Question

The length and width of this rectangular lawn were each side is an exact number of metres.

By decreasing the length by 3m and the width by 1m the area of the lawn was halved.
Find the area of the original lawn.

Answer 1

The area of the original lawn is 36 square meters.

Explanation:

Assume the original length of the rectangular lawn is "L" meters and the original width is "W" meters.

According to the problem, the length and width are each an exact number of meters, so both L and W are integers.

The area of the original lawn is given by:

Area = Length × Width

Area = L × W

Now, the problem states that by decreasing the length by 3 meters and the width by 1 meter, the area of the lawn was halved.

The new length becomes (L - 3) meters, and the new width becomes (W - 1) meters.

The new area after the decrease is given by:

New Area = (L - 3) × (W - 1)

According to the problem, the new area is half of the original area:

New Area = 1/2 * Area

Now, we can set up the equation and solve for the area of the original lawn:

(L - 3) × (W - 1) = 1/2 * L × W

Expand the left side of the equation:

LW - L - 3W + 3 = 1/2 * LW

Now, isolate the terms with LW on one side and the constant terms on the other side:

LW - 1/2 * LW = L + 3W - 3

Combine the terms with LW on the left side:

1/2 * LW = L + 3W - 3

Now, solve for LW:

LW = 2L + 6W - 6

Next, we know that L and W are both integers, and the original length and width are each an exact number of meters. Therefore, LW must also be an integer.

Since LW is equal to 2L + 6W - 6, and both L and W are integers, 2L and 6W are even integers. Thus, LW must be an even integer.

However, the product of two consecutive integers is always even. Therefore, LW cannot be an even integer unless one of L or W is an even integer and the other is an odd integer.

Let's consider the possible cases:

1. If both L and W are even, then LW is even. However, this would contradict the fact that LW is an odd integer (from the problem).
2. If both L and W are odd, then LW is odd. However, this would also contradict the fact that LW is an even integer (from the problem).
3. If one of L or W is even and the other is odd, then LW is even.

Since the first two cases are not possible, the third case must be true.

Therefore, one of L or W is even, and the other is odd.

Let's assume L is even and W is odd (the reverse case will yield the same result). So, L = 2a (where "a" is an integer) and W = 2b + 1 (where "b" is an integer).

Now, substitute L and W into the equation LW = 2L + 6W - 6:

(2a)(2b + 1) = 2(2a) + 6(2b + 1) - 6

Simplify the equation:

4ab + 2a = 4a + 12b + 6 - 6

Now, isolate the terms with "a" on one side:

4ab - 2a = 12b

Factor out "a" from the left side:

a(4b - 2) = 12b

Now, divide both sides by (4b - 2):

a = 12b / (4b - 2)

Since "a" and "b" are integers, the denominator (4b - 2) must be a factor of 12b. For simplicity, we can test for "b" values from 1 to 6.

For b = 1:

a = 12 * 1 / (4 * 1 - 2) = 12 / 2 = 6

For b = 2:

a = 12 * 2 / (4 * 2 - 2) = 24 / 6 = 4

For b = 3:

a = 12 * 3 / (4 * 3 - 2) = 36 / 10 = 3.6 (Not an integer)

For b = 4:

a = 12 * 4 / (4 * 4 - 2) = 48 / 14 = 3.428571 (Not an integer)

For b = 5:

a = 12 * 5 / (4 * 5 - 2) = 60 / 18 = 3.333333 (Not an integer)

For b = 6:

a = 12 * 6 / (4 * 6 - 2) = 72 / 22 = 3.272727 (Not an integer)

The only integer solution for "a" is when b = 1. So, a = 6, and b = 1.

Now, we can find the values of L and W:

L = 2a = 2 * 6 = 12

W = 2b + 1 = 2 * 1 + 1 = 3

Therefore, the original lawn's length (L) is 12 meters, and the width (W) is 3 meters.

Now, we can find the area of the original lawn:

Area = Length × Width

Area = 12 × 3

Area = 36 square meters

So, the area of the original lawn is 36 square meters.