Question

The area of rectangle whose length is five more than twice of breadth is 75 sq.cm. Find the length​

Answer 1

Answer:

Let the width of the rectangle = x units

Length = (2 x + 5) units

According to the question,

Area = x(2x + 5)

=> 75 = 2x2+5x

=> 2x2+5x−75=0

=> 2x2+15x−10x−75=0

=> x(2x+15)-5(2x+15)=0

=> (2x+15)(x-5)=0

=> x = 5 and −152

Width cannot be negative.

Width = 5 units

Length=2x+5 =2×5+5=15 units

Perimeter of the rectangle

= 2(15 + 5) = 40 units

Answer 2

ANSWER:

• Length of rectangle = 15 cm
• Breadth of rectangle = 5 cm

GIVEN:

• Area of rectangle = 75 cm²
• Length is five more than twice of breadth.

TO FIND:

• Length of rectangle.

SOLUTION:

Let the breadth of rectangle be x

Length of rectangle = 2x+5

Formula:

Area of rectangle = Length*Breadth.

=> 75 cm² = (2x+5)(x)

=> 2x²+5x = 75 cm²

=> 2x²+5x-75 = 0

=> 2x²+15x-10x-75 = 0

=> (2x²-10x)+(15x-75) = 0

=> 2x(x-5) +15(x-5) = 0

=> (x-5)(2x+15) = 0

Either (2x+15) = 0

=> 2x = -15

=> x = -15/2

This is not possible as breadth can't be negative.

Either (x-5) = 0

=> x = 5

Breadth = 5

Length = 2x+5

= 2(5)+5

= 15

Length of rectangle = 15 cm

Breadth of rectangle = 5 cm