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A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 02:18
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Competition Mode Question A student initially draws a rectangle ABCF. Later, he draws an adjacent trapezium FCDE in which side ED is parallel to side FC and drops perpendiculars EG and DH on FC. The dimensions of each side of the polygon ABCDEF, in centimeters, are shown on the figure above. What is the value of x? (1) The area of the polygon ABCHDEG is 332 square centimeters (2) If sides AE and BD are joined, the area of the quadrilateral ABDE will be 266 square centimeters Attachment:
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 05:34
Suppose FG = a and EG=DH= b
CH = 154a=11a
If we can find b, we can '11a' or 'a' and hence, we can find x. (Use pythagoras theorem)
Statement 1 
We know the area of polygon ABCHDEG and dimensions of rectangle ABCF. So, we can find the area of trapezium.
area of trapezium = \(\frac{1}{2} (ED+FC)*b\)
We know the area, ED and FC; we can find 'b'.
Sufficient
Statement 2
area of the quadrilateral ABDE = 1/2 * (sum of parallel sides) * altitude = \(\frac{1}{2} * (AB+ED) * (b+BC)\)
We know area, AB, ED and BC. We can find b.
Sufficient



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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 06:08
Image A student initially draws a rectangle ABCF. Later, he draws an adjacent trapezium FCDE in which side ED is parallel to side FC and drops perpendiculars EG and DH on FC. The dimensions of each side of the polygon ABCDEF, in centimeters, are shown on the figure above. What is the value of x? x = ? (1) The area of the polygon ABCHDEG is 332 square centimeters Area of ABCHDEG = Area of Rec. ABCF + Area of Rec. HDEG 332 = \(20*15 + 4*DH\) DH ≃ 8 In Triangle CHD \(10^2 = 8^2 + CH^2\) CH = 6 FG = 15  6  4 = 5 \(x = \sqrt{5^2 + 8^2}\) x = 9.43 SUFFICIENT. (2) If sides AE and BD are joined, the area of the quadrilateral ABDE will be 266 square centimeters Area of the quadrilateral ABDE = \(\frac{1}{2}\) * Perpendicular distance(⊥=EG+FA) * (4+15) \(\frac{266 * 2 }{ 19}\) = ⊥ ⊥ = 28 EG = HD = 8 CH = 6 Thus, FG = 15  6  4 = 5 x = 9.43 SUFFICIENT. Answer D.
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 07:31
Quote: A student initially draws a rectangle ABCF. Later, he draws an adjacent trapezium FCDE in which side ED is parallel to side FC and drops perpendiculars EG and DH on FC. The dimensions of each side of the polygon ABCDEF, in centimeters, are shown on the figure above. What is the value of x?
(1) The area of the polygon ABCHDEG is 332 square centimeters (2) If sides AE and BD are joined, the area of the quadrilateral ABDE will be 266 square centimeters solve the height of the trapezium and you find x by solving the distance between DC, then FG using pythagoras (1) sufic area rectangle + trapezium = 332 15*20+(4+15)/2*height=332 (2) sufic area trapezium = 266 (4+15)/2*(20+height)=266 Ans (D)



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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 11:59
IMO DLet DH = EG =h Stmt1: The area of the polygon ABCHDEG is 332 square centimeters Area of ABCHDEG = Area of rectangle ABCF + area of rectangle ABCF 332 = 20*15 + (4*h) h =8 In triangle DHC, \(10^2\) = \(DH^2\) + \(HC^2\) 100= \(8^2\) + \(HC^2\) HC = 6 FG = FC  (GH+HC) FG = 15 10 =5 In triangle EFG, \(x^2\) = \(FG^2\) + \(GE^2\) x= \(\sqrt{89}\) SUFFICIENTStmt2: If sides AE and BD are joined, the quadrilateral ABDE will be trapezium as AB  ED Area of ABDE = \(\frac{1}{2}\) * (AB + DE) * (h+20) = 266 This will give unique value of h. As solved in stmt 1, using value of h, x can be found. SUFFICIENT
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 13:12
IMO D.
A. In short this gives me the area of rectangle ABCF + Area of GHDE
I know area of ABCF. So area 332area of ABCF gives me area of GHDE. I already know Length of GHDE. I can find EG=DH height.
Now from triangle DHC, I can find HC as i know 2 sides.
Also, it helps me finding FG = 154HC.
Once I know EG and FG i can find X.
Sufficient.
B. Joining these sides results in a bigger trapezium ABDE. From its area and sum of sides given i can calculate the height of trapezium.
Also, trap heught = height of lower rectangle which is 20 + upper rectangle height.
Therefore I can find EG. Once I know EG, just like above in option A i can find x.
Sufficient.
Please correct me if i am wrong.
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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05 Jun 2020, 20:06
Quote: A student initially draws a rectangle ABCF. Later, he draws an adjacent trapezium FCDE in which side ED is parallel to side FC and drops perpendiculars EG and DH on FC. The dimensions of each side of the polygon ABCDEF, in centimeters, are shown on the figure above. What is the value of x? (1) The area of the polygon ABCHDEG is 332 square centimeters (2) If sides AE and BD are joined, the area of the quadrilateral ABDE will be 266 square centimeters IMO , D Statement (1) The area of the polygon ABCHDEG is 332 square centimeters 332 = Area of rectangle (15*20) + area of trapezium (0.5 {DH * (ED + FC )}) So we can find DH , then CH and FG and hence X . Sufficient . Statement (2) If sides AE and BD are joined, the area of the quadrilateral ABDE will be 266 square centimeters Again Area of trapezium EDBA =266 = (0.5 {perpendicular from ED on AB * (ED + AB )}) 266 = (0.5 {(DH + CB) * (ED + AB )}) So we can find DH , then CH and FG and hence X . Sufficient . Hence D .
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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06 Jun 2020, 15:16
IMO : Ans D
Both Options A and B are individually Sufficient to Find Value of Variable "x"
Lets assume side CH=y , so FH=11y (EC=15=EG+GH(4)+CH) Lets assume side EG=z , as EG is parallel to DH , so DH=z
Analysing Option 1 :
Area of ABCHDEF= Area of Trapezium ABCG + Area of Rectangle HGDE > 332= 0.5 * (15+4+y)*20+ 4*z >71=5y + 2z Lets name this equation as (1)
From Right angle Triangle DCH By Pythagoras Theorem
> 100= z^2 + y^2
But Range of Value of y will be between 0 and 11
By Replacing value of z from equation (1), we will have value of y and correspondingly value of FG=11y
We get value of z and y
In Right angle Triangle EFG We have x^2=z^2 + (11y)^2
Hence we can solve for x
Option1 is sufficient
Analysing Option 2 :
Area of Trapezium ABGE = 0.5* (15+4)* (z+20) >z=8
So In right angle Triangle DCH CH=y will be 6 (By Pythagoras theorem)
Hence FG will be 5
In Right angle Triangle, We know all other sides other than Hypotenuse
By using Pythagoras Theorem
We have one value of x
SO Option 2 is also Sufficient



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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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06 Jun 2020, 23:45
A student initially draws a rectangle ABCF. Later, he draws an adjacent trapezium FCDE in which side ED is parallel to side FC and drops perpendiculars EG and DH on FC. The dimensions of each side of the polygon ABCDEF, in centimeters, are shown on the figure above. What is the value of x?
(1) The area of the polygon ABCHDEG is 332 square centimeters (2) If sides AE and BD are joined, the area of the quadrilateral ABDE will be 266 square centimeters
1) ABCHDEG has 2 rectangles, so 2(15+20)+2(DH+4) = 332 DH can be found, from this we can find CH so we can find x Sufficient
2) ABDE forms a trapezium, so (15+4)*height/2=266 height can be found from this DH can be found x can be found sufficient
Ans D




Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen
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