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03 Apr 2020, 04:23
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C
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Difficulty:
75%
(hard)
Question Stats:
64% (02:59) correct
36%
(02:40)
wrong
based on 64
sessions
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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
Are You Up For the Challenge: 700 Level Questions
A. 1300
B. 1340
C. 1480
D. 1520
E. 1580
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 09:58
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Bunuel
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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04 Apr 2020, 09:26
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Given, the non-parallel sides are equal. Let the non-parallel sides be x cm each
x= √(12^2 + 5^2) = 13
So, we have 6 faces, out of which 2 are trapezoid faces and 4 are rectangular faces.
Area of trapezium = 1/2(sum of two parallel sides)(height)
Area of 2 trapeziums
= 2[(1/2)(12)(10+20)] = 360 cm^2
Area of rectangle = base*height
Area of 4 rectangles
= 2[13 × 20] + 20(20) + 10(20) = 1120 cm^2
Total area = 1120 + 360 = 1480 cm^2
Option (C)
x= √(12^2 + 5^2) = 13
So, we have 6 faces, out of which 2 are trapezoid faces and 4 are rectangular faces.
Area of trapezium = 1/2(sum of two parallel sides)(height)
Area of 2 trapeziums
= 2[(1/2)(12)(10+20)] = 360 cm^2
Area of rectangle = base*height
Area of 4 rectangles
= 2[13 × 20] + 20(20) + 10(20) = 1120 cm^2
Total area = 1120 + 360 = 1480 cm^2
Option (C)
