Find the area of the trapezium `PQRS` with height `PQ` given in the figure given below

Explanation

We have, trapezium `PQRS,` in which draw a line `RT` perpendicular to `PS.`
`" where, side"" "ST=PS-TP=12-7=5m. " "[because TP=PQ=7m]`

`"In right angled "triangleSTR," "(SR)^(2)=(ST)^(2)+(TR)^(2)" "["by using Paythagoras theorem"]`
`rArr" "(13)^(2)=(5)^(2)+(TR)^(2)`
`rArr" "(TR)^(2)=169-25`
`rArr" "(TR)^(2)=144`
`therefore" "TR=12m`
`["taking positive square root because length is alyas positive"]`
`"Now"" ""area of "triangleSTR=(1)/(2)xxTRxxTS" "[because"area of a rectangle=(base"xx"height)"]`
`=(1)/(2)xx12xx5=30m^(2)`
`"Now, area of rectangle "PQRT=PQxxRQ=12xx7" "[because"area of a rectangle=length"xx"breadth"]" "[becausePQ=TR=12m]`
`=84m^(2)`
`therefore" ""Area of trapezium=Area of DSTR+Area of rectangle PQRT"`
`=30+84=114m^(2)`
Hence, the area of trapezium is 114 `m^(2)`
Alternate Method
Find TR as in above method
`therefore" ""Area of trapezium"=(1)/(2)("Sum of parallel lines")xx"Distance between two points"`
`=(1)/(2)(PS+QR)xxTR=(1)/(2)xx(12+7)xx12" "["From Eq.(i)"]`
`=(1)/(2)xx19xx12=144m^(2)`
`"Hence, the area of trapezium is" 114m^(2)`.