# Thread: Can someone help me on this?

1. ## Can someone help me on this?

Hi! We just started doing a new lesson at mathes and I did pretty well for most homework we got, but I didn't get the three last questions we got here, so I'd love if someone would help me!
1-ABC is a right triangle (with A=90°). AB is 27 ft. The perimeter of this triangle is 108 feet. What is the length of AC and AB?
2-If x-y=-3 and x²-y²=6, what is x and y?
3-If x+y=20 and x²-y²=40, what is x and y?
4-A garden has a shape of a rectangle. If we remove 3 feet from the length of a garden, and add 6 feet to the width, the garden gets the shape of a square and the total area gains 78 sq ft. What is the length and width of the garden?

2. Originally Posted by Catherine
1-ABC is a right triangle (with A=90°). AB is 27 ft. The perimeter of this triangle is 108 feet. What is the length of AC and AB?
Let the other leg of the triangle (AC) be x and the hypotenuse (BC) be y. Then we know that
$\displaystyle 27^2 + x^2 = y^2$
and
$\displaystyle 27 + x + y = 108$

So from the second equation I get that
$\displaystyle y = 108 - 27 - x = 81 - x$

Inserting this into the first equation gives:
$\displaystyle 27^2 + x^2 = (81 - x)^2$

On simplifying I get that
$\displaystyle -162x + 5832 = 0$

or x = 36. Thus y = 45.

-Dan

3. Originally Posted by Catherine
2-If x-y=-3 and x²-y²=6, what is x and y?
$\displaystyle x - y = -3$
gives
$\displaystyle y = x + 3$

Inserting this into the other equation:
$\displaystyle x^2 - (x + 3)^2 = 6$

Expand and solve. I get x = -5/2. (And, of course, you need to find y.)

Problem 3 is done in exactly the same way.

-Dan

4. Originally Posted by Catherine
4-A garden has a shape of a rectangle. If we remove 3 feet from the length of a garden, and add 6 feet to the width, the garden gets the shape of a square and the total area gains 78 sq ft. What is the length and width of the garden?
Call the original length of the garden x and the original width y, and call A = xy the area of the original garden.

Then
$\displaystyle (x - 3)(y + 6) = A + 78 = xy + 78$

So
$\displaystyle (x - 3)(y + 6) = xy + 78$

Since the new shape is a square we also know that
$\displaystyle x - 3 = y + 6$

So solve this last equation for y and put the result into your area equation.

I get x = 23 and y = 14.

-Dan

5. Hello, Catherine!

The first three can be solved with the same trick . . .

1) $\displaystyle ABC$ is a right triangle (with $\displaystyle A=90^o$). .$\displaystyle AB$ is 27 ft.
The perimeter of this triangle is 108 feet.
What is the length of $\displaystyle AC$ and $\displaystyle BC$?
First, make a sketch . . .
Code:
    C *
|  *
|     *    a
b |        *
|           *
|              *
A * - - - - - - - - * B
27

Let $\displaystyle a = BC,\;b = AC$

The perimeter is 108: . $\displaystyle a + b + 27 \:=\:108\quad\Rightarrow\quad a + b \:=\:81$

Pythagorus tells us: .$\displaystyle a^2 \:=\:b^2 + 27^2\quad\Rightarrow\quad a^2-b^2 \:=\:729$

We have a system of equations: .$\displaystyle \begin{array}{cccc}a^2-b^2 &=&729 & {\color{blue}[1]}\\ a+b &=& 81 & {\color{blue}[2]}\end{array}$

Equation [1] is: .$\displaystyle (a-b)(a+b) \:=\:729$

Divide by [2]: .$\displaystyle \frac{(a-b)(a+b)}{a+b} \:=\:\frac{729}{81}\quad\Rightarrow\quad a-b\:=\:9\;\;{\color{blue}[3]}$

Add [2] and [3]: .$\displaystyle \begin{array}{ccc}a + b &=&81 \\ a-b &=&9\end{array}\quad\Rightarrow\quad 2a \:=\:90\quad\Rightarrow\quad a \:=\:45$

Substitute into [2]: .$\displaystyle 45 + b \:=\:81\quad\Rightarrow\quad b \:=\:36$

Therefore: .$\displaystyle AC = 36,\;BC = 45$ feet

2) If $\displaystyle x-y\:=\:-3$ and $\displaystyle x^2-y^2\:=\:6$, find $\displaystyle x$ and $\displaystyle y.$
Can you see that this system is identical to the system in #1?

We have: .$\displaystyle \begin{array}{cccc}x - y &=& \text{-}3 & {\color{blue}[1]}\\ (x-y)(x+y) &=& 6 & {\color{blue}[2]}\end{array}$

$\displaystyle \begin{array}{cccc}\text{Divide {\color{blue}[2]} by {\color{blue}[1]}:} & x + y &=&-2 \\ \text{Add {\color{blue}[1]}:} & x - y & =& -3\end{array}\quad\Rightarrow \quad 2x\:=\:-5\quad\Rightarrow\quad\boxed{x \:=\:-\frac{5}{2}}$

Substitute into [1]: .$\displaystyle -\frac{5}{2} - y \:=\:-3\quad\Rightarrow\quad\boxed{y \:=\:\frac{1}{2}}$

4) A garden has a shape of a rectangle.
If we remove 3 feet from the length of a garden, and add 6 feet to the width,
the garden gets the shape of a square and the total area gains 78 sq ft.
What is the length and width of the garden?
I approached this one "backwards".

Let $\displaystyle x$ = side of the square garden. .Its area is: $\displaystyle x^2$ ft²

The original garden was 3 feet longer: .$\displaystyle \text{Length} \:=\:x+3$
. . and was 6 feet narrower: .$\displaystyle \text{Width} \:=\:x-6$
The area of the original garden was: .$\displaystyle (x+3)(x-6)$ ft²

The square garden has 78 ft² more than the original garden.
. . . $\displaystyle x^2 \;=\;(x+3)(x-6) + 78\quad\Rightarrow\quad x \:=\:20$

Therefore, the original garden was $\displaystyle 23 \times 14$ ft.

6. Hi everyone! Thanks all for your answers! It seems so easy when you did it, sometimes I can really be stupid, I mean, how did I forget all this