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805+ (Hard)|   Geometry|      
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Bunuel
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A triangle shown above is divided into three smaller triangles, with a 17 Jun 2020, 03:56
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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)!

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95% (hard)

Question Stats:

37% (02:31) correct 63% (02:27) wrong based on 76 sessions

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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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C
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A triangle shown above is divided into three smaller triangles, with a 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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Attachment:
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Untitled (1).png [ 8.13 KiB | Viewed 6583 times ]
Suppose area of CEO = x and area of COD =y

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

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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nick1816
Attachment:
Untitled.png
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

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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nick1816.. I couldnot understand how OE : OB = 3:4 .. may you please help me on this .
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DasAshishAshutosh
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Draw an altitude from B on AD at X.

Area of BOA = 1/2 * BX * AO

Area of BOD = 1/2 *BX * DO


Area of BOA/Area of BOD = AO/DO

Since the altitude of both triangles is common from point B, ratio of their areas is same as that of their base.

If you still have any doubt, you can ask.
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DasAshishAshutosh
Attachment:
Untitled.png

Draw an altitude from B on AD at X.

Area of BOA = 1/2 * BX * AO

Area of BOD = 1/2 *BX * DO


Area of BOA/Area of BOD = AO/DO

Since the altitude of both triangles is common from point B, ratio of their areas is same as that of their base.

If you still have any doubt, you can ask.
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I got it . Thanks a lot nick1816. Indeed a tough one for me .
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nick1816 can you please let me know how you obtained:

Area of CEO : (Area of COD +Area of DOB) = 3:4

x : (y+2) = 3:4

I am not sure where the 'y' came from in the '(y+2)'.

Thanks for your help.
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A triangle shown above is divided into three smaller triangles, with a 19 Jun 2020, 04:39
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Hey rcgard

That was my bad. I assigned y for 2 distinct variables. I've edited the solution.

Area of CEO = x and Area of COD =y

Thanks!
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Area of a triangle = ½*base*height
Ratio of areas of triangles = base1 * height1 / base 2*height2
If height 1 = height 2, then ratio of areas = base 1 / base 2

With this understanding, lets divide the quadrilateral into two triangles with area A1 and A2 and name the vertices from A to F as in the diagram below.
To solve this question, we need to consider triangles with different bases and same height and equate the ration of bases to the ratio of areas.

Triangles AFB and BFC
B1/B2 = A2 / 3
Triangles ADB and BDC
B1/B2 = (A1 + A2 + 2) / 7
So, A2/3 = (A1 + A2 + 2)/7
-3A1 + 4A2 = 6

Triangles AFE and EFD
B3/B4 = A1 / 2
Triangles ACE and EFD
B3/B4 = (A1 + A2 + 3) / 6
So, A1/2 = (A1 + A2 + 3)/6
2A1 - A2 = 3

Solving the two equations, we get A1= 18/5 and A2= 21/5 and A+A2= 39/5, Ans C
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Can someone please clear my doubt....how do we know that those lines originating from vertices are altitude/height??

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Bunuel

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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Solution:

Whenever we have a triangle that is partitioned into 4 regions like the one shown, we can use the Ladder Theorem,

1/A + 1/Z = 1/(X + Z) + 1/(Y + Z)

where A is the area of the entire triangle, X is the area of the triangular region on the left, Y is the area of the triangular region on the right, and Z is the area of the triangular region at the bottom.

Therefore, we have:

1/A + 1/4 = 1/(3 + 4) + 1/(2 + 4)

1/A + 1/4 = 1/7 + 1/6

1/A = 1/7 + 1/6 - 1/4

1/A = 12/84 + 14/84 - 21/84

1/A = 5/84

A = 84/5

Therefore, the area of the quadrilateral region is:

84/5 - 3 - 2 - 4 = 84/5 - 9 = 84/5 - 45/5 = 39/5

Answer: C
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