PADSo, area of PBC=area of PADAssuming P to be the mid point of CDNow we know that each angle of rectangle is 90°C=90°AreaPBC=12×CB×PC  area of right =12×base×height=12×q2+4p×p2q-52=q2+4pp2q-54 sq. units.So, AreaPBC=Area of PAD=q2+4pp2q-54So, area of shaded region=Area of rectangle  ABCD- AreaPBC+Area of PAD=p2q-5q2+4p-q2+4pp2q-54+q2+4pp2q-54=p2q-5q2+4p-2q2+4pp2q-54=4q2+4pp2q-5-2q2+4pp2q-54=2q2+4pp2q-54=q2+4pp2q-52    sq. units.

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