Question

A rectangular field has an area of `1,764` `m^2`. The width of the field is `13` `m` more than the length. What is the perimeter of the field?

Answer 1

`170` m

Explanation:

If a rectangle has width `W` and length `L`, then the area `A = L*W` and the perimeter `P =2*(L +W)`

In this case `W = L +13` and `A = 1764`

Plugging this expression for W into the formula for area gives us
`1764 = L*(L+13)`
`L^2 +13L - 1764 = 0`

It is hard to establish the factors of a big number like `1764` visually so we use the quadratic formula
`L = (-13 +- sqrt(169 +7056))/2 = (-13+-sqrt(7225))/2 = (-13+-85)/2`
We can discount the negative root as the length of a real field cannot be negative. So `L = (-13+85)/2 =72/2 = 36`

`L = 36` give `W = 36 +13 = 49`

Therefore the perimeter `P =2*(36+49) =2*85 = 170` m

Answer 2

`170` `m`

Explanation:

Let the length of the rectangle be `l`, then the width will equal `13+l`

`color(blue)("Area of a rectangle"=l*w`

Where, `landw` are the length and width of the rectangle

So,

`rarr8*(13+l)=1764`

Use the distributive property `color(purple)(a(b+c)=ab+ac`

`rarr13l+l^2=1764`

`rarr13l+l^2-1764=0`

Write it in standard form

`rarrl^2+13l-1764=0`

Now, this is a quadratic equation. We solve it using the quadratic formula

`color(violet)(l=(-b+-sqrt(b^2-4ac))/(2a)`

Where `a,b and c` are the coefficients of the terms

Then,

`rarrl=(-13+-sqrt(13^2-4(1)(1764)))/(2(1))`

Solving this, we get

`rarrl=(-13+-85)/2`

`rarrl=((-13+85)/2,(-13-85)/2)`

`rarrl=(36,-49)`

As, length cannot be negative

The length of the rectangle is `36` and the width is `49`

We need to find the perimeter

`color(blue)("Perimeter of rectangle "=2(l+b)`

`rarr2(36+49)`

`color(green)(rArr170` `color(green)(m`