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How many rectangles with integral sides are possible where the area of the rectangle equals the perimeter of the rectangle?
(2014)
- a)One
- b)Three
- c)Two
- d)Infinitely many
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
How many rectangles with integral sides are possible where the area of...
Let the length and breadths of rectangle be a units and b units.
According to question,
ab = 2a + 2b
a = 2b / (b – 2)
Here, Two cases are possible
(i) a = 4, b = 4
(ii) a = 3 and b = 6
Thus, two different rectangles are possible.
According to question,
ab = 2a + 2b
a = 2b / (b – 2)
Here, Two cases are possible
(i) a = 4, b = 4
(ii) a = 3 and b = 6
Thus, two different rectangles are possible.
Most Upvoted Answer
How many rectangles with integral sides are possible where the area of...
There are two key pieces of information given in the question:
1. The area of the rectangle equals the perimeter of the rectangle.
2. The sides of the rectangle are integers.
To solve this question, we need to find the number of rectangles that satisfy both conditions.
Let's consider a rectangle with length L and width W. According to the given information, the area of the rectangle is LW, and the perimeter is 2(L + W).
From the first condition, we have LW = 2(L + W).
Rearranging this equation, we get LW - 2L - 2W = 0.
Using the Simon's Favorite Factoring Trick, we can rewrite this equation as (L - 2)(W - 2) = 4.
Now, we need to find all possible pairs of integers (L, W) such that their product is 4. These pairs are (1, 4), (2, 2), and (4, 1).
Let's analyze each pair separately:
1. For (1, 4): If L = 1 and W = 4, the perimeter is 2(1 + 4) = 10, which does not equal the area of the rectangle (1 * 4 = 4). Therefore, this pair does not satisfy the conditions.
2. For (2, 2): If L = 2 and W = 2, the perimeter is 2(2 + 2) = 8, which does equal the area of the rectangle (2 * 2 = 4). Therefore, this pair satisfies the conditions.
3. For (4, 1): If L = 4 and W = 1, the perimeter is 2(4 + 1) = 10, which does not equal the area of the rectangle (4 * 1 = 4). Therefore, this pair does not satisfy the conditions.
Therefore, there are two rectangles that satisfy the conditions: one with sides 2 and 2, and one with sides 2 and 2 (since it is a rectangle, the order of the sides does not matter).
Hence, the correct answer is option C: Two.
1. The area of the rectangle equals the perimeter of the rectangle.
2. The sides of the rectangle are integers.
To solve this question, we need to find the number of rectangles that satisfy both conditions.
Let's consider a rectangle with length L and width W. According to the given information, the area of the rectangle is LW, and the perimeter is 2(L + W).
From the first condition, we have LW = 2(L + W).
Rearranging this equation, we get LW - 2L - 2W = 0.
Using the Simon's Favorite Factoring Trick, we can rewrite this equation as (L - 2)(W - 2) = 4.
Now, we need to find all possible pairs of integers (L, W) such that their product is 4. These pairs are (1, 4), (2, 2), and (4, 1).
Let's analyze each pair separately:
1. For (1, 4): If L = 1 and W = 4, the perimeter is 2(1 + 4) = 10, which does not equal the area of the rectangle (1 * 4 = 4). Therefore, this pair does not satisfy the conditions.
2. For (2, 2): If L = 2 and W = 2, the perimeter is 2(2 + 2) = 8, which does equal the area of the rectangle (2 * 2 = 4). Therefore, this pair satisfies the conditions.
3. For (4, 1): If L = 4 and W = 1, the perimeter is 2(4 + 1) = 10, which does not equal the area of the rectangle (4 * 1 = 4). Therefore, this pair does not satisfy the conditions.
Therefore, there are two rectangles that satisfy the conditions: one with sides 2 and 2, and one with sides 2 and 2 (since it is a rectangle, the order of the sides does not matter).
Hence, the correct answer is option C: Two.
Community Answer
How many rectangles with integral sides are possible where the area of...
Let the length and breadths of rectangle be a units and b units.
According to question,
ab = 2a + 2b
a = 2b / (b – 2)
Here, Two cases are possible
(i) a = 4, b = 4
(ii) a = 3 and b = 6
Thus, two different rectangles are possible.
According to question,
ab = 2a + 2b
a = 2b / (b – 2)
Here, Two cases are possible
(i) a = 4, b = 4
(ii) a = 3 and b = 6
Thus, two different rectangles are possible.
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