Question

the length of a rectangular field in increased by 50% and its with is reduced by 50% we get a new rectangular field find the area of the new ground​

Answer 1

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The area of the new ground is $\sf{\dfrac{3xy}{4}}$

Change in area is 25%

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GiVeN :

• The length of a rectangular field is increased by 50%
• The width is reduced by 50%

To FiNd :

• Area of the new ground.

SoLuTioN :

Let the length of the rectangle be x cm.

Let the breadth of the rectangle be y cm.

We have the formula for area of the rectangle as follows :

$\bold{\large{\boxed{\sf{\red{Area\:of\:a\:rectangle\:=\:length\:\times\:breadth}}}}}$

Block in the values,

$\leadsto$ $\sf{x\:\times\:y\:}$

$\sf{\therefore{Area\:=\:xy\:--->\:(1)}}$

Now, the length is increased by 50% and the width is reduced by 50% a new rectangle is formed.

Let's find the length and breadth of this new rectangle.

Length :

$\leadsto$ $\sf{x\:+\:{\dfrac{50}{100}\:\times\:x}}$

$\leadsto$ $\sf{x\:+\:{\dfrac{50x}{100}}}$

$\leadsto$ $\sf{x\:+\:{\dfrac{1x}{2}}}$

$\leadsto$ $\sf{x\:+\:{\dfrac{x}{2}}}$

$\leadsto$ $\sf{\dfrac{2x\:+\:x}{x}}$

$\leadsto$ $\sf{\dfrac{3x}{2}}$ ----> 2

Breadth :

$\leadsto$ $\sf{y\:-\:{\dfrac{50}{100}\:\times\:y}}$

$\leadsto$ $\sf{y\:-\:{\dfrac{50y}{100}}}$

$\leadsto$ $\sf{y\:-\:{\dfrac{1y}{2}}}$

$\leadsto$ $\sf{y\:-\:{\dfrac{y}{2}}}$

$\leadsto$ $\sf{\dfrac{2y\:-\:y}{2}}$

$\leadsto$ $\sf{\dfrac{y}{2}}$ ----> 3

Now, we have the length and breadth of the new rectangle formed. Plug in the values in the formula for area of the rectangle to calculate the the area of the ground formed.

Let's begin!

$\leadsto$ $\sf{\dfrac{3x}{2}\:\times\:{\dfrac{y}{2}}}$

$\leadsto$ $\sf{\dfrac{3xy}{4}}$ ---> 4

Now the change in area, will be the difference between the initial area and the area of the new ground formed.

Change in area :

$\leadsto$ $\sf{xy\:-\:{\dfrac{3xy}{4}}}$

$\leadsto$ $\sf{\dfrac{4xy\:-\:3xy}{4}}$

$\leadsto$ $\sf{\dfrac{xy}{4}}$ ----> 5

Now finally, the area of the new ground will be :

$\leadsto$ $\sf{\frac{\frac{xy}{4}\:\times\:100}{\frac{xy}{1}}}$

$\leadsto$ $\sf{\frac{\frac{100xy}{4}}{\frac{xy}{1}}}$

$\leadsto$ $\sf{\frac{100xy}{4}}$ ÷ $\sf{\frac{xy}{1}}$

$\leadsto$ $\sf{\dfrac{100xy}{4}}$ $\times$ $\sf{\dfrac{1}{xy}}$

$\leadsto$ $\sf{\dfrac{100}{4}}$

$\leadsto$ $\sf{25}$

The change in the area of the ground is 25%

Answer 2

Answer:

The change in area of the ground is 25%