Question

If the area of a rectangle is 168 sq.cm and its diagonal is 25 cm, find its length, breadth and perimeter.

Answer 1

area=168 cm²
diagonal=25 cm
let length of the rectangle is x
bredth is y

then use Pythagorean theorem
x²+y²=25²
625=x²+y²
y²=625-x²...(1)
area(168)=x*y
168²=x² * y²
28224= x² * y²

we know y²=625-x²
then
x²(625-x²)28224
[tex] -x^{4} +625 x^{2} =28224 \\ [/tex]
multiply the whole equation with (-1)

then
$x^{4} -625 x^{2} = -28224$

we want to write this formula in (a-b)² type..
let a=x²,,,2ab=625x²
so we will get 2b=625
b=312.5

add (312.5) in two sides of the formula

[tex] x^{4} +312.5 ^{2} -625x= -28224+(312.25)^{2} \\ \\ ( x^{2} -312.5) ^{2}= 97656.25- 28224=69432.25 \\ \\ x^{2} -312.5= \sqrt{69432.25} =263.50 \\ x^{2} =312.5+263.5=576 \\ x^{2} =576 \\ x= \sqrt{576} =24 \\ length=24 \\ area=168 \\ length*bredth=168 \\
bredth=168/24=7 cm\\
 perimeter=2(length+bredth) =2(24+7)=62 cm[/tex]

Answer 2

l*b=168
l$^{2} *b^{2} =168 ^{2} =28224$
$l ^{2} +b ^{2} =[tex]25 ^{2}$=625
$l ^{2} =625 -b ^{2}$
$l ^{2} (625-l ^{2})=28224$
$625l ^{2} -l^{4} =28224$