Question

the area of square abfe is 49 square units. given ag = 9 units and ac = 10 units, what fraction of the area of rectangle ackg is represented by the shaded region? the fraction of the area of ackg that is represented by the shaded region is (simplify your answer.)

Explanation:

Step1: Find side length of square

The area of square \(ABFE\) is 49 square units. Since the area of a square is \(s^2\) (where \(s\) is the side length), we solve \(s^2 = 49\). Taking the square root, \(s=\sqrt{49} = 7\) units. So the side length of the square (e.g., \(AB = BF = FE = EA = 7\))? Wait, no, looking at the diagram, the square is \(ABFE\)? Wait, actually, the square has area 49, so side length is 7. Now, the rectangle \(ACIG\) has length \(AG = 9\) units? Wait, no, the problem says \(AG = 9\) units? Wait, no, the problem states: "Given \(AG = 9\) units and \(AC = 10\) units"? Wait, maybe I misread. Wait, the area of square \(ABFE\) is 49, so side length \(AB = 7\) (since \(7\times7 = 49\)). Now, the rectangle \(ACIG\): let's see, the length \(AC = 10\) units? Wait, no, maybe \(AG = 9\) and \(AC = 10\)? Wait, perhaps the rectangle \(ACIG\) has length \(AC = 10\) and width \(AG = 9\)? No, wait, the square \(ABFE\) has side 7, so maybe the white square at the bottom right is the square \(ABFE\)? Wait, no, the shaded regions: one horizontal and one vertical. Let's think again.

Wait, the area of square \(ABFE\) is 49, so side length \(s = 7\) (since \(s^2 = 49\)). Now, the rectangle \(ACIG\): let's find its area. Wait, the problem says \(AG = 9\) units and \(AC = 10\) units? Wait, maybe \(AC\) is the length and \(AG\) is the width? Wait, no, maybe the rectangle \(ACIG\) has length \(AC = 10\) and width \(AG = 9\)? Wait, no, the square \(ABFE\) has side 7, so the white square (the unshaded one) is \(ABFE\) with area 49? Wait, no, the shaded regions: the horizontal shaded rectangle and the vertical shaded rectangle. Let's find the area of the shaded regions.

First, the area of rectangle \(ACIG\) is length \(\times\) width. Wait, if \(AG = 9\) and \(AC = 10\), then area of \(ACIG\) is \(10\times9 = 90\)? No, that can't be. Wait, maybe \(AG = 9\) and the side of the square is 7, so the horizontal shaded rectangle: length is \(AG - 7 = 9 - 7 = 2\)? No, wait, maybe the square \(ABFE\) has side 7, so the vertical shaded rectangle: height is \(AC - 7 = 10 - 7 = 3\)? Wait, no, let's re-express.

Wait, the square \(ABFE\) has area 49, so side \(AB = 7\). Let's assume that \(AC = 10\) (length of the rectangle) and \(AG = 9\) (width of the rectangle). Then the area of rectangle \(ACIG\) is \(10\times9 = 90\)? No, that doesn't make sense. Wait, maybe the square is \(ABFE\) with side 7, so the white square (the small unshaded square) is \(EFHI\) or something, with side 7? Wait, no, the shaded regions: one is a horizontal rectangle (let's say \(DGH E\)) and one is a vertical rectangle (let's say \(B D F E\))? Wait, maybe the area of the shaded regions is the area of the two shaded rectangles. Let's calculate:

The square \(ABFE\) has area 49, so side 7. The rectangle \(ACIG\) has length \(AC = 10\) and width \(AG = 9\)? Wait, no, maybe \(AC\) is the height and \(AG\) is the length. Wait, perhaps the area of rectangle \(ACIG\) is \(AC \times AG = 10 \times 9 = 90\)? No, that seems too big. Wait, maybe the square is \(ABFE\) with side 7, so the horizontal shaded rectangle: length is \(AG - 7 = 9 - 7 = 2\), and height is 7? No, the vertical shaded rectangle: height is \(AC - 7 = 10 - 7 = 3\), and length is 7? Wait, no, the shaded area would be (horizontal shaded area) + (vertical shaded area) - (overlap, but since they overlap at the square, no, wait, the two shaded rectangles: one is \(DGH E\) (horizontal) with length \(AG - 7 = 9 - 7 = 2\) and height equal to the side of the square? No, maybe the area of the shaded region is (area of rectangle \(ACIG\) - area of square \(ABFE\) - area of the small white square? Wait, no, the problem says "the fraction of the area of rectangle \(ACIG\) is represented by the shaded region".

Wait, let's start over:

1. Area of square \(ABFE = 49\), so side length \(s = \sqrt{49} = 7\) units.

2. Area of rectangle \(ACIG\): let's find its length and width. From the diagram, \(AC\) is one side, \(AG\) is the other. Wait, the problem states \(AG = 9\) units and \(AC = 10\) units? Wait, maybe \(AC = 10\) (length) and \(AG = 9\) (width), so area of \(ACIG\) is \(10 \times 9 = 90\)? No, that can't be. Wait, maybe \(AG = 9\) and \(AC = 10\) are the length and width, so area is \(9 \times 10 = 90\). Then the square \(ABFE\) has area 49, so the white regions: the square \(ABFE\) and the small white square at the bottom right (which has side length \(9 - 7 = 2\) and \(10 - 7 = 3\)? No, that doesn't make sense. Wait, maybe the shaded area is (area of \(ACIG\) - area of square \(ABFE\) - area of the small white rectangle). Wait, the small white rectangle: length is \(9 - 7 = 2\), width is \(10 - 7 = 3\)? No, area would be \(2 \times 3 = 6\). Then total white area is \(49 + 6 = 55\), so shaded area is \(90 - 55 = 35\). Then the fraction is \(35/90 = 7/18\)? No, that doesn't seem right.

Wait, maybe I misread the problem. Let's check again: "The area of square \(ABFE\) is 49 square units. Given \(AG = 9\) units and \(AC = 10\) units, what fraction of the area of rectangle \(ACIG\) is represented by the shaded region?"

Wait, square \(ABFE\) has side 7 (since \(7^2 = 49\)). So in rectangle \(ACIG\), the length \(AC = 10\), width \(AG = 9\). The shaded regions: one is a vertical rectangle (let's say \(B D F E\)) with height \(AC - 7 = 10 - 7 = 3\) and width 7, and one is a horizontal rectangle (let's say \(DGH E\)) with length \(AG - 7 = 9 - 7 = 2\) and width 7? No, that would overlap. Wait, no, the shaded area is actually (area of vertical shaded) + (area of horizontal shaded) - (area of the overlapping square, but there is no overlapping). Wait, no, the two shaded rectangles: the vertical one is \(B D F E\) with dimensions \(7 \times (10 - 7) = 7 \times 3 = 21\), and the horizontal one is \(DGH E\) with dimensions \(7 \times (9 - 7) = 7 \times 2 = 14\). Then total shaded area is \(21 + 14 = 35\). The area of rectangle \(ACIG\) is \(10 \times 9 = 90\). Then the fraction is \(35/90 = 7/18\)? Wait, but that seems off. Wait, maybe the rectangle \(ACIG\) has length \(AG = 9\) and width \(AC = 10\), so area \(9 \times 10 = 90\). The square \(ABFE\) has area 49, and the small white rectangle has area \((9 - 7) \times (10 - 7) = 2 \times 3 = 6\). So total white area is \(49 + 6 = 55\), so shaded area is \(90 - 55 = 35\). Then \(35/90 = 7/18\). Wait, but let's check again.

Wait, maybe the square \(ABFE\) is inside the rectangle \(ACIG\), with \(AB = 7\), so the horizontal shaded region is from \(G\) to \(H\) (length \(9 - 7 = 2\)) and height equal to the side of the square? No, maybe the height of the horizontal shaded region is \(7\), and the width is \(9 - 7 = 2\), so area \(2 \times 7 = 14\). The vertical shaded region is from \(B\) to \(D\) (height \(10 - 7 = 3\)) and width \(7\), so area \(3 \times 7 = 21\). Then total shaded area is \(14 + 21 = 35\). Area of rectangle \(ACIG\) is \(10 \times 9 = 90\). So the fraction is \(35/90 = 7/18\). Wait, but let's confirm.

Alternatively, maybe the rectangle \(ACIG\) has length \(AC = 10\) and width \(AG = 9\), so area \(10 \times 9 = 90\). The square \(ABFE\) has area 49, and the small white rectangle has area \((10 - 7) \times (9 - 7) = 3 \times 2 = 6\). So total white area is \(49 + 6 = 55\), so shaded area is \(90 - 55 = 35\). Then \(35/90 = 7/18\). Yes, that makes sense.

Step1: Calculate side of square

Area of square \(ABFE = 49\), so side \(s = \sqrt{49} = 7\) units.

Step2: Area of rectangle \(ACIG\)

Length \(AC = 10\), width \(AG = 9\), so area \(A_{ACIG} = 10 \times 9 = 90\) square units.

Step3: Area of small white rectangle

Length of small white rectangle: \(AG - 7 = 9 - 7 = 2\) units.
Width of small white rectangle: \(AC - 7 = 10 - 7 = 3\) units.
Area of small white rectangle: \(2 \times 3 = 6\) square units.

Step4: Area of square \(ABFE\)

Given as 49 square units.

Step5: Total white area

Total white area = Area of square \(ABFE\) + Area of small white rectangle = \(49 + 6 = 55\) square units.

Step6: Area of shaded region

Shaded area = Area of \(ACIG\) - Total white area = \(90 - 55 = 35\) square units.

Step7: Fraction of shaded region

Fraction = \(\frac{\text{Shaded Area}}{\text{Area of } ACIG} = \frac{35}{90} = \frac{7}{18}\) (simplified by dividing numerator and denominator by 5).

Answer:

\(\boxed{\dfrac{7}{18}}\)