The area of rectangle ABCD is 72 square inches. A diagonal of rectangle ABCD is 12 inches and the diagonal of rectangle EFGH is 22 inches. Find the area of rectangle EFGH. Round to the nearest square inch if necessary.
it should be 11
72 (22/12)^2 if similar rectangles
242 in^2
To find the area of rectangle EFGH, we can use the relationship between the diagonals and the areas of the rectangles.
Let's start by examining the relationship between the diagonals and the areas of rectangles.
For any rectangle, the square of the diagonal is equal to the sum of the squares of its side lengths. This is known as the Pythagorean theorem:
diagonal² = length² + width²
Let's represent the length and width of rectangle ABCD as l and w, respectively. From the given information, we have:
diagonal of ABCD = 12 inches
area of ABCD = 72 square inches
Using the Pythagorean theorem, we get:
12² = l² + w²
144 = l² + w² ----(Equation 1)
Now, let's examine the relationship between the diagonals and the areas of rectangles ABCD and EFGH.
The lengths and widths of rectangles ABCD and EFGH are proportional, which means we can express them as a ratio:
l₁ : l₂ = w₁ : w₂
where l₁ and w₁ represent the length and width of ABCD, and l₂ and w₂ represent the length and width of EFGH.
We know the ratio of the diagonals:
diagonal₁ : diagonal₂ = 12 : 22
Substituting the values of the diagonals, we have:
12 : 22 = √(l₁² + w₁²) : √(l₂² + w₂²)
Now we can solve for the area of EFGH (area₂) in terms of the given area of ABCD (area₁).
Since the area of a rectangle is equal to its length multiplied by its width, we can write:
area₁ = l₁ * w₁
area₂ = l₂ * w₂
We can rearrange the equation above to solve for area₂:
area₂ = area₁ * (l₂/l₁) * (w₂/w₁)
Substituting the lengths and widths of ABCD and EFGH, we have:
area₂ = 72 * (l₂/l₁) * (w₂/w₁) ----(Equation 2)
Now we need to find the values of l₂ and w₂ in terms of l₁ and w₁.
Since the rectangles have the same ratio of lengths to widths, we can express this relationship as:
l₂ = k * l₁
w₂ = k * w₁
where k represents the scaling factor.
Substituting these values into equation 2, we get:
area₂ = 72 * (k * l₁ / l₁) * (k * w₁ / w₁)
= 72 * k²
So, the area of rectangle EFGH is given by 72 * k².
Now, since we don't know the value of k, we need to determine it.
To find k, we can use the relationship between the diagonals of the rectangles:
diagonal₁ : diagonal₂ = l₁² + w₁² : l₂² + w₂²
Substituting the values, we have:
12 : 22 = l₁² + w₁² : (k * l₁)² + (k * w₁)²
12 : 22 = l₁² + w₁² : k² * (l₁² + w₁²)
Cross-multiplying, we get:
12 * k² * (l₁² + w₁²) = 22 * (l₁² + w₁²)
12k²l₁² + 12k²w₁² = 22l₁² + 22w₁²
Simplifying, we have:
(12k² - 22)l₁² + (12k² - 22)w₁² = 0
Since l₁ and w₁ cannot be zero (otherwise, the rectangles would be degenerate), we can divide the equation by (12k² - 22) to get:
l₁² + w₁² = 0 / (12k² - 22)
l₁² + w₁² = 0
Since the sum of two non-negative numbers cannot be zero, this equation implies that 12k² - 22 = 0.
Solving for k², we find:
12k² - 22 = 0
12k² = 22
k² = 22/12
k² = 11/6
k = √(11/6)
Now that we have determined the value of k, we can substitute it into equation 2 to find the area of rectangle EFGH:
area₂ = 72 * (√(11/6))²
Calculating the expression inside the parentheses:
(√(11/6))² = (11/6)
Substituting back into equation 2:
area₂ = 72 * (11/6)
area₂ = 132 square inches
Therefore, the area of rectangle EFGH is 132 square inches.