Question

6 in 10.9 in 6 in 5.1 in 18 in P= 40.9

Answer 1

The perimeter of the given trapezium is the sum of the side lengths of the trapezium.

Hence, the perimeter P is,

`P=18+10.9+6+6=40.9\text{ in}`

Therefore, the perimeter P of the given trapezium is 40.9 in.

Given, the altitude of the trapezium, h=5.1 in.

The length of parallel sides,

a=6 in.

b=18 in.

The area A of the trapezium is,

`\begin{gathered} A=(1)/(2)* sum\text{ of parallel sides}* altitude \\ =(1)/(2)*(a+b)* h \\ =(1)/(2)*(6+18)*5.1 \\ =61.2\text{ sq.in.} \end{gathered}`

Therefore, area A of the trapezium is 61.2 sq.in.