Question

The length of a rectangle is 4 ya more than twice the width x. The area is 510 yd^2. Find the dimensions of the given shape

Answer 1

Answer:

The dimensions of the rectangle are

length = 34 yards

width = 15 yards

Step-by-step explanation:

Given:

The shape is of rectangle,

Width of rectangle = x

Area of rectangle = 510 yd²

According to the given condition

Length of rectangle = 4 + 2x

To Find:

length =?

Width = ?

Solution:

We know that,

$\textrm{Area of rectangle} = Length\times Width$

On substituting the given values we get

$510 = (4 + 2x)\times x\\510 = 4x+2x^{2}\ \textrm{using distributive property}\\\textrm{dividing throughout by two}\\255=2x+x^{2} \\\\x^{2} +2x -255 = 0$

Which is a quadratic equation so we will apply splitting the middle term that is factorization we get

∴ $x^{2}+ 17x -15x -255=0\\x(x+17)-15(x+17)=0\\(x-15)(x+17)=0\\\therefore x-15 =0\ or\ x+17=0\\\\\textrm{x cannot be negative i.e -17}\\\therefore x=15$

∴ Width  x = 15 yard

∴ Length = 4 + 2x = 4 +2 × 15 = 34 yard

The dimensions of the rectangle are

length = 34 yards

width = 15 yards