abcd is a square picture frame. efgh is a square inscribed within abcd as a space for a picture. the area of efgh (for the picture) is equal to the area of the picture frame (the area of abcd minus the area of efgh). if AB=6, what is the length of EF
efgh has half the area of abcd, so its side is 6/√2.
To find the length of EF, we need to use the information given in the problem and apply some geometric principles.
Let's break down the problem step by step:
1. We are told that ABCD is a square picture frame, and EFGH is a square inscribed within ABCD. This means that the vertices of EFGH touch the midpoints of the sides of ABCD.
2. We are also given that the area of EFGH is equal to the area of ABCD minus the area of EFGH. In other words, Area(EFGH) = Area(ABCD - EFGH).
3. If we let the length of a side of ABCD be x, then the area of ABCD is x^2.
4. The area of EFGH is equal to the square of the length of any side of EFGH, which we'll call y. So, Area(EFGH) = y^2.
5. We can now rewrite the equation for the area: y^2 = x^2 - y^2.
6. Rearranging the equation, we have: 2y^2 = x^2.
7. We are given that AB = 6. Since ABCD is a square, all sides are equal, so x = 6.
8. Substituting x = 6 into the equation, we have: 2y^2 = 6^2.
9. Simplifying, we get: 2y^2 = 36.
10. Dividing both sides by 2, we have: y^2 = 18.
11. Taking the square root of both sides, we find: y ≈ 4.24.
Therefore, the length of EF, which is equal to y, is approximately 4.24.