Question

The area of a rectangle gets reduced by 9 square units, if its length is reduced by
5 units and breadth is increased by 3 units. If we increase the length by 3 units and
the breadth by 2 units, the area increases by 67 square units.

Q.Find the dimensions
of the rectangle?
​

Answer 1

Answer:

$\large{\underline{\boxed{\sf \star \: Length = 17 \: units}}}$

$\large{\underline{\boxed{\sf \star \: Breadth = 9\: units}}}$

Step-by-step explanation:

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

Area of Rectangle = Length * Breadth

️ = (xy) unit²

Given that-

If the length is reduced by 5 units and breadth is increased by 3 units, area of a rectangle gets reduced by 9 square units.

New Length = (x - 5) units

New Breadth = (y + 3) units

Decreased area = (xy - 9) sq units.

⇒ New (l * b) = xy - 9

⇒ (x - 5)(y + 3) = xy - 9

⇒ xy + 3x - 5y - 15 = xy - 9

⇒ xy + 3x - 5y - 15 - xy + 9 = 0

⇒ 3x - 5y - 6 = 0 .... (i)

Again,

If we increase the length by 3 units and

the breadth by 2 units, the area increases by 67 square units.

New length = (x + 3) units

New breadth = (y + 2) units

New Area = (xy + 67) sq units

⇒ (x + 3)(y + 2) = xy + 67

⇒ xy + 2x + 3y + 6 = xy + 67

⇒ xy + 2x + 3y + 6 - xy - 67 = 0

⇒ 2x + 3y - 61 = 0 ... (ii)

We have got, two equations -

$\sf 3x - 5y = 6 \: \: ....(i)\\ \sf 2x + 3y = 61 \: \: ....(ii)$

Multiplying equation (i) by 2 and (ii) by 3, we get -

$\sf 6x - 10y = 12 \: \:...(iii) \\ \sf 6x + 9y =18 3 \: \: ....(iv)$

Subtracting equation (iii) from (iv) -

$\sf 6x + 9y - (6x - 10y) = 183 - 12 \\ \\ \sf 6x + 9y - 6x + 10y = 171 \\ \\ \sf 19y = 171 \\ \\ \sf y = \frac{171}{19} = 9$

Substituting the value of y in (i) -

$\sf 3x - 5y = 6 \\ \\ \sf 3x - 5 \times 9 = 6 \\ \\ \sf 3x - 45 = 6 \\ \\ \sf 3x =6 + 45 \\ \\ \sf 3x = 51 \\ \\ \sf x = \frac{51}{3} = 17$

Hence, the length of the rectangle is 17 units and breadth is 9 units.

Answer 2

Answer :-

$\boxed{\textbf{Length = 17 units}}$

$\boxed{\textbf{Breadth = 9 units}}$

Explanation :-

Let the Length of the rectangle be x and breadth be y

$\boxed{\sf{Area\:of\:Rectangle=Length \times Breadth}}$

Original Area of rectangle = x * y = xy units²

Decreased area:-

Length of the decreased area of the rectangle = Original length is reduced by 5 units = (x - 5) units

Breadth of the decreased area of the rectangle = Original breadth is increased by 3 units = (y + 3)

Decreased area of the rectangle = Original area of the rectangle is reduced by 9 units² = (xy - 9) units²

$\boxed{\sf{Area\:of\:Rectangle=Length \times Breadth}}$

$\sf{\implies{(x - 5)(y + 3) = xy - 9}}$

$\sf{\implies{x(y + 3) - 5(y + 3) = xy - 9}}$

$\sf{\implies{xy + 3x - 5y - 15 = xy - 9}}$

$\sf{\implies{3x - 5y - 15 + 9 = 0}}$

$\sf{\implies{3x - 5y -6 = 0}}$

$\sf{\implies{3x - 5y = 6..(1)}}$

Increased Area:-

Length of the increased area of the rectangle = Original length is increased by 3 units = (x + 3) units

Breadth of the increased area of the rectangle = Original breadth is increased by 2 units = (y + 2) units

Increased area of the rectangle = Original area of the rectangle is increased by 67 units² = (xy + 67) units²

$\boxed{\sf{Area\:of\:Rectangle=Length \times Breadth}}$

$\sf{\implies{(x + 3)(y + 2) = xy + 67}}$

$\sf{\implies{x(y + 2) + 3(y + 2) = xy + 67}}$

$\sf{\implies{xy + 2x + 3y + 6= xy + 67}}$

$\sf{\implies{2x + 3y + 6 - 67 = 0}}$

$\sf{\implies{2x + 3y -61= 0}}$

$\sf{\implies{2x + 3y = 61..(2)}}$

Multiplying (1) by 2 (2) by by 3 we have ,

$\sf{\implies{3x(2) - 5y(2) = 6(2)}}$

$\sf{\implies{6x - 10y = 12...(3)}}$

$\sf{\implies{2x(3) + 3y(3) = 61(3)}}$

$\sf{\implies{6x + 9y =183..(4)}}$

Now subtract eq(3) from eq(4)

$\sf{\implies{6x + 9y - (6x - 10y) = 183 - 12}}$

$\sf{\implies{6x + 9y - 6x + 10y = 171}}$

$\sf{\implies{19y = 171}}$

$\sf{\implies{y = \dfrac{171}{9} }}$

$\sf{\implies{y = 9}}$

Substitute y = 9 in eq(1) to get the value of x

$\sf{\implies{3x -5y = 6}}$

$\sf{\implies{3x -5(9)= 6}}$

$\sf{\implies{3x -45 = 6}}$

$\sf{\implies{3x = 6 + 45}}$

$\sf{\implies{3x = 51}}$

$\sf{\implies{x = \dfrac{51}{3} }}$

$\sf{\implies{x = 17}}$

$\boxed{\textbf{Length = 17 units}}$

$\boxed{\textbf{Breadth = 9 units}}$