Question

The figure below represents the plan and elevation of a square based prism of base 4 cm x 4 cm and height 20 cm. The prism is standing on the horizontal plane (HP) with its faces making an angle of 45 degrees with the vertical plane (VP). A cutting plane, perpendicular to VP, divides the prism as shown in the figure below.

The surface area (in cm\(^2\)) of the vertical faces of the prism below the cutting plane is (in integer).

Hint:

Read the cutting heights at edges 1 and 3 from the elevation (6 cm and 12 cm), average them for the middle edge (9 cm), then sum the four trapezoidal face areas (each width 4 cm).

Answer 1

Correct Answer: 144

Step 1: Understanding the Question.
A square prism of base 4 cm x 4 cm and height 20 cm stands vertically on HP, but is turned so its base faces make 45 degrees with VP. In plan, the square then appears as a diamond, with corners 1, 2, 3, 4 going around it. Corners 2 and 4 lie on the front-back diagonal (nearest to and farthest from VP), and corners 1 and 3 lie on the side diagonal. Since 1 and 3 sit on opposite sides of the centre, and 2, 4 sit exactly on the centre line between them, in the front view (elevation) edges 1 and 3 appear as the two outer vertical lines, and edges 2, 4 coincide as a single vertical line exactly midway between them. A cutting plane perpendicular to VP appears as a straight, sloping line in this elevation, and cuts the four vertical edges of the prism at different heights depending on their left-right position in the view.

Step 2: Read off the cutting heights from the figure.
From the figure, the sloping cutting-plane line crosses edge 1 (left, outermost) at a height of 6 cm and crosses edge 3 (right, outermost) at a height of 12 cm.

Step 3: Find the cutting height at the middle edge (2, 4).
Because the cutting plane trace is a straight line and edge (2,4) is exactly midway (in the left-right direction) between edges 1 and 3, its cutting height is the average of the two outer heights:
\[ h_{2,4} = \frac{6+12}{2} = 9 \text{ cm} \]
So the four vertical edges of the prism are cut at heights: edge 1 = 6 cm, edge 2 = 9 cm, edge 3 = 12 cm, edge 4 = 9 cm.

Step 4: Identify the shape of each face below the cutting plane.
The prism has 4 vertical faces, each a flat rectangle joining two adjacent edges (1-2, 2-3, 3-4, 4-1), each of width equal to the base edge, 4 cm. Since the two edges bounding a face are cut at different heights, the portion of each face below the cutting plane is a trapezium, with parallel sides equal to the two cut heights and the perpendicular distance between them equal to 4 cm (the base edge length).

Step 5: Find the area of each trapezoidal face.
Face 1-2 (heights 6 and 9):
\[ A_{1-2} = \frac{6+9}{2} \times 4 = 7.5 \times 4 = 30 \text{ cm}^2 \]
Face 2-3 (heights 9 and 12):
\[ A_{2-3} = \frac{9+12}{2} \times 4 = 10.5 \times 4 = 42 \text{ cm}^2 \]
Face 3-4 (heights 12 and 9):
\[ A_{3-4} = \frac{12+9}{2} \times 4 = 10.5 \times 4 = 42 \text{ cm}^2 \]
Face 4-1 (heights 9 and 6):
\[ A_{4-1} = \frac{9+6}{2} \times 4 = 7.5 \times 4 = 30 \text{ cm}^2 \]

Step 6: Add up all four face areas.
\[ \text{Total area} = 30+42+42+30 = 144 \text{ cm}^2 \]

Final Answer:
The surface area of the vertical faces of the prism below the cutting plane is 144 cm\(^2\). \[ \boxed{144 \text{ cm}^2} \]