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Total Surface Area of Trapezoidal Pillar

CAT 2017 Slot 2 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2017 Slot 2 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm10\text{ cm} and 20 cm20\text{ cm} while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm12\text{ cm}. If the height of the pillar is 20 cm20\text{ cm}, then the total area, in sq cm, of all six surfaces of the pillar is

  1. A.

    1300

  2. B.

    1340

  3. C.

    1480

  4. D.

    1520

Answer

C

Explanation

First, find the slant (non-parallel) sides of the trapezium base. The difference between parallel sides is 20−10=10 cm20 - 10 = 10\text{ cm}. Since it is an isosceles trapezium, each non-parallel side extends horizontally by 10/2=5 cm10 / 2 = 5\text{ cm}. Using Pythagoras theorem, non-parallel side length s=52+122=13 cms = \sqrt{5^2 + 12^2} = 13\text{ cm}.

Perimeter of trapezium base = 10+20+13+13=56 cm10 + 20 + 13 + 13 = 56\text{ cm}.

Area of trapezium base = 12×(10+20)×12=180 sq cm\frac{1}{2} \times (10 + 20) \times 12 = 180\text{ sq cm}.

Total surface area of the pillar = 2×(Base Area)+(Base Perimeter×Height)2 \times (\text{Base Area}) + (\text{Base Perimeter} \times \text{Height}) =2×180+56×20=360+1120=1480 sq cm.= 2 \times 180 + 56 \times 20 = 360 + 1120 = 1480\text{ sq cm}.

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