Pls solve the third one

Dear Student
3)
Area of hall = 288m²
Let the length=l
breadth = b

A r e a = l \times b 288 = l \times b b = \frac{288}{l} N o w, G i v e n, L e n g t h i s i n c r e a s e d b y 6 m B r e a d t h i s d e c r e a s e d b y 4 m n e w l e n g t h = l + 6 n e w b r e a d t h = b - 4 A r e a = (l + 6) (b - 4) 288 = (l + 6) (b - 4) 288 = (l + 6) (\frac{288}{l} - 4) (l + 6) (\frac{72}{l} - 1) = 72 (l + 6) (72 - l) = 72 l 72 l + 432 - 6 l - l^{2} = 72 l l^{2} + 6 l - 432 = 0 l^{2} + 24 l - 18 l - 432 = 0 l (l + 24) - 18 (l + 24) = 0 (l + 24) (l - 18) = 0 l = - 24 o r l = 18 l e n g t h o f t h e r e c \tan g l e c a n ’ t b e n e g a t i v e . H e n c e, t h e l e n g t h o f t h e g i v e n r o o m i s 18 m . R e g a r d s