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Question

Find the area of a quadrilateral ABCD in which AB=3 cm,BC=4 cm,CD=4 cm,DA=5 cm and AC=5 cm.

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Solution

Area of ABCD=Area of ΔADC+ Area of ΔABC
Area of ΔABC
=s(sa)(sb)(sc)
=a=3,b=4,c=5
s=a+b+c2=3+4+52=122=6cm
=6(63)(64)(65)
=6×3×2×1
=6×6
=6cm2
Area of ΔADC=s(sa)(sb)(sc)
a=5,b=4,c=5
s=5+4+52=142=7cm
=7(75)(74)(755)
=7×2×3×2
=2×2×3×7
=2×21
=2×4.58
=9.16
Area of ABCD=(6+9.16)cm2
=15.16cm2.

1184526_1242018_ans_9b30be8fec59452ea6695ac032ddce30.png

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