Question

what is the area of a cross section of the cube that contains points a, b, e and h?
diagram of a cube with side length 5 m, labeled with points a, b, c, d, e, f, g, h
enter your answer in the box. round your answer to the nearest square meter.
□ m²

Explanation:

Step1: Identify the shape of the cross - section

The cross - section containing points \(A\), \(B\), \(E\), and \(H\) is a rectangle? Wait, no. Let's analyze the cube. The side length of the cube is \(s = 5\) m. The cross - section \(ABEH\): \(AB\) is a side of the cube with length \(5\) m. To find the length of \(BE\) (or \(AH\)), we can use the Pythagorean theorem in the right triangle formed. In the cube, the distance from \(B\) to \(E\): the horizontal distance from \(B\) to \(C\) is \(5\) m and the vertical distance from \(C\) to \(E\) is \(5\) m? Wait, no. Wait, \(AB\) is along the base, length \(5\) m. The segment \(BE\): actually, the cross - section \(ABEH\) is a parallelogram? Wait, no, in a cube, the cross - section through \(A\), \(B\), \(E\), \(H\) is a rectangle? Wait, no, let's find the length of the diagonal of the face or the space diagonal? Wait, no. Let's consider the sides of the cross - section. \(AB\) has length \(5\) m. The length of \(BH\) (or \(AE\)): in the cube, the vector from \(B\) to \(H\): the \(x\) - component is \(0\), \(y\) - component is \(5\) (from \(B\) to \(G\) to \(H\)) and \(z\) - component is \(5\) (from \(B\) to \(C\) to \(E\) to \(H\))? Wait, no, let's use the Pythagorean theorem for the length of the side of the cross - section.

Wait, the cross - section \(ABEH\): \(AB\) is a side of the cube, length \(a = 5\) m. The other side: let's consider the triangle formed by \(AB\), \(BB_1\) (where \(B_1\) is a point) and the diagonal. Wait, actually, the cross - section \(ABEH\) is a rectangle? No, it's a parallelogram with sides \(AB = 5\) m and \(BE=\sqrt{5^{2}+5^{2}}=\sqrt{25 + 25}=\sqrt{50}=5\sqrt{2}\) m? Wait, no, maybe I made a mistake. Wait, the cross - section through \(A\), \(B\), \(E\), \(H\): let's look at the coordinates. Let's assign coordinates: let \(A=(0,0,0)\), \(B=(5,0,0)\), \(C=(5,5,0)\), \(D=(0,5,0)\), \(E=(0,5,5)\), \(H=(5,5,5)\), \(G=(5,0,5)\), \(F=(0,0,5)\). Then the coordinates of \(A=(0,0,0)\), \(B=(5,0,0)\), \(E=(0,5,5)\), \(H=(5,5,5)\). Now, the vector \(\overrightarrow{AB}=(5,0,0)\) and \(\overrightarrow{BE}=(- 5,5,5)\)? No, wait, from \(B\) to \(E\): \(E - B=(0 - 5,5 - 0,5 - 0)=(-5,5,5)\). Wait, no, that's not right. Wait, \(E\) is at \((0,5,5)\) and \(B\) is at \((5,0,0)\), so the distance between \(B\) and \(E\) is \(\sqrt{(0 - 5)^{2}+(5 - 0)^{2}+(5 - 0)^{2}}=\sqrt{25 + 25+25}=\sqrt{75}=5\sqrt{3}\)? No, that can't be. Wait, maybe the cross - section is a rectangle with length \(AB = 5\) and width equal to the length of \(BE\), where \(BE\) is the diagonal of a square face? Wait, no, let's re - examine the cube. The cross - section \(ABEH\): \(AB\) is along the bottom edge, \(BE\) is a space diagonal? No, maybe the cross - section is a parallelogram with base \(AB = 5\) and height equal to the length of the face diagonal. Wait, the face diagonal of a square with side \(5\) is \(5\sqrt{2}\). Wait, no, let's calculate the area of the cross - section.

Wait, the cross - section \(ABEH\) is a rectangle? Wait, no, in the cube, the four points \(A\), \(B\), \(E\), \(H\): \(AB\) is parallel to \(EH\) and \(BE\) is parallel to \(AH\). So it's a parallelogram. The area of a parallelogram is base \(\times\) height. Let's take \(AB\) as the base, length \(b = 5\) m. The height is the distance between the lines \(AB\) and \(EH\). To find this height, we can consider the right triangle formed. The vector perpendicular to \(AB\) in the plane of the cross - section. Alternatively, we can find the length of the other side. Wait, the length of \(BE\): in the cube, from \(B\) to \(E\), the horizontal (in the \(x - y\)) plane distance from \(B\) to \(E\) is \(\sqrt{(0 - 5)^{2}+(5 - 0)^{2}}=\sqrt{25 + 25}=\sqrt{50}=5\sqrt{2}\), and the vertical distance is \(5\)? No, no, \(E\) is at \(z = 5\) and \(B\) is at \(z = 0\), so the \(z\) - component is \(5\). Wait, no, coordinates: let's set \(A=(0,0,0)\), \(B=(5,0,0)\), \(C=(5,5,0)\), \(D=(0,5,0)\), \(E=(0,5,5)\), \(F=(0,0,5)\), \(G=(5,0,5)\), \(H=(5,5,5)\). Then vector \(AB=(5,0,0)\), vector \(BE=(-5,5,5)\). The area of the parallelogram is the magnitude of the cross - product of \(AB\) and \(BE\).

\(\overrightarrow{AB}=(5,0,0)\), \(\overrightarrow{BE}=(-5,5,5)\)

The cross - product \(\overrightarrow{AB}\times\overrightarrow{BE}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&0&0\\-5&5&5\end{vmatrix}=\mathbf{i}(0\times5 - 0\times5)-\mathbf{j}(5\times5 - 0\times(-5))+\mathbf{k}(5\times5 - 0\times(-5))=(0,- 25,25)\)

The magnitude of the cross - product is \(\sqrt{0^{2}+(-25)^{2}+25^{2}}=\sqrt{625 + 625}=\sqrt{1250}=25\sqrt{2}\approx35.355\)

Wait, but that seems complicated. Wait, maybe the cross - section is a rectangle with length \(5\) and width \(5\sqrt{2}\)? No, wait, another approach: the cross - section \(ABEH\) is a rectangle? Wait, no, \(AB\) is horizontal, \(BE\) is a diagonal of a rectangular face? Wait, no, in the cube, the cross - section through \(A\), \(B\), \(E\), \(H\) is a parallelogram. But maybe it's a rectangle with length \(AB = 5\) and the other side is the length of the diagonal of the square with side \(5\), i.e., \(5\sqrt{2}\). Then the area would be \(5\times5\sqrt{2}=25\sqrt{2}\approx35\). Wait, let's check again.

Wait, the side length of the cube is \(5\) m. The cross - section \(ABEH\): \(AB\) is length \(5\) m. The segment \(BE\): in the cube, from \(B\) to \(E\), we can consider the right triangle where one leg is the side of the cube (\(5\) m) and the other leg is the diagonal of the square face (\(5\sqrt{2}\) m)? No, that's not right. Wait, no, the cross - section is a rectangle? Wait, no, let's look at the figure. The cube has edges of length \(5\) m. The cross - section containing \(A\), \(B\), \(E\), \(H\): \(AB\) is a horizontal edge, \(BE\) is a vertical - horizontal edge? Wait, maybe the cross - section is a parallelogram with base \(AB = 5\) and height equal to the length of the face diagonal of the square face (since the height is the distance between the two parallel sides \(AB\) and \(EH\)). The face diagonal of a square with side \(5\) is \(5\sqrt{2}\), so the area is \(5\times5\sqrt{2}=25\sqrt{2}\approx35.36\), which rounds to \(35\) or \(35\) square meters. Wait, but let's do it correctly.

Wait, the cross - section \(ABEH\) is a rectangle? No, it's a parallelogram. But actually, in the cube, the four points \(A\), \(B\), \(E\), \(H\) form a rectangle? Wait, \(AB\) is parallel to \(EH\) and \(BE\) is parallel to \(AH\). Let's check the angles. The vector \(AB=(5,0,0)\) and vector \(BE=(-5,5,5)\). The dot - product of \(AB\) and \(BE\) is \(5\times(-5)+0\times5 + 0\times5=-25 eq0\), so they are not perpendicular. So it's a parallelogram. The area of a parallelogram is \(|\overrightarrow{AB}|\times|\overrightarrow{BE}|\times\sin\theta\), where \(\theta\) is the angle between them.

\(|\overrightarrow{AB}| = 5\), \(|\overrightarrow{BE}|=\sqrt{(-5)^{2}+5^{2}+5^{2}}=\sqrt{25 + 25+25}=\sqrt{75}=5\sqrt{3}\)

The dot - product \(\overrightarrow{AB}\cdot\overrightarrow{BE}=-25\)

\(\cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{BE}}{|\overrightarrow{AB}|\times|\overrightarrow{BE}|}=\frac{-25}{5\times5\sqrt{3}}=\frac{-1}{\sqrt{3}}\)

\(\sin\theta=\sqrt{1-\cos^{2}\theta}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\)

Then the area is \(5\times5\sqrt{3}\times\frac{\sqrt{6}}{3}=25\times\frac{\sqrt{18}}{3}=25\times\frac{3\sqrt{2}}{3}=25\sqrt{2}\approx35.36\), which rounds to \(35\) square meters.

Step2: Calculate the area

We found that the area of the cross - section (parallelogram) is \(25\sqrt{2}\approx35.36\), which rounds to \(35\) (or \(35\) square meters).

Answer:

\(35\)