Geometry questions that appear in the Quantitative Aptitude section of the CAT Exam broadly tests an aspirant on the concepts – Triangles, Circles, Quadrliaterals, Polygons & mixture of the above mentioned concepts. In CAT Exam, one can generally expect to get 4-6 questions from CAT Geometry. CAT Geometry is an important topic with lots of weightage in the CAT Exam. CAT Geometry questions are heavily tested in CAT exam. Make sure you master Geometry problems. Here is an interesting CAT level Geometry question with detailed solutions for the same. You might also want to check out this wonderful video on how to make the most of the remaining 4 months. Happy solving CAT-o-sapiens!

## Question

In a semicircle, the circular segment ADB is folded along AB to make the circular segment ATB. The point T is tangent to the diameter and divides the diameter into lengths of 4 and 2. What is the length of AB?

Before you look at the solution, give this question a try and then check out the solution of this beautiful question.

## Solution

Let’s consider AB as a mirror. Now the circular segment ADB is reflected about this mirror AB such that the resulting circular segment ATB is tangent at point T and divides the diameter into lengths of 4 and 2. So the the total length of the diameter is 4 + 2 = 6

Now, consider the center of this semi circle is P.

Now, we know the radius of a circle is = 1/2 * Diameter = 1/2 * 6 = 3

Now, from this we can conclude that, the length of PT is ( 3 – 2 ) = 1

Now, construct a circle with center as P‘ through the points A, T, and B. Since the original circular segment ADB is from a circle with center as P and with radius 3, the reflected circular segment ATB is also from a circle with center P‘ and with a radius of 3. ( If this doesn’t blow your mind, then you have no human emotions : P ).

Furthermore, P‘ is the reflected image of P about the chord AB since both centers are equidistant and a distance of 3 from A and B.

Finally construct P’T it will be perpendicular to the diameter of the circle with center as P, and the length of the P’T will be 3 (P’T is the radius of the circle with center p’). Now, we can only focus on the right angle triangle PTP’.

In right triangle PTP’, we can use the Pythagoras Theorem to find the length of PP’

(PP’)^{2} = 1^{2} + 3^{2}

PP’ = √10

Let C be the midpoint of PP’. Then PC = 0.5√10. Also notice C is half way between P and P’ it will also be along line segment AB.

Then construct radius PA = 3. In triangle PAC we can solve for AC using the Pythagoras Theorem again

(AC)^{2} = (PA)^{2} – (AC)^{2}

(AC)^{2} = 3^{2} – 10/4

AC = 0.5√26

Finally triangles PAC and PBC are congruent, so CB = AC. Thus we have

AB = AC + CB

AB = AC + AC

AB = 2(AC)

AB = 2(0.5√26)

AB = √26

Best wishes for CAT 2021!!

Abhishek Mukherjee works for 2IIM. Apart from solving interesting math questions he likes to eat biriyani and watch movies.

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