Question

a 2×2 grid of rectangles with areas: top - left 63, bottom - left 36, bottom - right 16, top - right blank. variables (e.g., x, y) are marked on horizontal/vertical sides of the grid.

Explanation:

Step1: Find the side length of the square with area 16

The area of a square is \( A = s^2 \), where \( s \) is the side length. For the square with area 16, we solve \( s^2 = 16 \). Taking the square root of both sides, \( s = \sqrt{16} = 4 \). So the horizontal side length (let's call it \( x \)) of the rectangle with area 36 and the square with area 16 is 4.

Step2: Find the vertical side length of the rectangle with area 36

The area of a rectangle is \( A = l \times w \), where \( l \) is length and \( w \) is width. For the rectangle with area 36 and width 4 (from Step 1), we have \( 36 = l \times 4 \). Solving for \( l \), we get \( l=\frac{36}{4}=9 \). So the vertical side length (let's call it \( y \)) of the rectangle with area 36 and the rectangle with area 63 is 9.

Step3: Find the area of the unknown rectangle

The unknown rectangle has length 9 (from Step 2) and width 4 (from Step 1). Using the area formula for a rectangle \( A = l \times w \), we get \( A = 9\times4 = 36 \)? Wait, no, wait. Wait, the rectangle with area 63: let's check again. Wait, the rectangle with area 63 has one side as 9 (from the rectangle with area 36, since they share the vertical side) and the other side (let's call it \( z \)) such that \( 63 = 9\times z \), so \( z=\frac{63}{9}=7 \)? Wait, maybe I misassigned. Wait, the figure is a 2x2 grid of rectangles/squares. Let's label the columns and rows. Let the bottom row have two rectangles: left with area 36, right with area 16. The top row: left with area 63, right unknown. Let the width of the bottom left rectangle be \( a \), bottom right be \( b \), height of bottom row be \( h \), and height of top row be \( k \). So for bottom left: \( a\times h = 36 \), bottom right: \( b\times h = 16 \). Top left: \( a\times k = 63 \), top right: \( b\times k =? \). From bottom right: \( b\times h = 16 \), so \( b = \frac{16}{h} \). From bottom left: \( a = \frac{36}{h} \). Top left: \( \frac{36}{h} \times k = 63 \), so \( \frac{k}{h}=\frac{63}{36}=\frac{7}{4} \). Then top right: \( b\times k=\frac{16}{h}\times k = 16\times\frac{k}{h}=16\times\frac{7}{4}=28 \). Wait, maybe my initial step was wrong. Let's do it with ratios. The ratio of the areas of the top left and bottom left rectangles is \( \frac{63}{36}=\frac{7}{4} \), which should be the ratio of their heights (since they share the same width). Then the top right and bottom right rectangles also share the same width, so their areas should have the same ratio of heights. So the area of the top right is \( 16\times\frac{7}{4}=28 \). Alternatively, find the side lengths. The bottom right square (wait, 16 is a square, so its side is 4. So the width of the bottom right is 4, so the width of the bottom left is also 4? No, wait, 36: if the height is, say, 9 (since 49=36), then the bottom left is 9x4, bottom right is 4x4 (since 44=16). Then the top left is 9x7 (since 9*7=63), so the top right is 4x7=28. Yes, that makes sense. So the unknown area is 28.

Answer:

28