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Math Expert V
Joined: 02 Sep 2009
Posts: 65194
A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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10 00:00

Difficulty:   75% (hard)

Question Stats: 41% (02:18) correct 59% (03:03) wrong based on 44 sessions

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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: 1.png [ 18.54 KiB | Viewed 686 times ]

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DS Forum Moderator V
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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1
1
Suppose FG = a and EG=DH= b

CH = 15-4-a=11-a

If we can find b, we can '11-a' 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
Sloan MIT School Moderator V
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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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.

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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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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)
IESE School Moderator S
Joined: 11 Feb 2019
Posts: 308
Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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IMO D

Let 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= $$\sqrt{89}$$
SUFFICIENT

Stmt2: 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  [#permalink]

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IMO D.

A. In short this gives me the area of rectangle ABCF + Area of GHDE

I know area of ABCF. So area 332-area 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 = 15-4-HC.

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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Stern School Moderator S
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Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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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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Intern  B
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Posts: 45
Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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IMO :- Ans D

Both Options A and B are individually Sufficient to Find Value of Variable "x"

Lets assume side CH=y , so FH=11-y (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=11-y

We get value of z and y

In Right angle Triangle EFG
We have x^2=z^2 + (11-y)^2

Hence we can solve for x

Option-1 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
Manager  G
Joined: 23 Jan 2020
Posts: 177
Re: A student initially draws a rectangle ABCF. Later, he draws an adjacen  [#permalink]

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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   [#permalink] 06 Jun 2020, 23:45

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