Question

4) The length of Lawiectangle exceeds ita bovadth
by scom 9f the perimeter of the rectangles 25cm, find the dimensions of the rectangles​

Answer 1

Step-by-step explanation:

Let the breadth be b and length be b+9

Perimeter=2(l+b)

25=2(b+b+9)

25=4b+18

4b=7

b=7/4

Breadth=4/7 cm

Length=4/7+9 = 67/7 cm

Answer 2

• Correct Question

The length of a rectangle exceeds its breadth by 9 cm. The perimeter if the rectangle is 25 cm. Find the dimensions of the rectangle.

Answer ⇒

• Breadth = 1.75 cm
• Length = 10.75 cm

• Given

• Length of rectangle exceeds by 9 cm of its breadth.
• Perimeter of rectangle = 25 cm

• To find

• Dimensions of the rectangle (length, breadth)

• How to solve?

⇒

• Firstly, we will let the breadth = x. then as given in the question (Length of rectangle exceeds its breadth by 9cm). So, add 9 cm in its breadth. Now we have the length.

• We have the value of perimeter of rectangle. So, by using its formula we will find the length, breadth of rectangle.

Some important formulae related to rectangle -

• Perimeter = 2(l + b)
• Area = length × breadth
• Diagonal = $\sf{\sqrt{l^2 + b^2}}$

• Solution

Let the breadth be x and length be x + 9.

Perimeter = 2(l + b)

Substitute the given values,

⇒ 25 = 2(x + 9 + x)

⇒ 25 = 2(2x + 9)

⇒ 25 = 4x + 18

⇒ 25 - 18 = 4x

⇒ 7/4 = x

The value of x is 7/4.

Dimensions of rectangle -

• Breadth = x = 7/4 = 1.75 cm
• Length = x + 9 = 7/4 + 9 = 10.75 cm

━━━━━━━━━━━━━━━━━━━━━━

Let's verify -

Perimeter = 2(l + b)

⇒ 2(10.75 + 1.75)

⇒ 2(12.5)

⇒ 25 cm

Hence, verified.