Question

find the
dimension of a rectangle whose area is
48 cm sq. and its perimeter is 28 cm​

Answer 1

Answer:

Given :- Perimeter of rectangle = 28

2 (l + b ) = 28 

l + b = 14 

l = 14 - b --- (i)

area = 40 

l × b = 40 

from (i)

( 14 -b ) × b = 40 

(14 - b²) = 40

b² - 14b + 40 = 0

After spilliting the zeros you will get...

(length) = 10

breadth = 4

Answer 2

Answer:

perimeter=28cm

equation - 1

perimeter=2(l+b)

28=2(l+b)

28/2=l+b

14=l+b

l=14 &b=_______1

area=48

l×b= 48

from EQ 1

(14-b)×b=48

$(14 - {b}^{2} ) = 48$

${b}^{2} - 14b + 48 = 0$

after splitting the zero

${b}^{2} - 4b - 10b + 48 = 0$

$b(b - 4) - 10(b - 4) = 0$

$b - 4 = 0b - 10 = 0$

b=4 and b=10