Question

please answer this question if you know the answer then only answer ch linear equations​

Answer 1

Answer: The answer is Follow

Step-by-step explanation:

let the length of rectangle be x+9

and breadth be x

if the length and breadth increased by 3cm,

area of rectangle = (x+9)( x) - 84cm²

(x+12) (x+3)= x²+9x + 84 cm²

x²+12+ 3x +36 = x² +9x+ 84cm²

15x -9x = 84-36

6x = 48

x= 8

breadth = 8cm

length = 8+ 9

           = 17cm

plz mark me brillients

Answer 2

$\huge\rm\underline\purple{GIVEN:}$

• $\large\rm{Length\:exceeds\:Breadth\:by\:9cm}$
• $\large\rm{The\:Area\:of\:rectangle\:when\:length\:and\:breadth\:are\:increased\:by\:3cm\:is\:84cm^2}$

$\huge\rm\underline\purple{TO\:FIND:}$

• $\large\rm{Length\:and\:Breadth\:of\:rectangle}$

$\huge\rm\underline\purple{SOLUTION:}$

$\huge\rm\underline{CASE-I:}$

$\large\rm{Let\:the\:breadth\:of\:rectangle\:be\:'b'}$

$\large\rm{Given\:Length\:exceeds\:breadth\:by\:9cm}$

$\large\rm{\implies{Length\:of\:rectangle\:=\:b+9}}$

$\large\rm\bold{\boxed{Area\:of\:rectangle\:=\:Length\times\:Breadth}}$

$\large\rm{\implies{Area\:=\:b(b+9)\:=\:b^2+9b--eq(1)}}$

$\huge\rm\underline{CASE-II:}$

$\large\rm{Again\:length\:and\:breadth\:are\:increased\:by\:3cm}$

$\large\rm{\implies{Length\:=\:b+3+9\:=\:b+12cm}}$

$\large\rm{\implies{Breadth\:=\:b+3cm}}$

$\large\rm{Area\:of\:the\:rectangle\:in\:this\:case=\:(b+12)(b+3)\:=\:b^2+15b+36}$

$\large\rm{Given\:Area\:of\:rectangle\:in\:case-I\:+84=\:Area\:of\:rectangle\:in\:case-II}$

$\large\rm{\implies{(b^2+9b+84)\:=\:(b^2+15b+36)}}$

$\large\rm{\implies{9b\:+84=\:15b+36}}$

$\large\rm{\implies{84-36\:=\:15b-9b}}$

$\large\rm{\implies{48\:=\:6b}}$

$\large\rm{\implies{b\:=\:\frac{48}{6}\:=\:8cm}}$

$\large\rm{l\:=\:b+9\:=\:8+9\:=\:17cm}$

$\large\rm{\therefore{Length\:of\:Rectangle\:=\:17cm}}$

$\large\rm{Breadth\:of\:Rectangle\:=\:8cm}$