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Updated on: 09 Jun 2019, 07:19
Originally posted by CareerGeek on 08 Jun 2019, 22:23.
Last edited by CareerGeek on 09 Jun 2019, 07:19, edited 1 time in total.
Last edited by CareerGeek on 09 Jun 2019, 07:19, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
52% (02:21) correct
48%
(02:31)
wrong
based on 27
sessions
History
Date
Time
Result
Not Attempted Yet
Find the area of the trapezium ABCD if AB ll CD
A. 27
B. 21.6
C. 22.8
D. 24.2
E. 25.6
A. 27
B. 21.6
C. 22.8
D. 24.2
E. 25.6
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19.png [ 7.26 KiB | Viewed 1708 times ]
08 Jun 2019, 23:11
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Dillesh4096
Be prepared to visualize a figure in different form
(I) Let us modify the trapezium first
draw the perpendicular AE and BF as shown. They are also equal to height or h
Now we have
a) A rectangle AEFB with side 7 and 4
b) Two triangles CAE and BDF
(II) Place AE over BF, so we have a triangle CAD with sides 3-4-(a+b).
Now a+b is nothing but 12-7=5 as a+7+b=12, so CAD becomes a triangle with sides 3-4-5, which is a right angled triangle.
3-4-5 is a triplet for a right angles triangle
(III) Area of CAD=\((\frac{1}{2})*3*4=6\)
But if we take base as 5 and altitude as h, the area becomes \((\frac{1}{2})*(CE+FD)*AE=(\frac{1}{2})*5*AE\)..
Now, both these areas should be equal, so \((\frac{1}{2})*5*AE=6... AE=12/5=2.4\)
(IV) Area of rectangle AEFB=\(7*h=7*2.4=16.8\)
(V) Thus, Total area = area of AEFD+area of CAD=16.8+6=22.8
C
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chetan.png [ 68.1 KiB | Viewed 1611 times ]
09 Feb 2022, 09:58
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Hi,
we create a rectangle and two triangles like in chetan's first picture.
We know:
(1) h^2+b^2=16 -> h^2=16-b^2
(2) h^2+a^2=9 -> h^2=9-a^2
(3) a+b=5 -> a=5-b
Plugging in (3) into (2) and then equating (1) and (2), we get:
16-b^2=9-(5-b)^2
7-b^2=-25+10b-b^2
10b=32
b=3.2 -> a=1.8
Using the pythagorean on the left, smaller triangle (with side length 3), we get:
root(3^2-1.8^2)=h
root(9-3.24)=h
root(5.76)=h, we know that root(576)=24, so
2.4=h
Area of a Trapezoid:
((Length of Base 1+Length of Base 2)/2)*h -> ((7+12)/2)*2.4=9.5*2.4=22.8 -> (C)
we create a rectangle and two triangles like in chetan's first picture.
We know:
(1) h^2+b^2=16 -> h^2=16-b^2
(2) h^2+a^2=9 -> h^2=9-a^2
(3) a+b=5 -> a=5-b
Plugging in (3) into (2) and then equating (1) and (2), we get:
16-b^2=9-(5-b)^2
7-b^2=-25+10b-b^2
10b=32
b=3.2 -> a=1.8
Using the pythagorean on the left, smaller triangle (with side length 3), we get:
root(3^2-1.8^2)=h
root(9-3.24)=h
root(5.76)=h, we know that root(576)=24, so
2.4=h
Area of a Trapezoid:
((Length of Base 1+Length of Base 2)/2)*h -> ((7+12)/2)*2.4=9.5*2.4=22.8 -> (C)
