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

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

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

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

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

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

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.

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.

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