|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
08 Sep 2022, 04:22
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (03:12) correct
68%
(03:20)
wrong
based on 84
sessions
History
Date
Time
Result
Not Attempted Yet
The median CD to the hypotenuse AB cuts a right triangle ABC into two smaller triangles with the perimeters of 36 and 50 cm, respectively. What is the value of AC + BC ?
A. 24
B. 26
C. 32
D. 34
E. 36
A. 24
B. 26
C. 32
D. 34
E. 36
|
Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
14 Sep 2022, 12:01
Kudos
Bookmarks
The median CD to the hypotenuse AB cuts a right triangle ABC into two smaller triangles (isosceles) with the perimeters of 36 and 50 cm, respectively.
We need to find value of AC + BC =?
Property of median of right angle triangle, \(CD = AD = BD = \frac{AB}{2} i.e. AB=2*CD \)
\(AC+CD+AD = 36\)
\(AC + AB = 36\).....(1)
\(BC+CD+BD = 50\)
\(BC + AB = 50\) ........(2)
Subtract equation 1 from 2 we get \(BC - AC = 14.\).....(3)
Also from Pythagoras theorem, \(AC^2 + BC^2 = AB^2\)
subtract \(2*AC*BC\) from both sides and re write the equation as
\(AB^2 - 2* AC*BC = (AC - BC)^2 \)
substitute the value of AC and BC from equation 1 and 2 and value AC-BC from equation 3 in above equation
\(AB^2 - 2 (36-AB)* (50-AB) = (-14)^2\)
solving the equation we get
\(AB^2 - 172*AB + 3796 = 0\)
=> \((AB-26)*(AB-146) = 0\)
AB = 26 or 146 (Very high value so can be ignored)
Now substitute the value of AB in equation 1 and 2 to get value of AC = 10 and BC = 24
Therefore, \(AC + BC = 34\)
Answer: D
PS: At 0030 hrs with a drained mind I could workout this solution which is pretty lengthy. Will request the experts to post a short and crisp solution
We need to find value of AC + BC =?
Property of median of right angle triangle, \(CD = AD = BD = \frac{AB}{2} i.e. AB=2*CD \)
\(AC+CD+AD = 36\)
\(AC + AB = 36\).....(1)
\(BC+CD+BD = 50\)
\(BC + AB = 50\) ........(2)
Subtract equation 1 from 2 we get \(BC - AC = 14.\).....(3)
Also from Pythagoras theorem, \(AC^2 + BC^2 = AB^2\)
subtract \(2*AC*BC\) from both sides and re write the equation as
\(AB^2 - 2* AC*BC = (AC - BC)^2 \)
substitute the value of AC and BC from equation 1 and 2 and value AC-BC from equation 3 in above equation
\(AB^2 - 2 (36-AB)* (50-AB) = (-14)^2\)
solving the equation we get
\(AB^2 - 172*AB + 3796 = 0\)
=> \((AB-26)*(AB-146) = 0\)
AB = 26 or 146 (Very high value so can be ignored)
Now substitute the value of AB in equation 1 and 2 to get value of AC = 10 and BC = 24
Therefore, \(AC + BC = 34\)
Answer: D
PS: At 0030 hrs with a drained mind I could workout this solution which is pretty lengthy. Will request the experts to post a short and crisp solution
General Discussion
08 Sep 2022, 11:49
Kudos
Bookmarks
I have a little doubt in the question itself. Median from the point on the hypotenuse bisects the hypotenuse. It also happens to be the circumcentre of the triangle meaning thereby the triangles ACD and BCD should have equal areas. This fact contradicts the data given in the question. Kindly explain.
