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Home » If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1unit and the breadth increased by 2units, the area increases by 33square units. Find the area of the rectangle.
If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1unit and the breadth increased by 2units, the area increases by 33square units. Find the area of the rectangle.
CBSE | Class 10 | Exercise 3.11 | Maths | Maths | Pair of Linear Equations in Two Variables | Pair of Linear Equations In Two Variables | RD Sharma
If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1unit and the breadth increased by 2units, the area increases by 33square units. Find the area of the rectangle.

Solution:

Let the length and breadth of the rectangle be x units and y units respectively.

The area of rectangle = xy sq.units

Case 1:

Length is increased by 2 units

⇒ The new length is

    \[x+2\]

units

Breadth is reduced by 2 units

 ⇒ The new breadth is

    \[y-2\]

units

Given in the question the area is reduced by

    \[28\]

sq.units. =

    \[xy\text{ }\text-{ }28\]

Calculation:

    \[~\left( x+2 \right)\left( y-2 \right)\text{ }=\text{ }xy\text{ }-\text{ }28\]

    \[\Rightarrow xy\text{ }-\text{ }2x\text{ }+\text{ }2y\text{ }\text{ }- 4\text{ }=\text{ }xy\text{ }-\text{ }28\]

    \[\Rightarrow -2x\text{ }+\text{ }2y\text{ }\text{ }4\text{ }+\text{ }28\text{ }=\text{ }0\]

    \[\Rightarrow -2x\text{ }+\text{ }2y\text{ }+\text{ }24\text{ }=\text{ }0\]

    \[\Rightarrow 2x\text{ }-\text{ }2y\text{ }\text{ }- 24\text{ }=\text{ }0\text{ }\ldots \ldots \ldots \text{ }(i)\]

Length is reduced by 1 unit

⇒ the new length is

    \[x-1\]

units

Breadth is increased by 2 units

 ⇒ The new breadth is

    \[y+2\]

units

And, it’s given that the area is increased by 33 sq.units =

    \[xy\text{ }+\text{ }33\]

    \[\left( x-1 \right)\left( y+2 \right)\text{ }=\text{ }xy\text{ }+\text{ }33\]

    \[\Rightarrow xy\text{ }+\text{ }2x\text{ }\text{ }- y\text{ }\text{ }- 2\text{ }=\text{ }x\text{ }+\text{ }33\]

    \[\Rightarrow 2x\text{ }\text{ }- y\text{ }-\text{ }2\text{ }-\text{ }33\text{ }=\text{ }0\]

    \[\Rightarrow 2x\text{ }\text{ }- y\text{ }-35\text{ }=\text{ }0\text{ }\ldots \ldots \ldots ..\text{ }\left( ii \right)\]

By cross multiplication method (i) & (ii)

R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.11 - 1

x=\frac{46}{2}

x=23

And,

y=\frac{22}{2}

    \[y\text{ }=\text{ }11\]

The length of the rectangle =

    \[23\]

units.

The breadth of the rectangle is =

    \[11\]

units.

The area of the rectangle = length x breadth,

= x×y

=

    \[23\times 11\]

=

    \[253\]

sq. units

\therefore The area of rectangle is

    \[253\]

sq. units.

i

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