# Find the dimension of a rectangle A rectangle is 10ft longer than it's width. The width of the surrounding deck is 3inches on each side. If the area is 1824sq.ft, find the dimension of the outside...

Find the dimension of a rectangle

A rectangle is 10ft longer than it's width. The width of the surrounding deck is 3inches on each side. If the area is 1824sq.ft, find the dimension of the outside edge of the deck.

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We are given that the length of rectangle is 10ft more that the width.

So let us say that width of rectangle = x

Therefore length of lectangle = x+10

Given the area of rectangle=1824sq.ft

Hence we have,

x(x+10)=1824

```x^2+10x`

`x^2+10x-1824=0`

`(x-38)(x+48)=0`

x=38, -48

Since x cannot be negetive we consider only x=38ft

Hence length = x+10 = 48ft

Also given that the width of the surrounding deck is 3inches = 0.25ft

We are asked to find the dimension of the outer edge of the deck.

See the figure. It is clear that the outer dimension will be 38.5ft`` 48.5 ft respectively.

It's not possible to answer this question accurately without resolving the ambiguity about area. Is 1824 sq ft the area of the rectangle, or of the deck?

Let the width of the deck be x *f*

Then the length becomes (x+10) *f*

Area of the rectangle = *l * b*

`x * (x+10) = 1824`

`x^2 + 10x - 1824=0`

`x^2 - 38x + 48x - 1824 = 0`

`(x+48)(x-38) =0`

since* x* cannot be negative,

`therefore x = 38f`

`x+10 = 48f`

length and breadth of the deck are 38 *f* and 48 *f*

Also given that width of the surrounding deck is 3 inch = 0.25 *f*

Dimensions of Outer edge are 38+2*0.25= 38.5 *f* and

48 + 2*0.25= 48.5 *f*

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the rectangle is 38 ft wide and 48 ft long.

Let the width of the rectangle be x. Then the length is x+10.

Since the area of the rectangle is 1824sq.ft, we have `x*(x+10)=1824`

`x^2+10x=1824`

`x^2+10x-1824=0`

`(x+48)(x-38)=0`

` `

`x=-48 or x=38`

Since x cannot be negative, x=38ft, and so length=x+10=58ft

Be careful when calculating dimensions of the deck, since the extra width is given in inches, not feet like before.

So 3 inches = 0.25 feet

Then the dimensions of the deck are **38.25** and **48.25** respectively.

dimension is length x breadth

dimension is length x breadth

work it out from that :)

**Question**

The area of a rectangle is 45 square cm. If the length is 4 cm greater than the width, what is the dimensions of the rectangle?

**Answer**

The picture below shows the rectangle with the area of 45cm2. Now, let the width be **w**. Since, the length is 4cm greater than the width, the length will be **w+4** cm.

Before we can find the dimensions of the rectangle, we need find **w** first. Here's how: **1) Write an equation that relates 45cm2, w+4 and w.**

To do so, we know that the area of the rectangle, 45cm2 can be found by multiplying

**w**with

**w+4**. Hence, we have:

To continue, we need to remove the bracket and simplify the equation. This is shown below:

base times height or bxh

dimension is length x breadth

Going through the discussions I put forth my view to solve the problem with the kind of situations that arose among we mathematics friends:

Assume that the area of the rectangle without the deck =1824. Then the length and breadth without deck is a solution of product of breadth, x and length,x+10 = 1824 : x(x+10)=x^2+10x=1824. Or x^2+10x-1824=0. or (x-38)(x+48)=0. So, x=38 feet is breadth and x=38+10=48 feet is the length of the rectangle.

The area of the deck = Area of outer rectangle -area of inner rectangle:

[38'+2*3'') ][48'+2*3"]-1824 sq ft = 38.5'*48.5'-1824 sq ft.=[1867.25 -1824] sq ft= 43.25 sq ft.

Therefore, the outer dimensions of the deck are: breadth 38.5' , length 48.5' and inner dimensions:38 ft and 48 ft and area of the deck 43.75 sq ft.

If you assume that 1824 sq ft is the area of the rectangle inclusive of the 3" border, then the area of the deck around the rectangle = [1824-(38-0.5)(48-0.5) ] sq ft=[1824-37.5*47.5] sq ft= [1824-1781.25] sq ft. = 42.75 sq ft.

Deck dimension : 38' by 48' outer . 37.5' by 47.5' inner and the deck area =42.75 sq ft.

If you assume the deck area =1824 sq ft, then also the deck area = outer area-inner area of the rectangles:

(x+0.5)(x+10.5)-x(x+10)=1824. giving x=1824-5.25. So, the length and breadth of deck (both outer measurements) = 1818.25 feet and 1828.25 feet.

If the deck is consired within edge of rectangle, then x(x+10)-(x-0.5)(x+9.5) = 1824 . So, x=1824-4.75=1819.25feet.And x+10=1829.25 feet.

To solve this problem we must first find out the dimension of the rectangle.

To do this let us assume that

width of the rectangle is X ft.

Then the length will be = X + 10

Then area of the rectangle = X * (X + 10) = 1824

Therefore

X^2 + 10X = 1824

adding 25 on both sides of equation we get

X^2 + 10x + 25 = 1849

Therefore

(X + 5)^2 = 1849

(X + 5) = 43

X = 43 - 5 = 38

Therefore the rectangle is 38 ft wide and 48 ft long.

As the deck is 3 inch (0.25 ft) wide the dimensions of the outer edge of deck are:

38.5 ft wide by 48.5 ft long.