ABCD is a parallelogram. P is any point on AD. CT is drawn perpendicular to PB. If PB=12 cm and CT=5cm determine the area of the parallelogram ABCD.
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Consider P is the mid point of AD,and a line is drawn from P to B.
CT is the perpendicular drawn from C to line PB. ( assumed T is the centre of PB)
Area of paralellogram ABCD = bh sq.units
base = BC,,,,,,Height = AB
In triangle BTC ,,, CT=5,TB=6
From pythagorous theorm
BC^2= CT^2+TB^2
BC^2=5^+6^2
BC^2=61
BC=7.8
In Parallelogram the parallel sides are equal;
therefore BC=AD
In triangle ABP ,A/c to pythagorus theorm
BP^2=AP^2+AB^2 (AP=AD/2,AP=7.8/2),,,,,Given BP=12
12^2=(7.8/2)^2+AB^2
AB^2=144-15
AB^2=129
AB=11.3
Area of parallelogram = base*height
=BC*AB
=7.8*11.3
Area of parallelogram=88.14 cm^2
Hope it helps!!!!
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