CAT 2017 Set-2 | Question: 82

Question

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

• A. $1300$
• B. $1340$
• C. $1480$
• D. $1520$

Answer 1

Let's first draw the diagram.

Let the length of the non-parallel side be $x$ cm.

Let $\text{DF} = \text{EC} = a\;\text{cm}.$

$\Rightarrow 2a = 20-10$

$\Rightarrow 2a = 10$

$\Rightarrow \boxed{a = 5\;\text{cm}}$

In $\triangle \text{AFD}$, apply the Pythagoras theorem

$(\text{AD})^{2} = (\text{AF})^{2}+(\text{DF})^{2}$

$\Rightarrow x^{2} = 12^{2}+5^{2}$

$\Rightarrow x^{2} = 144+125$

$\Rightarrow x^{2} = 169$

$\Rightarrow \boxed{x = 13\;\text{cm}}$              

The diagram below represents the cross-section of the pillar.

The total surface area of all the six surfaces $=$  Area of $4$ vertical surfaces with height $20$ cm and $2$ horizontal surfaces with the area as the area of trapezium.

$\qquad = \text{(Perimeter of the trapezium)} \times \text{Height of the pillar} + 2 \times \text{Area of trapezium}$

$\qquad  = (10+13+20+13) \times 20+2\times\frac{1}{2}(10+20)\times12$

$\qquad  = 56 \times 20 + 30 \times 12$

$\qquad = 1120 + 360$                    

$\qquad  = 1480\;\text{ cm}^{2}$          

$\therefore$ The total area, in sq cm, of all six surfaces of the pillar, is $1480.$

Correct Answer $:\text{C}$