GMAT Club World Cup 2022 (DAY 5): In a quarter of a circle, a right an

Answer is C

To answer this question, you need to draw out the whole half circle. --> to point E so that AE is the diameter of the circle
We have AB = 24 and BC =7, so we can get AC = 25, so that CE = 25.
Then, BE is 25 + 7 = 32, and we know that angle ABE is 90 degree, so that AE = 40 which means the radius is 20.
Then if we calculate OC which is 15, thus we got DC = 20-15 = 5

Extend the figure to the right by drawing the other quarter of the circle and extending BC until it intersects with the circle at point E. AB is 24.
Starting with triangle ABE, we have right triangle with a side that looks like it'll make a tidy 3-4-5. Let's see if we can backdoor our way to the solution.
AB is 24 and is the shortest side of triangle ABE. If that is the "3" in a 3-4-5, the other leg will be 32 and the hypotenuse will be 40. The hypotenuse is 2r, so r=20. We are given that BC is 7, so that would make CE 25 and CO is 15. That makes DC = 5.
ABE and COE are similar and ABE is a 3-4-5, so COE should be, too. It is, so everything checks out.

Answer choice C.

Hello everyone,

The correct answer is C.
The detailed explanation of the solution in the attached pic.

Correct answer : Choice C

I am guessing.

A. 3
B. 4
C. 5
D. 7
E. 15

Steps -
1. Find AC using the Pythagorean theorem.
AC = 25
2. \(\triangle ACE\) is an isosceles triangle. Therefore, \(CE = AC = 25\).
3. \(BE = 25 + 7 = 32\)
6. From Right angled \(\triangle ABE\), AE = 40. Hence, Radius of Circle is 20.
7. From Right angled \(\triangle AOC\), CO = 15. Hence, \(DC = 20 - 15 = 5\).

Hence, the answer is C.

We can quickly spot that the hypotenuse of the \(triangle ABC \) to be 25 as the other legs 7,24 with 25 form a Pythagorean Triplets

Now since the 90-degree angle is inscribed in the quarter circle, the line BC will extend to meet the diametrical opposite of point A say A'. This will form a new Right Angle triangle ABA' and isosceles triangle ACA'. Refer sketch in the spoiler

From the isosceles triangle AC = CA' = 25

The right triangle ABA' will have legs of the following lengths \(AB = 24\), \(BA' = BC + CA' --> 7+25 = 32\), \(AA' = 2r\)

Using Pythagorean Therom

\(A'A^2 = AB^2 + A'B^2\)
\( (2r)^2 = 32^2 + 24^2\)
\( (2r)^2 = (30+2)^2 + (25-1)^2\)
\( (2r)^2 = 900 + 4 + 120 + 625 + 1 - 50 \)
\( (2r)^2 = 1600 \)
\( (2r) = 40 \)
\( r = 20 = AO\)

Now to find CO; we again use Pythagorean Therom on the Triangle AOC
\(AC^2 = AO^2 + CO^2\)
\(25^2 = 20^2 + CO^2\)
\(25^2 - 20^2 = CO^2\)
\((25 - 20)(25+20) = CO^2\)
\((5)(45) = CO^2\)
\((225) = CO^2\)
\(15 = CO\)

To Find CD
\(CD = DO - CO\)
\(CD = 20 - 15\)
\(CD = 5 \)

Ac = 25
AO = 20, OC = 15, so CD = 15

Answer C

Answer: C

Right angle ABC is inscribed in a quarter of a circle, as shown above. What is the length of line segment DC ?

extend the arc ABD (from point D) to draw a semi-circle. Extend radius AO to have a diameter of the semi-circle. Let the diameter be AE.
Now if we extend BC, it will intersect at point E. Since, the angle inscribed in a semi-circle is always 90 degrees.

Let the radius AO = r, then diameter AE = 2r
Now we have multiple right angled triangles.
Triangle ABC, Triangle ABE, Triangle AOC

Now, AC^2 = AB^2 + BC^2 = 24^2 + 7^2
=> AC = 25.

Also, AE^2 = AB^2 + BE^2
=> (2r)^2 = 24^2 + BE^2 = 24^2 + (7 + CE)^2
CE = 25 (since, AOC will be congruent to EOC)
=> 4r^2 = 24^2 + 32^2
solving, we get r = 20

Also, AC^2 = AO^2 + OC^2
=> 25^2 = r^2 + OC^2
=> OC^2 = 25^2 - 20^2
=> OC = 15

Now, DC = OD - OC
= r - 15
= 20 - 15
= 5

Please refer to the attachment for the solution.

[quote="Bunuel"]
Right angle ABC is inscribed in a quarter of a circle, as shown above. What is the length of line segment DC ?

A. 3
B. 4
C. 5
D. 7
E. 15

Answer C. Solution attached

The correct answer choice is C. Refer to the image attached.

Not sure if I am missing any geometrical property here.
AC is 25
AO < AC
OC <AC
AO^2 + OC^2 =AC^2
= 625
we need to find a pair of square whose sum is 625
20 and 15 are a good match
Hence radius is 20 and CD= 5
ANS : C

From the given data
AC^2 = 24^2 + 7^2 = 25^2
AC = 25 =CE (Extend BC to meet semi circle at E such that AE is dia of the circle as angle ABC =90)

From triangle ABE
AE^2 = (2AO)^2 = 24^2 + (7+CE)^2
= (2AO)^2 = 24^2 + (7+25)^2 = 1600
therefore OA = 20

OA =OD= OC+CD =20 ......eqn 1
Now from triangle AOC
OC^2 = 25^2-20^2 = 15^2

substituting OC in eqn 1 we get CD = 5
Answer: C

Theory used - angle subtended by a diameter/semi circle on any point of circle is 90° right angle

Pythagorean triplet: 7-24-25

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Lets imagine the BC line extending, it will touch the circle at E. Now triangle ABE is a right angle triangle, so line AE is the diameter.

so AC2 = CO2 + AO2 = CE2

Now lets imagine the line CD is a, radius is r

As per the figure, line AC = 25 (7, 24, 25 right angle triangle)

hence line CE is also 25. Total BE is 32.

For triangle ABE, 2 sides are 24 and 32, so the third side opposite right angle will be 40. ( 6*4, 8*4, 10*4)

Diameter is 40, so the radius is 20 = OD

now triangle AOC, 2 sides are 25, 20 so the third side will be 15 (4*5, 5*5, 3*5), and OC = 15

DC = OD - OC = 20-15 = 5

AC will be 25 (pythagorean triplet).
AO=DO as both are radii
DC=AO-CO
One quarter circle is symmetrical to the other quarter circle.
Draw imaginary other quarter circle to the right. Say AE is the diameter.
Since symmetrical, AC=CE=25 and BE=7+25=32
So, AE=40 and AO=20
CO will be 15
So, DC=5
Answer C.

Let's complete the semi-circle and extend the line BC till it meets the extended line AO.
Now the extended line AO is the diameter of the semi-circle
Let the radius be R

For the triangle ABC we have AC = 25 basis of the lengths of AB and BC

For the new triangle created AB(X) formed by extending AO and the semicircle we have the following
(2R)^2 = 24^2 + (25+7)^2
(2R)^2 = 40^2
=> R = 20

Now we have from the quarter circle
25^2 = 20^2 + (20 - DC)^2
=> 15^2 = (20 - DC)^2
=> DC = 5

IMHO Option C