Finding dimensions using quadratic formula

Question

A 52 m long fence is constructed on three sides of a housing block with area 240 m^2. The fourth side facing the road is left open. Find the dimensions of the block.

Also here's another question I don't quite understand:

ABDC is a rectangle in which AB = 21 cm. The square is AXYD and the remaining rectangle has area 80 cm^2. Find the length of [BC]

Here's the rectangle in the second question:

---

                                           D__________Y____C
                                           |          |    |
                                           |          |    |
                                           |          |    |
                                           |          |    |
                                           |__________|____|
                                           A          X    B

Answer 1

(1)

First lets equate the total length (perimeter)

2x+y = 52

And the area equation is

xy= 240

Therefore x = 240/y

Now subsitute this as x in the first equation we found (perimeter equation)

2(240/y) + y = 52

Multiply each term by y

120 + y^2 = 52y

y^2 - 52y + 120 = 0 Now solve for y and then put that y value to

x = 240/y in order to get the value of x

There you go , these are the dimensions

(2)

Hint : Use Similar Figures

Answer 2

The housing block has $4$ sides and out of which $3$ sides have fence whereas $1$ sides is without fence.

It is not given whether the block is square or rectangular, so we consider it to be a rectangular block (because square is a particular case of rectangle).

Let $a,b$ be the sides of the housing block and let the side that is not fenced has length $b$

So $ \ 2a+b=52 \ $ and $ \ a\times b=240$, solve them to get the answer.

For the second problem, let $x$ be the length of $XB$, therefore length of $AB=21-x$

Area of $AXYD$ + Area of $XBCY$ = Area of $ABCD$

$(21-x)^2+80=21(21-x)$

$(21-x)^2-21(21-x)+80=0$

Substitute $(21-x)^2=m$, we get $m^2-21m+80=0$

Answer 3

(1) Let x and y be the two side's of the rectangle. Then we have,

$xy = 240 \Rightarrow x = \frac{240}{y}$

$2x + y = 52 \Rightarrow 2\left(\frac{240}{y}\right) + y = 52 \\ y^2 - 52y + 480 = 0 \\ y = {40 \quad \& \quad 12} \Rightarrow x = {6 \quad \& \quad 20}$