Let length of rectangle be x cm and breadth be y cm. From first condition – 2(x + y) = 40 x + y = 20 . . . (I) From 2nd condition – x = 2y + 2 x – 2y = 2 . . . (II) Let’s solve eq. (I), (II) by determinant method x + y = 20 x – 2y = 2 Read more on Sarthaks.com – https://www.sarthaks.com/368788/perimeter-rectangle-length-rectangle-more-than-double-its-breadth-find-length-and-breadth Reply
Let length of rectangle be x cm and breadth be y cm. From first condition – 2(x + y) = 40 x + y = 20 . . .
(I) From 2nd condition – x = 2y + 2 x – 2y = 2 . . .
(II) Let’s solve eq. (I),
(II) by determinant method x + y = 20 x – 2y = 2 Read more on Sarthaks.com – https://www.sarthaks.com/368788/perimeter-rectangle-length-rectangle-more-than-double-its-breadth-find-length-and-breadth
Step-by-step explanation:
2(l+b)=40
l+b=20
b=20-l
the answer is 20-l
i hope it helps