Question

Given a rectangle of length $a$ and breadth $b$:


In another rectangle, the length is $2$ units more than the length and breadth is $3$ units less than the breadth of the given rectangle.
Represent the area of another rectangle as an algebraic expression.

• A. $ab-3a-2b+6$
• B. $ab-3a+2b-6$
• C. $ab$
• D. $ab+2b-6$

Answer 1

The correct option is B $ab-3a+2b-6$

We are given with a recatngle as shown:


Length of another rectangle is $2$ units more than the length of the given rectangle.
Length of another rectangle $=a+2$

Breadth of another rectangle is $3$ units less than the breadth of the given rectangle.
Breadth of another rectangle $=b-3$

Another rectangle can be shown as:



Area of another rectangle
$=$ Length $\times$ Breadth
$=(a+2)\times (b-3)$

This area can be evaluated using distributive property as:
$(a+2)\times (b-3)$
$=a(b-3)+2(b-3)$
$=(a\times b)-(a\times 3)$
$+(2\times b)-(2\times 3)$
$=ab-3a+2b-6$

The algebraic expresion of area of other rectangle is $ab-3a+2b-6$. Therefore, option (b.) is the correct answer.