The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is
- 1300
- 1340
- 1480
- 1520
The Correct Option is C
Solution and Explanation
The problem involves calculating the total surface area of a vertical pillar with a trapezoidal base. We start by finding the area of the trapezium, which is the base of the pillar.
The area \(A\) of a trapezium with bases \(a\) and \(b\), and height \(h\) is given by:
\(A = \frac{1}{2} \times (a+b) \times h\)
Here, \(a = 10 \, \text{cm}\), \(b = 20 \, \text{cm}\), and \(h = 12 \, \text{cm}\).
Substituting the values:
\(A = \frac{1}{2} \times (10+20) \times 12 = \frac{1}{2} \times 30 \times 12 = 180 \, \text{sq cm}\)
Next, calculate the lateral surface area of the pillar. The pillar is 20 cm high, making the lateral surface area equal to the perimeter of the trapezium times the height of the pillar.
Find the length of the non-parallel sides of the trapezium using the Pythagorean theorem.
Let \(l\) be the length of the non-parallel sides. Since bases are offset, the triangular height calculated for the right triangles formed by non-parallel sides is:
\((20-10)/2 = 5 \, \text{cm}\)
Using the Pythagorean theorem, we have:
\(l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \text{cm}\)
The perimeter \(P\) of the trapezium is:
\(P = 10 + 20 + 13 + 13 = 56 \, \text{cm}\)
Thus, the lateral surface area is:
\(L = P \times 20 = 56 \times 20 = 1120 \, \text{sq cm}\)
The total surface area is the sum of the base area, the top area, and the lateral surface area. Since top and base are identical trapeziums:
Total surface area \(S\) is:
\(S = 2 \times 180 + 1120 = 360 + 1120 = 1480 \, \text{sq cm}\)
Therefore, the total area of all six surfaces of the pillar is: 1480 sq cm
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