Find the area of a rectangle with length 4 and diagonal 5...
I thought this was impossible because the diagonal must be longer than the length of a side so i'm not sure what to do. Please help!
did you draw a diagram. always draw a diagram if possible--especially for a geometry question. if we consider just a half of the rectangle, that is, the portion on either side of the diagonal, we realize that it forms a right-angled triangle with a hypotenuse of 5 and a base of 4 (draw a diagram to see this). just find the area of that triangle and multiply it by two. if you are still having problems, get back to me.
Hello,
1. The area of a rectangle can be calculated by: A = length * width.
You know already: l = 4
2. The diagonal d divides the rectangle into 2 right triangles. The diagonal is the hypotenuse of the triangle. Use Pythagorean theorem to calculate the missing side w:
d² = l² + w² ==> w² = d² - l² ==> w = √(d² - l²)
Therefore w = √(5² - 4²) = √(25 - 16) = √(9) = 3
3. Plug in the values into the formula: A = 4 * 3 = 12 square units