find area of a quadrilateral in which AB=7cm BE=6cm cd=12cm da=15cm and ac=9cm
Answers
Answered by
2
Answer:
area of quadrilateral ABCD= Area of triangle ABC + Area of triangle ADC
Area of trianlge= sqrt(s(s-a)(s-b)(s-c))
FOR area of triangle ABC
s=(7+6+9)/2=11
area of triangle ABC=sqrt(11(11-7)(11-6)(11-9))
=sqrt(440)
= 20.9761
similarly,
area of triangle ADC=54
therefore area of quadrilateral ABCD=20.9761+ 54
=74.9761
Answered by
3
ar(ABCD) = ar(ADC) + ar(ABC)
Now,
In ∆ADC,
s = 15+12+9/2 = 18cm
by Heron's Formula,
area = √s(s-a)(s-b)(s-c)
= √18(18-15)(18-12)(18-9)
= √3×3×2×3×3×2×3×3
= 3×2×3×3
= 54cm² _____(i)
In ∆ABC,
s = 7+6+9/2 = 11cm
by Heron's Formula,
area = √s(s-a)(s-b)(s-c)
= √11(11-7)(11-6)(11-9)
= √11×2×2×5×2
= 2√11×2×5
= 2√110
= 2 × 10.49
= 20.98cm² _____(ii)
Now,
Area of quadrilateral = (i) + (ii)
= 54 + 20.98
= 74.98cm²
Now,
In ∆ADC,
s = 15+12+9/2 = 18cm
by Heron's Formula,
area = √s(s-a)(s-b)(s-c)
= √18(18-15)(18-12)(18-9)
= √3×3×2×3×3×2×3×3
= 3×2×3×3
= 54cm² _____(i)
In ∆ABC,
s = 7+6+9/2 = 11cm
by Heron's Formula,
area = √s(s-a)(s-b)(s-c)
= √11(11-7)(11-6)(11-9)
= √11×2×2×5×2
= 2√11×2×5
= 2√110
= 2 × 10.49
= 20.98cm² _____(ii)
Now,
Area of quadrilateral = (i) + (ii)
= 54 + 20.98
= 74.98cm²
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