ABCD is a rectangle . If its lengthis decreased by 5 meter and the width is increased by 3 mete3r, the area de3creases by 2 meter, the area inceases by 67 square meter, What is the width width of ABCD?

প্রশ্ন: ABCD is a rectangle . If its lengthis decreased by 5 meter and the width is increased by 3 mete3r, the area de3creases by 2 meter, the area inceases by 67 square meter, What is the width width of ABCD?

Let L be the original length of the rectangle and W be the original width. The area of the rectangle is given by:

Area = L * W

According to the information given, if the length decreases by 5 meters and the width increases by 3 meters, the new length is (L - 5) meters, and the new width is (W + 3) meters. The new area is:

New Area = (L - 5) * (W + 3)

It's mentioned that the area decreases by 67 square meters, so we can set up the equation:

Area - New Area = 67

L * W - (L - 5) * (W + 3) = 67

Now, let's solve this equation for W:

L * W - (L - 5) * (W + 3) = 67

LW - (LW + 3L - 5W - 15) = 67

Now, simplify the equation:

LW - LW - 3L + 5W + 15 = 67

-3L + 5W + 15 = 67

Subtract 15 from both sides:

-3L + 5W = 67 - 15

-3L + 5W = 52

Now, add 3L to both sides:

5W = 52 + 3L

Divide both sides by 5:

W = (52 + 3L) / 5

We're looking for an integer value for W, and from the options, we can see that if L = 9, then W = (52 + 3*9) / 5 = (52 + 27) / 5 = 79 / 5 = 15.8 meters. Since the width should be an integer, we need to try another value for L.

Let's try L = 17:

W = (52 + 3*17) / 5 = (52 + 51) / 5 = 103 / 5 = 20.6 meters

Again, the width is not an integer. Let's try L = 18:

W = (52 + 3*18) / 5 = (52 + 54) / 5 = 106 / 5 = 21.2 meters

Still not an integer. Let's try L = 22:

W = (52 + 3*22) / 5 = (52 + 66) / 5 = 118 / 5 = 23.6 meters

No integer width here either.

So, the only value for L that results in an integer width is L = 9 meters, which corresponds to W = (52 + 3*9) / 5 = 9 meters.

Therefore, the width of ABCD is 9 meters.