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Intern  Joined: 21 Jan 2013
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ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 63% (02:09) correct 37% (02:30) wrong based on 237 sessions

HideShow timer Statistics 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

Attachment: 1796929.png [ 6.54 KiB | Viewed 189459 times ]

Originally posted by 005ashok on 18 Jan 2014, 07:52.
Last edited by Bunuel on 20 Feb 2019, 03:27, edited 1 time in total.
Renamed the topic and edited the question.
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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005ashok wrote:
ABC is a right angled triangle with BC=6cm 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

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
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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005ashok wrote:
ABC is a right angled triangle with BC=6cm 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

detailed explanation is given here: http://mathforum.org/library/drmath/view/54670.html
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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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

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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Hi All,

My work as follows.

The area of the right triangle should be same which ever be the base.

1/2 * 8 *6= 1/2 * 10 * h (h= the altitube drawn to the hypotenuse)

h= 4.8.

h= BO+ OP.

B0 = sqrt 2 * r (diagnoal of a square)

sqrt 2 * r + r= 4.8== r = 1.92

Please let me know whether I am correct.

005ashok wrote:
Attachment:
1796929.png
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

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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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The formula for r is when circle is inscribed in right triangle is -

Perpendicular+base-hypotenuses =2 r

Let me know if someone wants to know the derivation. Current Student D
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GRE 1: Q169 V154 Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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There are two ways as far as i know
i will use the area formula as that is more likely to hit my mind first
although the rule states that for any circle to be inscribed in an 90 triangle => the radius => P+B -H/2 where P,B,H are three sides of the triangle and H being the largest
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GMAT 1: 640 Q44 V35 Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Quote:
Hi All,

My work as follows.

The area of the right triangle should be same which ever be the base.

1/2 * 8 *6= 1/2 * 10 * h (h= the altitube drawn to the hypotenuse)

h= 4.8.

h= BO+ OP.

B0 = sqrt 2 * r (diagnoal of a square)

sqrt 2 * r + r= 4.8== r = 1.92

Please let me know whether I am correct.

I don' think you're completely correct .
you assumed that the altitude drawn from B to AC , and the bisector of angle A are the same line .
This is not always true , unless ABC is a isosceles triangle ( in this problem it is not ) .

Where does that come from ?
the bisectors of angles of any triangle will intercept in one point ,O , which is the center of the inscribed circle .
its radius r is not necessarily equals the distance between O and A , or B or C .
Now , if the triangle is isosceles ,then the altitude and the angle bisector will be the same ( only the one drawn to the third side, not the legs )
you have assumed this case .

Hope it is clear .
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Originally posted by foryearss on 03 May 2018, 05:18.
Last edited by foryearss on 03 May 2018, 05:33, edited 1 time in total.
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GMAT 1: 640 Q44 V35 Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Quote:
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

I tried to solve it with another method , but had two values for r , why?
these triangles are congruent : MOC ,MNC , the area of NOC is r(6-r)/2 . the area of MONC is r(6-r)
these triangles are congruent MOA , AOP , the area of OAP is r(8-r)/2 .the area of MOPA is r(8-r)

the are of the big triangle ABC consists of : MONC , MOPA , and the square shape ONBP which has a side of r .
r^2 + r(6-r) + r(8-r) = 6*8/2
I solved and gor two values for r
r=2
r=12
Why am i getting two values one of them is 12 ?
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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005ashok wrote:
Attachment:
1796929.png
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

As sides BC=6 am and AB = 8 cm, we can derive AC=10 cm. Let BM =h.

Area of triangle ABC = $$\frac{1}{2}$$*10*h=$$\frac{1}{2}$$*6*8 -> h=4.8 cm.

I used simple logic that 2r+something = 4.8 cm so r<2.4 cm . Looking at the answer choices only 2 cm makes sense.

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GMAT 1: 710 Q44 V41 Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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stonecold wrote:
There are two ways as far as i know
i will use the area formula as that is more likely to hit my mind first
although the rule states that for any circle to be inscribed in an 90 triangle => the radius => P+B -H/2 where P,B,H are three sides of the triangle and H being the largest

That is an interesting formula even though you forgot to mark the brackets I think. It should be $$\frac{P+B-H}{2}$$ - so in this case $$\frac{6+8-10}{2}= 2$$

I will try to remember this formula, thank you!
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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The easiest way to deal with this question is 6-8-10 is pythagorean triplet.
Area of the trinagle = 1/2*6*8 = 24 and also when a circle is inscribed inside a triangle area of the triangle = RS, where R= inradius and S= semi-perimeter
So, 24= r* (6+8+10)/2
r=2.
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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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. 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
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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To calculate incentre of right triangle

We have formula

R=( sum of other two sides - hypotenuse)/2

= AB +BC-AC/2
=8+6-10/2= 2

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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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005ashok wrote: 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
Attachment:
1796929.png

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
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GMAT 1: 650 Q47 V33 Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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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
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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005ashok wrote: 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

Attachment:
1796929.png

$$100 = (14-2r)^2$$

$$r = 2cm$$
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Area = S*r where S=(a+b+c)/3 and r is inner radius.
S= 8+6+10/3= 8
A= 24= 1/2*8*6
Hence, r= 2 mm
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Quick elimination tip

BC = 6

BC > BM

BM > 2r

Therefore 3 > r

Eliminate C, D, E, guess and move on.
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Re: ABC is a right angled triangle with BC = 6 cm and AB = 8 cm. AC is  [#permalink]

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Is the concept of in-radius, circumcenter etc in the syllabus of GMAT ?
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