Question

Q.3(A) complete the activity. (any 1) (3 Marks) 1) The perimeter of a rectangle is 40 cm. The length of the rectangle is more than double its breadth by 2. Find length and breadth. Solution : Let length of rectangle bo x cm and breadth be y cm. From first condition - 2(x + y) = 40 x+y=20... (1) From 2nd condition - x = 2y + 2 .: V.2y=2... (II) Let's solve eq. (I), (II) by determinant method x+y= 20 x 2y= 2 |-(1x1)=-2-1=-3 RAM (les 120 Dx 1-(1x2)=-40-2=-42 D=-11- 2 29 -Bixan 1 201 = Dy = -(20x1)=2-20=-18 X and y= DY X= ..X= x=and y=0 ...x=14, y=0 : Length of the rectangle is 14 cm and breadth is 6 cm. bong from 1 to 150​

Answer 1

Answer:

6 cm

Step-by-step explanation:

Perimeter of rectangle =40 cm (given)

condition :- length = 2× breadth +2

⇒l=2b+2

perimeter p=2(l+b)=40

⇒2(2b+2+b)=40

⇒2(3b+2)=40

⇒6b=40−4=36

∴b=6cm

l=2×6+2=12+2=14cm

∴ length =14 cm, breadth =6 cm