Consider the rectangular box shown below. Draw the views as seen from X and Y. Find the: total surface area, volume, diagonal HA, angle the diagonal makes with the base.

a)
The rectangular box has dimensions: Length ($L$) = 6 m, Width ($W$) = 7 m, Height ($H$) = 8 m.

• The view from X (front view) shows the face ABEH. This is a rectangle with dimensions $L \times H$, which is $6 \text{ m} \times 8 \text{ m}$.
• The view from Y (right side view) shows the face BCFE. This is a rectangle with dimensions $W \times H$, which is $7 \text{ m} \times 8 \text{ m}$.

b)(i)
The formula for the total surface area ($SA$) of a rectangular box is $SA = 2(LW + LH + WH)$.

$$SA = 2((6 \text{ m})(7 \text{ m}) + (6 \text{ m})(8 \text{ m}) + (7 \text{ m})(8 \text{ m}))$$ $$SA = 2(42 \text{ m}^2 + 48 \text{ m}^2 + 56 \text{ m}^2)$$ $$SA = 2(146 \text{ m}^2)$$ $$SA = \boxed{292 \text{ m}^2}$$

b)(ii)
The formula for the volume ($V$) of a rectangular box is $V = LWH$.

$$V = (6 \text{ m})(7 \text{ m})(8 \text{ m})$$ $$V = (42 \text{ m}^2)(8 \text{ m})$$ $$V = \boxed{336 \text{ m}^3}$$

b)(iii)
The question asks for "diagonal HA". In the provided diagram, HA is a vertical edge of the box, with length 8 m. However, the term "diagonal" usually refers to a line segment connecting two non-adjacent vertices. Given part (iv) asks for the angle this diagonal makes with the base, which is typically a non-90 degree angle for a space diagonal, we will assume "diagonal HA" is a mislabeling and refers to the space diagonal of the rectangular box (e.g., from A to F, the opposite vertex).
The length of the space diagonal ($d$) is given by $d = \sqrt{L^2 + W^2 + H^2}$.
Here, $L = 6$ m, $W = 7$ m, $H = 8$ m.

$$d = \sqrt{(6 \text{ m})^2 + (7 \text{ m})^2 + (8 \text{ m})^2}$$ $$d = \sqrt{36 \text{ m}^2 + 49 \text{ m}^2 + 64 \text{ m}^2}$$ $$d = \sqrt{149 \text{ m}^2}$$ $$d = \boxed{\sqrt{149} \text{ m} \approx 12.21 \text{ m}}$$

b)(iv)
We need to find the angle ($\theta$) that the space diagonal (e.g., AF) makes with the base (ABCD).
Consider the right-angled triangle formed by the space diagonal (AF), the height of the box (CF = H), and the diagonal of the base (AC).
First, calculate the length of the base diagonal (AC).

$$AC = \sqrt{L^2 + W^2}$$ $$AC = \sqrt{(6 \text{ m})^2 + (7 \text{ m})^2}$$ $$AC = \sqrt{36 \text{ m}^2 + 49 \text{ m}^2}$$ $$AC = \sqrt{85 \text{ m}^2}$$ $$AC = \sqrt{85} \text{ m}$$

Now, in the right triangle formed by A, C, and F, we have:
Opposite side to $\theta$ is CF (height) = 8 m.
Adjacent side to $\theta$ is AC (base diagonal) = $\sqrt{85}$ m.
The tangent of the angle $\theta$ is given by $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$.

$$\tan \theta = \frac{H}{AC}$$ $$\tan \theta = \frac{8 \text{ m}}{\sqrt{85} \text{ m}}$$ $$\theta = \arctan\left(\frac{8}{\sqrt{85}}\right)$$ $$\theta \approx \arctan\left(\frac{8}{9.2195}\right)$$ $$\theta \approx \arctan(0.8677)$$ $$\theta \approx \boxed{41.0^{\circ}}$$