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Bunuel
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Can the experts explain the question in detail with the diagram, it is very confusing to imagine and understand.

Bunuel
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Quote:
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
E. 1580

hemantbafna
Hope the images clarify the form.

So Basically,
The question is asking for
Face1 + Face2 + Face3 + Face4 + Face5 + Face6
=400+200+260+260+180+180
=1480

Answer C
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BSTEPHANOS
Quote:
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
E. 1580

hemantbafna
Hope the images clarify the form.

So Basically,
The question is asking for
Face1 + Face2 + Face3 + Face4 + Face5 + Face6
=400+200+260+260+180+180
=1480

Answer C


Appreciate your hard work friend, awesome work on the figure!
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How did you get DE = 5?


Dillesh4096
Bunuel
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
E. 1580

In the isosceles trapezium ABCD, ADE is a right angled triangle.
--> \(AD^2 = AE^2 + ED^2 = 12^2 + 5^2 = 169\)
--> \(AD = EC = 13\)

Note that all the sides of the trapezium forms 4 vertical walls of the pillar with height 20 cm and 2 horizontal surfaces with area as the area of trapezium

Area of vertical surfaces = \(10*20 + 13*20 + 13*20 + 20*20 = 56*20 = 1120\) sq.cm
Area of horizontal surfaces = \(2[\frac{1}{2}*(10 + 20)*12] = 30*12 = 360\) sq.cm

--> Total area = \(1120 + 360 = 1480\) sq. cm

Option C
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