The width of a rectangle is 1 cm more than half the length. when both the length and width are increased by 1 cm, the area increases by 20 cm squared. find the original perimeter of the rectangle.

38 cm.

Step-by-step explanation:

Let us assume that W and L are the width and length respectively of the given rectangle.

Hence, as the width of the rectangle is 1 more than half the length, so, we can write that,

W= 1+L/2, ⇒2W =2+L, ⇒ L=2W-2 (1)

Again, the given condition is that, if both L and W are increased by 1 cm, the area increases by 20 cm².

Hence, (W+1)(L+1) - WL =20, ⇒W+L =19 (2)

Now, solving equations (1) and (2) by substitution method, we get

W+(2W-2) =19, ⇒ 3W=21, ⇒W =7 cm.

Therefore, from equation (1), we get, L= 2W-2 =2×7-2=12 cm.

Hence, the original perimeter of the rectangle= 2(L+W) =2(7+12) =38 cm. (Answer)

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24

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