Question

The breadth of a rectangle is 3 less than the length. If both length and breadth are reduced by
3 units, the area of the rectangle reduces by 90 sq. units. Find the dimensions of the original
rectangle.
[HOT
The length of a rectangle is 3 more than twice the breadth of the rectangle. If the length is reduced​

Answer 1

Answer:

Answer is 2+7+8+9+0+× 0

Step-by-step explanation:

2+7 = 11

11+8 = 190

190+9 = 20

20+0 = 0

0 × 0 = 000000

Answer 2

Answer:

• Length and breadth of the rectangle are 18 units and 15 units respectively.

Step-by-step explanation:

Given that:

• The breadth of a rectangle is 3 less than the length of the rectangle.
• If both length and breadth are reduced by 3 units, the area of the rectangle reduces by 90 sq. units.

To Find:

• Dimensions of the original rectangle.

Let us assume:

• Length of the rectangle be x.

Then,

• Breadth of the rectangle is (x - 3).

Now,

As we know that:

• Area of rectangle is (l × b) sq. units.

Substituting the values,

Area of the rectangle = x × (x - 3)

= x² - 3x

If both length and breadth are reduced by 3 units:

• Length = x - 3
• Breadth = x - 3 - 3 = x - 6

Now,

Area of rectangle with reduced dimensions = (x - 3) × (x - 6)

= x(x - 6) - 3(x - 6)

= x² - 6x - 3x + 18

= x² - 9x + 18

According to the question:

$\sf{\longrightarrow(x^2-3x)-(x^2-9x+18)=90}$

Opening the brackets,

$\sf{\longrightarrow x^2-3x-x^2+9x-18=90}$

Arranging the numbers,

$\sf{\longrightarrow x^2-x^2-3x+9x-18=90}$

Solving further,

$\sf{\longrightarrow 6x-18=90}$

Transposing 18 from LHS to RHS and changing its sign,

$\sf{\longrightarrow 6x=90+18}$

Adding the numbers,

$\sf{\longrightarrow 6x=108}$

Transposing 6 from LHS to RHS and changing its sign,

$\sf{\longrightarrow x=\dfrac{108}{6}}$

Dividing the numbers,

$\sf{\longrightarrow x=18}$

Hence, x = 18.

Therefore,

• Length of the rectangle = x = 18 units
• Breadth of the rectangle = (x - 3) = (18 - 3) = 15 units