Question

The length of a rectangle exceeds its breadth by 4cm. If the length is increased by 3cm and the breadth is increased by 2cm, the new area exceeds the original area by 79 sq cm. Find the dimensions of the given rectangle.​

Answer 1

Given:

➠ Let the breadth of the rectangle be x

➠ then, the length = (x+4)

Solution:

⟼ Area of the rectangle = l×b

⟼ Area = (x+4)(x) = x²+4x

If

⟼ length = (x+4)+3 = x+7

⟼ breadth = x+2

then,

⟶ Area of the rectangle = (x+2)(x+7)

⟼ Area = x²+9x+14

which is also equal to:

⟼ x²+9x+14 = x²+4x+79

⟼ x²-x²+9x-4x = 79-14

⟼ 5x = 65

⟼ x = 65/5

∴ x = 13

so,

⟼ length = (x+4) = 13+4 = 17cm

⟼ breadth = x = 13 cm

⟼ Area of the rectangle = x²+4x

= 169+52 = 221 cm²

Answer 2

Answer:

The length is 17 cm and breadth is 13 cm.

Step-by-step explanation:

Length exceeds breadth by = 4 cm

Length when increased by 3 cm and breadth increased by 3 cm, the new area exceeds the original area by = 79 cm².

Dimensions of the rectangle = ??

Let the length be x and breadth be y.

Length exceeds the breadth by 4 cm.

$\longrightarrow$ x - y = 4

$\longrightarrow$ x = 4 + y ---- (Equation I)

Area of rectangle = Length × Breadth

Original area =

$\longrightarrow$ x × y

$\longrightarrow$ xy

According to the Question,

• Length = x + 3
• Breadth = y + 2

$\longrightarrow$ xy + 79 = (x + 3)(y + 2)

$\longrightarrow$ xy + 79 = xy + 2x + 3y + 6

$\longrightarrow$ 79 - 6 = 2x + 3y

$\longrightarrow$ 73 = 2x + 3y ----- (Equation II)

By elimination method,

Eq. I → 2x + 3y - 73 = 0

Eq. II → x - y - 4 = 0

Equation II is rewritten as 2x - 2y - 8 = 0.

Subtract Equation I from Equation II.

$\longrightarrow$ 2x - 2y - 8 - (2x + 3y - 73) = 0

$\longrightarrow$ 2x - 2y - 8 - 2x - 3y + 73 = 0

$\longrightarrow$ -5y + 65 = 0

$\longrightarrow$ -5y = -65

$\longrightarrow$ -y = -65/5

$\longrightarrow$ y = 13 cm

Breadth = 13 cm

Place the value of y in equation I,

$\longrightarrow$ x = 4 + 13

$\longrightarrow$ x = 17 cm

Length = 17 cm

Therefore, the length is 17 cm and breadth is 13 cm.