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Question

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


Solution

In the quadrilateral, AC is the diagonal which divides the figure into two triangles 
Now in Δ ABC, AB = 3 cm, BC = 4cm, AC = 5cm 

s=a+b+c2=3+4+52=122=6


 Area ofΔABC=s(sa)(sb)(sc)


v=6(63)(64)(65)


=6×3×2×1=36=6cm2


In Δ ADC, AC= 5 cm, AD= 5cm, CD=4cm


s=a+b+c2=5+5+42=142=7


Area of ΔADC=s(sa)(sb)(sc)


=7(75)(75)(74)=7×2×2×3


=221=2×4.58=9.16cm2


Total area of quadrilateral ABCD


 = 6+9.16cm2=15.16cm2


Mathematics
RD Sharma
Standard IX

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