Question

A rectangular lot is 140 meters long and 100 meters wide. Give the length and width of another rectangular lot that has the same perimeter but a larger area.

Answer 1

Step-by-step explanation:

Length of rectangle is 140 m and its breadth is 100 m

Perimeter = sum of all sides

For rectangle, P = 2(l+b)

P = 2(140+100)

P = 480 m

Area = length × breadth

For a rectangle, A = lb

A = 140 × 100

A = 14000 m²

We need to write the length and breadth such that it has the same perimeter but a larger area.

Let the length is 110 m and breadth is 130 m.

Perimeter, P = 2(110+130)

P = 480 m

Perimeter is same as previous

Area, A = 110 × 130

A =  14300 m²

Area is more than the initial area.

Hence, this is the required solution.

Answer 2

answer:

it's $\boxed{L=130, \W=110}$

explanation:

$\boxed{L=140}\\\boxed{W=100}\\P=2(L+W)\\P=2(140+100)\\\boxed{P=480m}\\A=L*W\\A=140*100\\\boxed{A=14000m^{2}}$

It's easy to get a Length and Width with an equal Perimeter as this rectangle, how? the key is in the Perimeter formula for rectangle, it is literally 2 times the sum of the Length and Width, so $2(L+W)=2(sum)$, that means we can change the value of both the Width and the Length as long as the sum of both the Length and the Width is equal;

$\boxed{L=140}\\\boxed{W=100}\\L+W=140+100=\boxed{240}$

to change these values and still have a perimeter equal to 480 meters, and an area larger than 14,000 meters square, we gotta have a length and a width that have a sum of 240 and still maintain a hundred place value, for example 130 and 110, not 150 and 90, because $150*90=13500$ which is less than our goal; while $130*110=14300$, note that as long as one of the values doesn't fall below 100 you're solid.