Question

The length of a rectangle is five more than thrice its breadth find the area if this rectangle whose perimeter is 490 m​

Answer 1

Step-by-step explanation:

Let the width be x, then length = 2x

∴Perimeter=x+2x+x+2x=24

⇒x=4

Hence width = 4 and length =8

∴ Required area=8×4=32sq. cm

Answer 2

length (l ) is thrice width(w) +6

length (l ) is thrice width(w) +6l =3 w+6

length (l ) is thrice width(w) +6l =3 w+6perimeter =100

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =50

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =50

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6w=44/4

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6w=44/4w= 11

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6w=44/4w= 11l=3(11)+6

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6w=44/4w= 11l=3(11)+6l=33+6

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6w=44/4w= 11l=3(11)+6l=33+6l=39

length (l ) is thrice width(w) +6l =3 w+6perimeter =1002l +2w = 100l+w= 50(3w+6) +w =504w+6 =504w=50–6w=44/4w= 11l=3(11)+6l=33+6l=39Length of rectangle is 39 and width is 11.

I hope helpful answer.