Kindly refer to the figure, AB is the tangent to the circle with cente

Question

GMATBusters’ Quant Quiz Question -7
For past quiz questions, click here

Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5

Show Spoiler

Attachment:

QQQ.JPG [ 11.83 KiB | Viewed 4527 times ]

Show Answer

Official Answer

C

gurmukh · Answer

1/2×3×4=1/2×5×r
r =2.4

Posted from my mobile device

ashwini2k6jha · Answer

answer C
Let D be point radius from O touches tangent
triangle OAD and triangle OAB are simmilar
hence corresponding ratio of sides equal
radius/4=3/5
radius=2.4

uc26 · Answer

If we join the center of the circle with the point of tangency, we get a right angle. Let the point of tangency be T and the line be OT. OT is the radius of the circle.
We can see the AB=5 (3-4-5 pythagorean triplet)
Area of triangle AOB= 1/2 * AO* AB= 1/2* 3*4= 6
Also, Area= 1/2 * OT* AB
so 6= 1/2 * OT* 5
OT= 12/5= 2.4.
Answer: C

Archit3110 · Answer

given OB is 3 cm so radius of circle has to be < 3cm D &E are out 
now for 
∆ AOB ; side AB = 5 AO = 4 and BO = 3
∆ AOD OD = r and AO = 4
OD/OA= OB/AB 
=> r/4 = 3/5
=> r= 12/5 ; 2.4
OPTION C

Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5

Kinshook · Answer

Given: Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees.
Asked: If OA =4 cm, and OB = 3cm. What is the radius of the circle?

Attachment:

Screenshot 2020-05-24 at 8.52.15 AM.png [ 27.06 KiB | Viewed 4092 times ]

Since \(\triangle\) AOB & \(\triangle\) AOD are similar; Angle A & \(90^0\) are common.
\(\frac{OD}{AO} = \frac{r}{4} = \frac{OB}{AB} = \frac{3}{5}\)
Radius of circle = \(r = \frac{12}{5} = 2.4\)

IMO C

MayankSingh · Answer

IMO C

AB is the tangent to the circle with center O, 
OA =4 cm, and OB = 3cm.

Lets Draw radius OD from O at point of contact of tangent to circle.(Shown in figure)
Tangent is perpendicular to radius at point of contact.
So ∠ ADO= 90

In triangle AOB & ADO
∠AOB=∠ADO=90
∠OAB=∠DAO (Common)
Both triangles are similar
OB/AB= OD/AO
OD= OB x AO/AB = 3 X 4 /5 = 2.4

C. 2.4

tyson1990 · Answer

In triangle AOB, <AOB = 90 deg
OA = 4, OB = 3
using rule of AB^2 = OA^2 + OB^2
AB = 5

Now radius of circle is perpendicular to tangent
Hence considering rule of triangle area

1/2 X OA X OB = 1/2 X R X AB
R = (OA X OB) / AB = 3X4/5 = 2.4

shashwat91 · Answer

ab^2 = 4^2+3^2
ab=5
area ABO(considering base BO) = 1/2*3*4=6
area ABO(considering base AB) = 1/2*5*h=6 (h=point where circle and ab intersect/meet)
therefore h=12/5
and h=r
so r= 2.4(ans)

karan12345 · Answer

Area of triangle AOB =  \frac{1}{2} * base * height   
 = \frac{1}{2}  * 3 * 4 = 6 sq. cm

We know , radius of circle which joins centre to tangent makes right angle with circumference.

AB =  \sqrt{AO^2 + OB^2}  =  \sqrt{25}  = 5 cm (by pythagoras theorem)
 
Also, Area of triangle AOB =  \frac{1}{2}  * AB * radius of circle
=> 6 = 2.5 * radius of circle
=> Radius of circle =  \frac{6}{2.5}  = 2.4 cm

Hence, OA is C.

Codebug4it · Answer

Answer: OPTION C - 2.4

As AB is a tangent and radius of a circle is always perpendicular on a tangent  , hence we can draw a perpendicular from O to AB, as OD.

Using Pythagoras theorem,

AB ^ 2 = (OA) ^ 2 + (OB) ^ 2  = (4 ^ 2) + (3 ^ 2)
=>  AB = 5

Area of traingle = 1/2 * (base * height)

As triangle OAB is a right triangle and has two perpendiculars so,

=>  1/2 * (OA * OB) = 1/2 * (AB * OD) 
 =>  1/2 * 4 * 3 = 1/2 * 5 * OD 
 =>  OD = 12 / 5  =  2.4

Crank513 · Answer

Let's say that the points where the tangent is touching the circle be x. 
Angle OXB = 90, and angle XBO is 45. Sin 45 = perpendicular/hypotenuse. 
Hypotenuse is 3, so 1/√2 = ox/3, OX = 3/√2 
Which is 2.08 so 2.1

Posted from my mobile device

urshi · Answer

Answer : Option C ( 2.4cm )

Posted from my mobile device

Hulikunte · Answer

C .

Let point of intersection of circle with Tangent AB be ' D'. 
Also, AB = 5 for Right Angle Triangle AOB with legs OA = 4 & OB = 3

Area of Triangle AOB = 1/2 x OB x OA = 1/2 x AB x OD ( For Base AB , OD is the altitude )
Radius = OD = 12 /5 = 2.4 cm

adityasuresh · Answer

Equate the areas.
1/2*4*3 =6
Consider 5 as the base and draw a radius perpendicular to the base 5
Equate the areas.

Kindly refer to the figure, AB is the tangent to the circle with cente

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

Kindly refer to the figure, AB is the tangent to the circle with cente

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