Question

The area of a rectangle whose length is five more than twice its width is 75 cm2. Calculate its
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+5x2x2+5x

=> 2x2+5x−75=02x2+5x−75=0

=> 2x2+15x−10x−75=02x2+15x−10x−75=0

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

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

=> x = 5 and −152−152

Width cannot be negative.

Width = 5 units

Length=2x+5 =2×5+52×5+5=15 units

Perimeter of the rectangle

= 2(15 + 5) = 40 units

Answer 2

★ Answer :

Let the breadth of the rectangle is x cm.

So, Length of rectangle be will be (2x + 5) cm.

We know that,

$\Large{\star{\boxed{\sf{Area = Length \times Breadth}}}}$

__________[Putting Values]

→ 75 = (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

→ 2x + 15 = 0

→ 2x = -15

→ x = -15/2

OR

→ x - 5 = 0

→ x = 5

As, length and breadth can't be negative. So, x = 5.

Length = 2x + 5 = 2(5) + 5 = 10 + 5 = 15 cm

Breadth = x = 5 cm