Question

An iron wire of length 46cm is bent into a rectangle and its diagonal is 17cm. find rectangle breadth and length?​

Answer 1

Answer:

let the length of the rectangle=xcm

breadth=ycm

so, perimeter=2(x+y)=46

=> x+y=46/2

=> x+y=23

again , diagonal if the rectangle is√(x^2+y^2)=17

x^2+y^2=17^2

=> (x+y)^2-2xy=17^2

=> 23^2-2xy=17^2

=> 2xy=23^2-17^2

=> 2xy=(23+17)(23-17)

=> 2xy=40×6

=> 2xy=240

again,

x^2+y^2=17^2

=> (x-y)^2+2xy=17^2

=> (x-y)^2=17^2-2xy

=> (x-y)^2=289-240

=> (x-y)^2=49=7^2

=> x-y=7

=> x=7+y

putting this value in the equation,

x+y=23

=> 7+y+y=23

=> 2y=23-7

=> 2y=16

=> y=16/2

=> y=8

so, x=7+8=15

so, length of the rectangle is 15 cm

breadth of the rectangle is 8 cm