If ABCD is a square with area 625, and CEFD is a rhombus wit

Question

rumbs.JPG If ABCD is a square with area 625, and CEFD is a rhombus with area 500, then the area of the shaded region is A. 125 B. 175 C. 200 D. 250 E. 275

KarishmaB · Accepted Answer

Area of shaded region = Area of square - (Area of rhombus - Area of triangle MFD)

Attachment:

Ques2.jpg [ 8.55 KiB | Viewed 56036 times ]

Area of square = 625 so side = 25
Area of rhombus = 500. So altitude = 500/25 = 20
MF = root (25^2 - 20^2) = 15
Area of triangle MFD = (1/2) * 15 * 20 = 150

Area of shaded region = 625 - (500 - 150) = 275

USCTrojan2006 · Answer

Nice explanation.

kinjiGC · Answer

Kudos to Karishma. Failed to apply some basic theorems. Went to calculation of diagonals of rhombus

GMAT questions are lot easier than you think !!

Nunuboy1994 · Answer

VeritasPrepKarishma

The only thing that threw me off in this problem is the fact that when I read, on a different website, it said that the figure is not drawn to scale? I had initially made the assumption that the base of the rhombus was equal to the length of the square and derived the height using that- but then I second guessed myself and try to work backwards from the area of the rhombus in order to find the lengths of the diagonals and found myself in a mess. In this problem we can make the assumption that the base of the rhombus is equal to a side length of the square? That seems to be a valid takeaway from initially glancing at the problem

pclawong · Answer

Dear Karishma,

how do you get "MF = root (25^2 - 20^2) = 15"?

KarishmaB · Answer

It is not an assumption from the figure. You are given that ABCD is a square so CD is a side of the square. You are also given that CEFD is a rhombus so CD is a side of the rhombus too. Hence the base of the rhombus is equal to the length of the side of the square.

KarishmaB · Answer

Area of rhombus is Base*Height = 500
Height = 500/25 = 20 = MD

Now MFD is right angled triangle so using pythagorean theorem, 
MD^2 + MF^2 = FD^2
MF^2 = FD^2 - MD^2
MF = sqrt(25^2 - 20^2)
MF = sqrt(625 - 400)
MF = sqrt(225)
Mf = 15

profileusername · Answer

I don't understand. Is there a different perspective with which we are looking at the diagram in this solution? The one in question is from a different angle, right?

arvind910619 · Answer

The area of the rhombus is given by base*height 
Since area =500=25*h therefore h=20
now we can find the hypotenuse of the triangle =15 
area of the triangle is =150 
So 500-150=350
The area of the shaded region =625-350=
275

amargad0391 · Answer

Hi Karishma

Can you please tell How did we arrive at FD=25 ?

Thank you

testcracker · Answer

hi I am not Karishma mam the great, but I will try to help you please remember that the sides of a rhombus are always equal here you can see the side "CD", which is a side of both the square and the rhombus so, as the side of the square is 25, the side of the rhombus is also 25 hope this helps! thanks

siddharthfrancis · Answer

Area of square = 625 so side = 25
Area of rhombus = 500. So altitude = 500/25 = 20
MF = root (25^2 - 20^2) = 15
Area of triangle MFD = (1/2) * 15 * 20 = 150

Area of shaded region = 625 - (500 - 150) = 275

in this why are we separately subtracting the area of the triangle.. ?? is it not already a part of the rhombus??

Fdambro294 · Answer

(1st) notice that the the bottom side CD is shared by both the rhombus and the square

(2nd) we can find the side of the “square” that would be hidden behind the rhombus by taking the square root of the area = 625

Side CD = sqrt(625) = 25

(2nd) we are given that the area of the rhombus = 500

Area of rhombus = (Base side) * (Perpendicular Height)

We can draw 2 perpendicular height lines:

(1) drawn from vertex E perpendicular to side CD ——> call the intersection point X

(2) drawn from vertex D to side EF of the rhombus ———> call the intersection point Y

The area of the shaded portion = area of square - (area of right triangle CED) - (area of rectangle EXDY)

(3rd) find the length of the perpendicular heights = EX and DY ——> call H

Area of rhombus = 500 = (side CD) * (H)

500 = 25H

H = 20 = EX = DY

(4th)we can find the length of CX and XD, which together sum to equal side CD of 25, using Pythagorean Theorem

Let the leg CX = l

Using right triangle CEX:

(25)^2 = (20)^2 + (l)^2

(l))^2 = 225

l = 15 ——> which means

CX = 15

And

DX = CD - CX

DX = 25 - 15

DX = 10

(4th) we can find the area of the right triangle CEX and the rectangle EXYD now:

Area right triangle CEX = (1/2) (15) (20) = 150

Area of rectangle EXYD = (10) (20) = 200

150 + 200 = 350

350 is the area covering what would be the square ABCD

Area of shaded region = 625 - 350 = 275

(E)275

Posted from my mobile device

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

If ABCD is a square with area 625, and CEFD is a rhombus wit

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If ABCD is a square with area 625, and CEFD is a rhombus wit

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards