Question

the area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth is increased by 3 units. The area is increased by 67 square units of length is increased by 3 units and breadth is increased by 2 units. Find the perimeter of the rectangle.​

Answer 1

Answer:

1st case;

let length =X and breadth=y

(x-5)(y+2)=xy-9...............(1)

2nd case;

(X+3)(y+2)=xy+67...........(2)

by solving (1) and (2),we get X=17 and y=9

thus, perimeter= 26

Answer 2

Answer:

Let the Length of the Rectangle be : L

Let the Breadth of the Rectangle be : B

★ Area of Rectangle = [Length × Breadth]

Given : Area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth is increased by 3 units.

$:\implies\sf (Length-5) \times (Breadth+3)=Area-9\\\\\\:\implies\sf (L - 5)(B + 3) = LB - 9\\\\\\:\implies\sf LB + 3L - 5B - 15 = LB - 9\\\\\\:\implies\sf 3L - 5B - 15 = -\: 9\\\\\\:\implies\sf 3L - 5B = 15 - 9\\\\\\:\implies\sf 3L - 5B = 6\qquad....eq.\:(1)$

$\rule{100}{0.8}$

Given : The area is increased by 67 square units of length is increased by 3 units and breadth is increased by 2 units.

$:\implies\sf (Length+3) \times (Breadth+2)=Area+67\\\\\\:\implies\sf (L +3)(B + 2) = LB+67\\\\\\:\implies\sf LB + 2L +3B +6 = LB+67\\\\\\:\implies\sf 2L+3B+6 = 67\\\\\\:\implies\sf 2L+3B =67-6\\\\\\:\implies\sf 2L+3B=61\qquad....eq.\:(2)$

$\rule{150}{1.5}$

☯ Multiplying eq. ( 1) by 3 & eq. ( 2) by 5 and Adding both of them :

$\dashrightarrow\sf\:\:9L-15B=18\\\\\dashrightarrow\sf\:\:10L+15B=305\\\dfrac{\qquad\qquad\qquad\qquad\qquad}{}\\\dashrightarrow\sf\:\:(9L+10L)=(18+305)\\\\\\\dashrightarrow\sf\:\:19L=323\\\\\\\dashrightarrow\sf\:\:L=\dfrac{323}{19}\\\\\\\dashrightarrow\sf\:\:L=17\:units$

⠀

☯ Substituting value of L in eq. ( 1) :

$\dashrightarrow\sf\:\:3L-5B=6\\\\\\\dashrightarrow\sf\:\:3(17)-5B=6\\\\\\\dashrightarrow\sf\:\:51-5B=6\\\\\\\dashrightarrow\sf\:\:51-6=5B\\\\\\\dashrightarrow\sf\:\:45=5B\\\\\\\dashrightarrow\sf\:\:\dfrac{45}{5}=B\\\\\\\dashrightarrow\sf\:\:B=9\:units$

$\rule{200}{2}$

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.5mm}\put(7.7,3){\large\sf{A}}\put(7.1,2){\sf{\large{9 units}}}\put(7.7,1){\large\sf{B}}\put(9.2,0.7){\sf{\large{17 units}}}\put(11.1,1){\large\sf{C}}\put(11.1,3){\large\sf{D}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\end{picture}$

$\underline{\bigstar\:\textsf{Perimeter of the Rectangle :}}$

$\longrightarrow\sf Perimeter=2(Length+Breadth)\\\\\\\longrightarrow\sf Perimeter=2(L+B)\\\\\\\longrightarrow\sf Perimeter=2(17\:units+9\:units)\\\\\\\longrightarrow\sf Perimeter=2 \times 26\:units\\\\\\\longrightarrow\underline{\boxed{\sf \sf Perimeter=52\:units}}$

⠀

$\therefore\:\underline{\textsf{Hence, Perimeter of Rectangle is \textbf{52 units}}}.$