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17 Jun 2020, 03:56
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Difficulty:
95%
(hard)
Question Stats:
37% (02:31) correct
63%
(02:27)
wrong
based on 76
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A triangle shown above is divided into three smaller triangles, with areas of 2, 3 and 4, and a quadrilateral. What is the area of the quadrilateral?
A. 6
B. 7
C. 39/5
D. 8
E. 41/5
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Attachment:
2020-06-17_1439.png [ 48.87 KiB | Viewed 7257 times ]
yashikaaggarwal
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17 Jun 2020, 06:09
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Let the triangle be ABC
Area of triangle = A
Area 3 = X
Area 4 = Y
Area 2 = Z
Using Ladder theorem ;
1/A + 1/y = 1/x+y + 1/y+z
1/A + 1/4 = 1/3+4 + 1/4+2
1/A + 1/4 = 1/7 + 1/6
1/A + 1/4 = 6+7/42
1/A + 1/4 = 13/42
1/A = 13/42 - 1/4
1/A = 52-42/168
1/A = 10/168
A = 16.8
Area of triangle = Triangle 2 + Triangle 3 + Triangle 4 + Quadrilateral
16.8 - 2 - 3 - 4 = Area of quadrilateral
7.8 = Area if Quadrilateral
=> 78/10 => 39/5
Answer is C
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Area of triangle = A
Area 3 = X
Area 4 = Y
Area 2 = Z
Using Ladder theorem ;
1/A + 1/y = 1/x+y + 1/y+z
1/A + 1/4 = 1/3+4 + 1/4+2
1/A + 1/4 = 1/7 + 1/6
1/A + 1/4 = 6+7/42
1/A + 1/4 = 13/42
1/A = 13/42 - 1/4
1/A = 52-42/168
1/A = 10/168
A = 16.8
Area of triangle = Triangle 2 + Triangle 3 + Triangle 4 + Quadrilateral
16.8 - 2 - 3 - 4 = Area of quadrilateral
7.8 = Area if Quadrilateral
=> 78/10 => 39/5
Answer is C
Posted from my mobile device
General Discussion
17 Jun 2020, 05:04
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Untitled (1).png [ 8.13 KiB | Viewed 6413 times ]
OE : OB = 3:4 (Ratios of the areas of AEO and AOB)
Hence, Area of CEO : Area COB = 3:4
Area of CEO : (Area of COD +Area of DOB) = 3:4
x : (y+2) = 3:4
4x=3y+6 ........(1)
AO : OD = 4:2 (Ratios of the areas of AOB and DOB)
Hence, Area of AOC : Area DOC = 4:2
(Area of AOE + Area of COE) : Area of DOB= 4:2
(3+x) : y = 2:1
3+x=2y.......(2)
From (1) and (2)
x= 21/5 ; y=18/5
x+y = 39/5
Bunuel
