Question

if in a rectangle,the length is increased and breadth decreased by 10m,the area is reduced by 200m^2.but if the length is decreased by 5m and the breadth is increased by 10m,the area is increased by 200m^2,find the length and breadth​

Answer 1

Answer:

$\bigstar{\bold{Length\:of\:the\:rectangle=40\:m}}$

$\bigstar{\bold{Breadth\:of\:the\:rectangle=30\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• If the length of a rectangle is increased and breadth decreased by 10m, the area is reduced by 200 m²
• If the length is decreased by 5m and breadth increased by 10 m, the area is increased by 200 m²

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The length and breadth of the rectangle

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let the length of the rectangle be l m

→ Let the breadth of the rectangle be b m

→ Area of a rectangle is given by

 Area of a rectangle = length × breadth

→ Hence the area of the rectangle is given by

  Area of the rectangle = l × b

→ By the first case,

  (l + 10) × (b - 10) = lb - 200

   lb - 10l + 10b - 100 = lb - 200

→ Cancelling lb on both sides and simplifying,

   10b - 10l = -100

→ Dividing the whole equation by 10

  b - l = -10

  b = -10 + l---(1)

→ In the second case,

  (l - 5) × (b + 10) = lb + 200

  lb + 10l - 5b -50 = lb + 200

  10l - 5b = 250

→ Substitute the value of b from equation 1

  10l - 5(-10 + l) = 250

  10l + 50 -5l = 250

   5l = 200

     l = 200/5

     l = 40 m

$\boxed{\bold{Length\:of\:the\:rectangle=40\:m}}$

→ Now substitute the value of l in equation 1

  b = -10 + 40

  b = 30 m

→ Hence the breadth of the rectangle is 30 m.

$\boxed{\bold{Breadth\:of\:the\:rectangle=30\:m}}$

$\Large{\underline{\underline{\bf{Verification:}}}}$

→  (l + 10) × (b - 10) = lb - 200

   (40 + 10) × (30 - 10) = (40 × 30) - 200

    50 × 20 = 1200 - 200

    1000 = 1000

→   (l - 5) × (b + 10) = lb + 200

    (40 - 5) × (30 + 10) = 40 × 30 + 200

     35 × 40 = 1200 + 200

     1400 = 1400

→ Hence verified.