The sides of a quadrangular field, taken in order are `26 m ,\ 27 m, 7 m\ a n d\ 24 m` respectively. The angle contained by the last two sides is a right angle. Find its area.

Explanation

The sides of a quadrangular field, taken in order are `26 m ,\ 27 m, 7 m\ a n d\ 24 m` respectively.
Diagonal AC is joined
Now`/_\ADC,AC^2=AD^2+CD^2`
`AC^2=24^2+7^2`
`AC^2=625=25m`
Now area of`/_\ABC,S=1/2(AB+BC+CA)`
`=>1/2(26+27+25)=78/2=39m`
Area of`/_\ABC=sqrt(39(39−26)(39−27)(39−25))`
`=sqrt39xx13xx12xx14`
`=>sqrt85179=291.85m^2`
Now area of `/_\ADC,S=1/2(AD+CD+AC)`
`=>1/2(25+24+7)=56/2=28 m`
Area of`/_\ADC=sqrt(S(S−AD)(S−DC)(S−CA))`(heron's formula)
`=> sqrt(28(28−24)(28−7)(28−25))`
`=>sqrt(28xx4xx21xx3)=sqrt7056=84 m^2`
hence, total area is,`375.85 m^2`