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.
Lets say , a is length and b is breadth of rectangle
Perimeter = 2(a+b) = 46
a+b = 23
b = 23 - a
Diagonal = $\displaystyle \sqrt{a^2 +b^2 }=17 $
$\displaystyle 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
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For an equation
lx^2 +mx +n = 0
Multiplication of roots = n/l
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So here it will be = 120
Hi rickymylv,
Let's assume the sides are integers.
$\displaystyle 2l + 2w = 46$
[1] $\displaystyle l + w = 23$
[2] $\displaystyle l^2 + w^2 = 289$
Solve the above system.
[1] $\displaystyle l = 23-w$
[2] $\displaystyle (23-w)^2+w^2=289$
$\displaystyle 529-46w+w^2+w^2-289=0$
$\displaystyle 2w^2-46w+240=0$
$\displaystyle w^2-23w+120$
$\displaystyle (w-8)(w-15)=0$
$\displaystyle w=8 \ \ or \ \ w=15$
Therefore, $\displaystyle l=8 \ \ or \ \ l=15$
The width is 8 and the length is 15.