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Find the area of a quadrilateral A B C...

Find the area of a quadrilateral `A B C D` is which `A B=3c m ,\ B C=4c m ,\ C D=4c m ,\ D A=5c m\ a n d\ A C=5c m`
`S=1/2(a+b+c)=1/2(3+4+5)=6`
`ar \of\ ABC`=sqrt(s(s-a)(s-b)(s-c)`
`ar ABC=`sqrt(6(6-3)(6-4)(6-5))`
`=>sqrt(6xx3xx2xx1)=6cm^2`
Now in `/_\ADC,S=(a+b+c)/2`
`=>s = (5 + 4 + 5)/2=14/2=7cm`
By using Heron’s formula
Area of `/_\ADC=sqrt(s(s - a)(s - b)(s - c))`
`=sqrt(7(7 - 5)(7 - 4)(7 - 5))`
`=sqrt(7 × 2 × 3 × 2)`
`= 2sqrt21 cm^2`
Area of`/_\ADC=9.2 cm^2` (approx.)
Area of the quadrilateral ABCD = Area of `/_\ADC + Area of /_\ABC= 9.2cm^2+ 6cm^2= 15.2 cm^2`
Thus, the area of the quadrilateral,` ABCD is 15.2 cm^2`.

Text Solution

Verified by Experts

`S=1/2(a+b+c)=1/2(3+4+5)=6`
`ar \of\ ABC`=sqrt(s(s-a)(s-b)(s-c)`
`ar ABC=`sqrt(6(6-3)(6-4)(6-5))`
`=>sqrt(6xx3xx2xx1)=6cm^2`
Now in `/_\ADC,S=(a+b+c)/2`
`=>s = (5 + 4 + 5)/2=14/2=7cm`
By using Heron’s formula
Area of `/_\ADC=sqrt(s(s - a)(s - b)(s - c))`
`=sqrt(7(7 - 5)(7 - 4)(7 - 5))`
`=sqrt(7 × 2 × 3 × 2)`
`= 2sqrt21 cm^2`
Area of`/_\ADC=9.2 cm^2` (approx.)
Area of the quadrilateral ABCD = Area of `/_\ADC + Area of /_\ABC= 9.2cm^2+ 6cm^2= 15.2 cm^2`
Thus, the area of the quadrilateral,` ABCD is 15.2 cm^2`.
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