Let's assume the width of the rectangle is x inches.
According to the problem, the length of the rectangle is 2 inches more than the width, so the length would be x + 2 inches.
We can use the Pythagorean theorem to find the length of the diagonal:
(x + 2)^2 + x^2 = (20 - √2)^2
Simplifying the equation, we get:
2x^2 + 4x + 4 = 416 - 40√2 + 2
2x^2 + 4x - 414 = -40√2
We can ignore the negative solution since length cannot be negative. So we have:
2x^2 + 4x - 414 = 40√2
x^2 + 2x - 207 = 20√2
To solve this equation, we can use the quadratic formula:
x = (-2 ± √(2^2 - 4(1)(-207))) / 2(1)
x = (-2 ± √(4 + 828)) / 2
x = (-2 ± √(832)) / 2
x = (-2 ± 4√13) / 2
x = -1 ± 2√13
Since x = width, we can ignore the negative solution. Therefore, the width is -1 + 2√13 inches.
The length would be x + 2, so the length is -1 + 2√13 + 2 = 1 + 2√13 inches.
The area of the rectangle is given by the formula: Area = length × width
Area = (1 + 2√13)(-1 + 2√13)
Area = -1 + 2√13 + 2√13 - 4(√13)^2
Area = -1 + 4√13 - 4(13)
Area = -1 + 4√13 - 52
Area = -53 + 4√13
Therefore, the area of the rectangle is -53 + 4√13 square inches.
The length of a rectangle is 2 inches more than the width. If the diagonal is 20−−√ inches, find the area of the rectangle.
Responses
A 4 in24 in 2
B 8 in28 in 2
C 16 in216 in 2
D 32 in2
1 answer