Area of rectangle

**rickymylv** · April 3rd 2009, 06:51 AM

The diagonal of a rectangle is 17c.m. long and the perimeter of the rectangle is 46 c.m. Find the area of the rectangle.

Please tell me how to solve this problem.

**ADARSH** · April 3rd 2009, 07:08 AM

Lets say , a is length and b is breadth of rectangle

Perimeter = 2(a+b) = 46
a+b = 23
b = 23 - a
Diagonal = $\sqrt{a^2 +b^2 }=17$

$a^2 + b^2 = 289$

a^2 + a^2 + 529 - 46a = 289

2a^2 - 46a + 240 = 0

a^2 - 23a + 120 = 0

The value that we will get as a can be either length or breadth (as even if you solved for b you would have recieved same equaion)

Hence area = Lengt*Breadth = Multiplication of roots of this quadratic equation

-----------------------
For an equation

lx^2 +mx +n = 0

Multiplication of roots = n/l

-----------------

So here it will be = 120

**masters** · April 3rd 2009, 07:20 AM

Originally Posted by rickymylv

The diagonal of a rectangle is 17c.m. long and the perimeter of the rectangle is 46 c.m. Find the area of the rectangle.

Please tell me how to solve this problem.

Hi rickymylv,

Let's assume the sides are integers.

$2l + 2w = 46$

[1] $l + w = 23$

[2] $l^2 + w^2 = 289$

Solve the above system.

[1] $l = 23-w$

[2] $(23-w)^2+w^2=289$

$529-46w+w^2+w^2-289=0$

$2w^2-46w+240=0$

$w^2-23w+120$

$(w-8)(w-15)=0$

$w=8 \ \ or \ \ w=15$

Therefore, $l=8 \ \ or \ \ l=15$

The width is 8 and the length is 15.

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