# Thread: Easiest Sum for You'll

1. ## Easiest Sum for You'll

A rectangle has perimeter 92cm and the lenght of its diagnol is 34cm...Find it's area.....I'm studin in class 8 so......
The answer which Iam getting is 480cm2 but the correct answer is 240...Can any1 solve it and tell me how you get it....Thx in advance

3. Originally Posted by avincool
A rectangle has perimeter 92cm and the lenght of its diagnol is 34cm...Find it's area.....I'm studin in class 8 so......
The answer which Iam getting is 480cm2 but the correct answer is 240...Can any1 solve it and tell me how you get it....Thx in advance
It's a rectangle hence the 2 opposite side will have the same length. Lets call them $\displaystyle x$ and $\displaystyle y$.

The Perimeter is the total length of the rectangle hence the opposite sides multiplied by 2 therefore:

$\displaystyle 2x +2y = 92 \ \ \ --(1)$

The diagonal is 34cm and by Pythagoras' theorem, we know that it is equal to the length squared hence:

$\displaystyle x^2 + y^2 = 34^2 \implies x^2 + y^2 = 1156 \ \ \ --(2)$

The equation (1) rearranges to give:

$\displaystyle x = 46 - y$

Substituting this $\displaystyle x$ value into equation (2) gives:

$\displaystyle (46 - y)^2 + y^2 = 1156$
$\displaystyle y^2 - 92y + 2116 + y^2 = 1156$
$\displaystyle 2y^2 - 92y + 960= 0$
$\displaystyle y^2 - 46y +480 = 0$
$\displaystyle (y - 30)(y - 16) = 0$
$\displaystyle \therefore y = 30, \ y=16$

$\displaystyle y= 30 \implies x = 16, \ y=16 \implies x=30$

Therefore the rectangle is of length and width of $\displaystyle 30$cm and $\displaystyle 16$cm respectfully.

Area of rectangle = $\displaystyle l \times w = 30 \times 16 = 480$$\displaystyle \text{cm}^2$

4. Hello, avincool!

A rectangle has perimeter 92cm and the lenght of its diagonal is 34cm.
Find it's area.

The answer which I am getting is 480 cm². . . . . Right!

Let $\displaystyle x$ = length, $\displaystyle y$ = width.

As Air pointed out, the perimeter is 92 cm.
. . So we have: .$\displaystyle 2x + 2y \:=\:92\quad\Rightarrow\quad x + y \:=\:46\;\;{\color{blue}[1]}$

The diagonal is 34.
. . So we have: .$\displaystyle x^2+y^2\:=\:34^2\quad\Rightarrow\quad x^2+y^2\:=\:1156\;\;{\color{blue}[2]}$

Sqaure [1]: .$\displaystyle (x+y)^2 \:=\:46^2\quad\Rightarrow\quad x^2 + 2xy + y^2 \:=\:2116$

$\displaystyle \text{So we have: }\;2xy + \underbrace{x^2+y^2}_{\text{This is 1156}\;{\color{blue}[2]}} \:=\:2116$

Hence: .$\displaystyle 2xy + 1156 \:=\:2216\quad\Rightarrow\quad 2xy \:=\:960$

Therefore, the area of the rectangle is: .$\displaystyle xy \:=\:480\text{ cm}^2$