ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is

Question

https://gmatclub.com/forum/download/file.php?id=27154 ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is hypotenuse. A circle with centre O and radius x has been inscribed. What is the value of x. A. 2.4 cm B. 2 cm C. 3.6 cm D. 4 cm E. 3 cm 1796929.png

GMATPill · Accepted Answer

Once you have circle inscribed into that triangle -- you'll notice you won't have the traditional tips to solve this question. This is a 3-4-5 triangle but the angles are not the typical 30-60-90 degree triangle. Instead you have so solve for r - radius.

The way to solve this is to draw a line from radius to all 3 vertices - so you have 3 separate lines. This creates OA, OB, and OC and 3 smaller triangles within the larger 6-8-10 triangle. THe sum of the 3 smaller triangles should equal the sum of the large triangle.

The reason we care about the smaller triangles is because we know their "height" is going to be "r". So 1/2 * base * height, the height will be r.

1/2 * 6* r + 1/2 * 8 * r + 1/2 * 10 * r = 1/2 * 8 * 6
12r = 24
r = 2

manishkhare · Answer

This is a right angled triangle with sides 6 and 8 .Third side will be 10(Pythagorean triplets) .
The circle is inscribed in the triangle . So all the three sides are at tangent to the circle .A line drawn from center of the circle will be perpendicular to the sides of the triangle

Area of triangle :6*8=48 
The larger triangle can be broken into three smaller triangles with base =8,height=r ,base=10,height=r ,base 6 height=r .
Sum of areas of these three triangle :8r+10r+6r=24r .24r=48 ,hence r=2 .Option B

Regards,
Manish Khare

Princ · Answer

OA: B
It is a right angled triangle and 6,8 and 10 are Pythagorean triplet.
So AC=10
Radius of Incircle of Right angled triangle is given  r =\frac{(a+b−c)}{2} 
Where c is hypotenuse, a and b are other two sides.
 https://gmatclub.com/forum/download/file.php?id=40863

Reason for AE=AD and EC=CF: length of tangents drawn from an external point to a circle are equal

here putting c = 10, a=6 , b= 8, we get  r =\frac{(6+8−10)}{2}=2 
OA=B

bunny12345 · Answer

Option B it is.
I have a different way of solving it. Please give kudos if it seems interesting to you.

Its obvious AC is 10. Now, if a draw a perpendicular from B on AC (Let the length of perpendicular be P), then that line will coincide with the diameter of incircle(with r radius) extended to point B. 
Now, equate the area of triangle ABC

1/2*AB*BC = 1/2*AC*P 
From this we get P = 4.8

P can also be written as r + r*root2.

Hence, when r = 2, then 2 + 2*root2 gives us 2 + 2*1.4 = 2 + 2.8 =  4.8

chandra004 · Answer

This may be a useful shortcut:-
AP = 8-r; And since we know that length of tangents from a single point are equal, hence, AM = 8-r
Similarly, CM = 6-r
Also by Pythagoras theorem, AC = 10
So AP + CM = AC
=> 8-r + 6-r = 10
Hence r = 2. et voilà

GmatKnightTutor · Answer

Knowing the in-radius rule for this sort of thing can be helpful, but I wonder if simply trying to visualize that last step may be good, too.

3 of the answer choices can be quite quickly eliminated off the bat.

Fdambro294 · Answer

First, we have a multiple of a 3-4-5 right triangle ——-> hypotenuse = AC = 10 cm

Since we are given the radius = r from point O to M at the hypotenuse, it becomes an exercise of setting the hypotenuse equal to 2 expressions

Rule: When the radius of a circle is drawn to the point of tangency, it creates a 90 degree angle with the tangent line

Thus the radius OP is perpendicular to side AB

As well as the radius ON is perpendicular to side BC

This creates a square OPBN of side r

The remaining length from point N to C = NC = 6 - r

The remaining length from point P to A = PA = 8 - r

(2) rule: when 2 lines are drawn from the same exterior point and are tangent to the same circle, the distance from the exterior point to the point of tangency will be equal

Since we know that ON is the radius of the inscribed circle, OM will be drawn perpendicular to the hypotenuse.

The inscribed circle is tangent to the Right Triangle at Points: M , P, and N

(From exterior point C)

CM and NC will be equal tangent lines drawn from the same exterior point

We found the value of NC above

NC = 6 - r = CM

(From exterior point A)

DA and AM will be equal tangent lines drawn from the same exterior point

DA = 8 - r = AM

Finally, the hypotenuse = AM + CM = 10

Which means:

(6 - r) + (8 - r) = 10

14 - 2r = 10

4 = 2r

r = 2

Posted from my mobile device

NoelGEM · Answer

This question was part of 10th board exam in Indian CBSE board and I somehow remebered it.Nostalgia

WheatyPie · Answer

By far, the easiest method to answer this is via approximation. Comparing BC to BN, for example by putting your fingers or pen or paper to the screen, copying the length of BN and placing it repeatedly along BC, BN seems to be almost exactly one third of BC. Answer B.

The problem though is that this is not actually a GMAT question - so it's not really to scale. If this were a real GMAT question, this would be the technique to use.

GSV · Answer

From Pythagorean theorem, AC = 10 
Area of ABC = Area of AOB + Area of BOC + Area of COA 
0.5 * 8 * 6 = 0.5*8r + 0.5*6r +0.5*10r 
48 = 24 r 
r =2

ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is

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ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is

Prep Toolkit

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