Chapter ISG, Problem 8.2.7P

To determine

To write and simplify expression

Answer to Problem 8.2.7P

  $28s^2+3s$ cubic unit

Explanation of Solution

Given information:

An enclosed area.

Formula used:

Area of rectangle= $length\times breadth$

Calculation:

To find the area, divide the area into sub-areas and add the areas of the figure of together.

  

Solving rectangle with length 6s - 1 and breadth s

Substituting values in formula,

Area= $(6s-1)(s)$

Multiplying,

Area= $6s^2-s$

(1)

Again, solving area with length 5s and breadth 4s (breadth= 4s because 6s -1-( 2s -1) = 4)

Substituting values in formula,

Area= $(5s)(4s)$

Solving,

Area= $(5)(4)s^2$

So,

Area= $20s^2$

(2)

Now, solving area with length 2s and breadth s +2

Substituting values in formula,

Area= $2s(s+2)$

Solving,

Area= $2s(s)+2s(2)$

So,

Area= $2s^2+4s$

(3)

Adding (1), (2) and (3)

Area= $6s^2-s$ + $20s^2$ + $2s^2+4s$

Adding like terms,

Area= $(2s^2+6s^2+20s^2)+(4s-s)$

Hence, total area of the shape is $28s^2+3s$ cubic unit.