What is the area of a rectangle with 4.5 for diagonals and 4 for width
You need to use the pythagorean theorem to find length of the base. c2 = a2 +b2
Then use (bxh)/2 to find the area
back from my break.
I'm not sure if I put the values in their correct spots in the Pythagorean theorem correctly but,
4^2 + 4.5^2 = 36.25
(bxh)/2
(36.25 * 4) / 2 = 72.5
h^2 + W^2 = (4.5)^2.
W = 4.
h = ?
Area = h * W.
To find the area of a rectangle, you need to know the length and width of the rectangle. In this case, we are given the width of the rectangle, which is 4.
To find the length of the rectangle, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. In a rectangle, the diagonals are also the hypotenuses of two right triangles formed by the width and length sides.
In this case, the diagonal of the rectangle is given as 4.5 units. Let's assume the length of the rectangle is L. Then we have the following equation using the Pythagorean theorem:
4.5^2 = 4^2 + L^2
20.25 = 16 + L^2
Subtracting 16 from both sides:
L^2 = 20.25 - 16
L^2 = 4.25
Taking the square root of both sides:
L = √4.25 ≈ 2.06
Now that we know both the width and length of the rectangle, we can find the area. The area of a rectangle is given by the formula:
Area = Length × Width
Substituting the values:
Area = 2.06 × 4 ≈ 8.23 square units
Therefore, the area of the rectangle is approximately 8.23 square units.