Last visit was: 04 Sep 2026, 06:51 It is currently 04 Sep 2026, 06:51
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 Sep 2026
Posts: 113,128
Own Kudos:
839,054
 [1]
Given Kudos: 111,327
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,128
Kudos: 839,054
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
CounterSniper
Joined: 20 Feb 2015
Last visit: 14 Apr 2023
Posts: 609
Own Kudos:
Given Kudos: 74
Concentration: Strategy, General Management
Posts: 609
Kudos: 871
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
PKN
Joined: 01 Oct 2017
Last visit: 11 Oct 2025
Posts: 809
Own Kudos:
Given Kudos: 41
Status:Learning stage
WE:Supply Chain Management (Energy)
Posts: 809
Kudos: 1,685
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,794
Own Kudos:
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,794
Kudos: 6,344
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Attachment:
image031.jpg
image031.jpg [ 9.79 KiB | Viewed 2024 times ]

In the figure, triangle AXD is a right-angled triangle with hypotenuse 25 and one of the sides 24
This is because BCDX is a rectangle and BC = XD(since we have drawn DX to be a perpendicular line)

Using this, we can find the length of the third sides using Pythagoras theorem - \(AD^2 = AX^2 + DX^2\)
Re-writing, we get \(AX^2 = AD^2 - DX^2\) -> \(AX^2 = 25^2 - 24^2 = 625 - 576 = 49\) -> AX = \(\sqrt{49} = 7\)

The area of the triangle is \(\frac{1}{2} * 24 * 7 = 12 * 7 = 84\) | Area of the rectangle = \(24 * 7 = 168\)

Therefore, the area of the quadrilateral ABCD is the sum of the individual areas, which is 252(Option D)
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 04 Sep 2026
Posts: 6,129
Own Kudos:
Given Kudos: 164
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 6,129
Kudos: 6,088
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

What is the area of quadrilateral ABCD?


A. 84
B. 168
C. 172
D. 252
E. 336


Attachment:
image031.jpg

Quadrilateral ABCD can be split into lower rectangle and upper triangle.

Height of upper triangle = \(\sqrt{25^2 - 24^2 } = 7\)

Area of quadrilateral ABCD = 24*7 + 1/2 * 24 * 7 = 24*7 * 3/2 = 252

IMO D
Moderator:
Math Expert
113125 posts