Area of PQRS = Area of PQR + Area of ΔPQS = (6+9.166)๐๐2=15.166๐๐2
The sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m are 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.
Solution
The sides of a quadrilateral field taken order as AB = 26m
BC = 27 m
CD = 7m and DA = 24 m
Diagonal AC is joined
Now ΔADC
By applying Pythagoras theorem
⇒`AC^2=AD^2+CD^2`
⇒`AC=sqrt(AD^2 +CD^2)`
⇒`AC=sqrt(24^2+7^2)`
⇒`AC=sqrt625=25 m`
Now area of ΔABC
`S=1/2(AB+BC+CA)=1/2(26+27+25)=78/2=39 m`
By using heron’s formula
Area (ΔABC) = `sqrt(S(S-AD)(S_BC)(S-CA))`
`=sqrt(39(39-26)(39-21)(39-25))`
`sqrt(39xx14xx13xx12xx1)`
`=291.849cm^2`
Now for area of ΔADC
`S=1/2(AD+CD+AC)`
`=1/2(25+24+7)=28cm`
By using heron’s formula
`∴ Area of ΔADC = sqrt(S(S-AD)(S-DC)(S-CA))`
`=sqrt(28(28-24)(28-7)(28-25))`
`=84m^2`
∴ Area of rectangular field ABCD = area of ΔABC + area of ΔADC
= `291.849+84`
= `375.8m^2`
