Question

length of a rectangle is 6cm more than its breadth. areas is 1216sq.cm find its length​

Answer 1

Given :-

Length of a rectangle is 6cm more than its breadth and its areas is 1216sq.cm

To find :-

Length of rectangle

Solution :-

• Area of rectangle = 1216 cm²

As we know that

→ Area of rectangle = l × b

Where " l " is length and " b " is breadth.

• According to the given condition

Let the breadth be x then length be x + 6

→ Area of rectangle = 1216 cm²

→ l × b = 1216

→ (x + 6)x = 1216

→ x² + 6x = 1216

→ x² + 6x - 1216 = 0

Apply quadratic formula

• a = 1
• b = 6
• c = - 1216

→ D = b² - 4ac

→ D = (6)² - 4 × 1 × (-1216)

→ D = 36 + 4864

→ D = 4900

Now,

→ x = - b ± √D/2a

→ x = - 6 ± √4900/2 × 1

→ x = - 6 ± 70/2

Either

→ x = - 6 + 70/2

→ x = 64/2

→ x = 32

Or

→ x = - 6 - 70/2

→ x = - 76/2

→ x = - 38

• Length or breadth never in negative

Hence,

• Length of rectangle = x + 6 = 32 + 6 = 38 cm

• Breadth of rectangle = x = 32 cm

Answer 2

Answer:

Question :-

• length of a rectangle is 6cm more than its breadth. areas is 1216sq.cm. find its length

Answer :-

• Length = 32 + 6 = 38 cm
• Length = 32 + 6 = 38 cm Breadth = 32 cm

Given:-

• Length = x+6 cm
• Breadth = x cm
• Area = 1216sq.cm

Solution:-

$\boxed{ \large{area = length \times breadth}}$

$\implies \: (x + 6) \times x = 1216 \\ \\ \implies {x}^{2} + 6x = 1216 \\ \\ \implies {x}^{2} + 6x - 1216 = 0$

By splitting the middle term

$\implies \: {x}^{2} + 38x - 32x - 1216 = 0 \\ \\ \implies \: x(x + 38) - 32(x + 38) = 0 \\ \\ \implies(x + 38) \: (x - 32) = 0$

Either,

$\implies \: x + 38 = 0 \\ \\ \\ \implies \: x = - 38$

Or,

$\implies \: x - 32 = 0 \\ \\ \implies \: x =32$

Length or breadth can't be negative

• Length = 32 + 6 = 38 cm
• Breadth = 32 cm