The circle above is inscribed inside right triangle ABC. What is the

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Competition Mode Question https://gmatclub.com/forum/download/file.php?id=56021 The circle above is inscribed inside right triangle ABC (angle B is a right angle). What is the radius of the circle if BC = 8? A. \frac{\sqrt{8}}{4} B. \frac{\sqrt{2}}{\sqrt{2}+1} C. 8-4\sqrt{2} D. 4\sqrt{2} E. \frac{4\sqrt{2}}{\sqrt{2}-1} download (7).png

exc4libur · Accepted Answer

radius of circle inscribed in a right triangle: 
(a+b-hyp)/2 or area/semi-perimeter

since one angle is 45 this is an isosceles right triangle
sides proportion = x:x:x√2 = 8:8:8√2

radius: (8+8-8√2)/2=16-8√2/2=8(2-√2)/2=8-4√2

ans (C)

***other formulas

radius circle INSCRIBED equilateral:
(side of triangle)/2√3=(side√3)/6

radius circle CIRCUMscribed equilateral:
(side of triangle)/√3=(side√3)/3

GMATWhizTeam · Answer

Solution 
 • Circle is inscribed in the triangle ABC. 
 o Let us assume that the radius of the given circle is r. 
• Triangle ABC is right angled at B.
• ∠BCA = 45
 o So, triangle ABC is a 45-45-90 triangle.
  Hence, ∠BCA = ∠CAB = 45
 And, BC = AB = 8
 • Therefore, AC  = 8\sqrt{2}    
• Based on the above information let us re-draw the given diagram as below:
  https://www.dropbox.com/s/ykyd5aqg7e1phzy/16_07_tri.jpg?dl=1 
• From the above diagram,
 o Area of triangle ABC = Area of triangle AOB + Area of triangle BOC + Area of triangle COA

⟹ \frac{1}{2}*BC*AB = \frac{1}{2}*AB*OF + \frac{1}{2}*BC*OD = \frac{1}{2}*AC*OE

⟹ \frac{1}{2}*8*8 = \frac{1}{2}*8*r +\frac{1}{2}*8*r + \frac{1}{2}*8 \sqrt{2}*r

⟹ 2r + \sqrt{2}r = 8

⟹ r = \frac{8}{2 +\sqrt{2}} = \frac{8*(2 -\sqrt{2})}{(2 +\sqrt{2})(2 -\sqrt{2})} = \frac{8*(2 -\sqrt{2})}{2} = 8 - 4\sqrt{2}

Thus, the correct answer is  Option C.

MayankSingh · Answer

IMO C

Refer attached image.

GMATinsight · Answer

Radius of Incircle = ∆/s

∆ = Area of Triangle 
s = Semiperimeter  = \frac{a+b+c}{2}  where a, b and c are sides of triangle

∆ = Area of Triangle =  (\frac{1}{2})8*8 = 32 
s = Semiperimeter  = \frac{8+8+8√2}{2}

i.e. Radius of Incircle = (1/2)*8*8/ = \frac{32*2}{8(2+√2)} = \frac{8}{(2+√2)} = \frac{8(2-√2)}{2} = 8-4√2

\frac{8}{(2+√2)} = \frac{8*(2-√2)}{(2+√2)*(2-√2)} = \frac{8*(2-√2)}{(2^2-√2^2)} = = \frac{8*(2-√2)}{(4-2)} = 8-4√2

Answer: Option C

nick1816 · Answer

Legs of isosceles right angle triangle= 8

Hypotenuse = 8√2

Incenter of a right angle triangle  =\frac{ 8+8 - 8√2 }{2} = 8 - 4√2

unraveled · Answer

Image
The circle above is inscribed inside right triangle ABC (angle B is a right angle). What is the radius of the circle if BC = 8?

A.  \frac{√8}{4}

B.  \frac{√2}{√2 + 1}

C. 8 − 4√2

D. 4√2

E.  \frac{4√2}{√2-1} 
AB = a, BC = b and AC = c
Here AB = BC = 8 since triangle ABC has angles(eventually sides) in ratio 1:1:√2.
Hence AC = 8√2
Radius of circle,  r = \frac{a + b - c }{2} 
 r = \frac{8 + 8 - 8√2}{2} 
r = 8 - 4√2

Answer C.

GMATinsight · Answer

SOLUTION-2

https://www.youtube.com/watch?v=9Ps3jJEQvos

Answer: Option C

Shubham1213372 · Answer

There is direct formula to calculate radius if circle is inscribed in right traingle and that is :

R= (a+b-c)/2.

We know given traingle is right traingle whose angles are 90°,45°,45°
Let AC =c
BC =b
AB =a

BC(b) = AB(a) = 8.
By hypotenuse formula,

AC(c) = √8^2 + 8^2 
AC(c) = 8√2

We get a,b, and c values.
Put in the formula.

=(a+b-c)/2
=(8+8-8√2)/2
=4+4-4√2
=8-4√2

Hence, correct answer should be C

Posted from my mobile device

Shubham1213372 · Answer

There is direct formula to calculate radius if circle is inscribed in right traingle.
 R= (a+b-c)/2

We know given traingle is right traingle whose angles are 90°,45°,45°
Let AC =c ,BC =b, and AB =a
Given: BC = 8 
BC and AB angles are equals hence there sides will also be equal.
Therefore, BC(b) = AB(a) = 8.
By hypotenuse formula,

AC(c) = √8^2 + 8^2 
AC(c) = √64+64
AC(c) = √128
AC(c) = 8√2

We get a,b, and c values.Put in the formula.

=(a+b-c)/2
=(8+8-8√2)/2
=4+4-4√2
=8-4√2

Hence, correct answer should be C

Expecting Kudos, if it helps.

Archit3110 · Answer

The circle above is inscribed inside right triangle ABC (angle B is a right angle). What is the radius of the circle if BC = 8?
Radius of incircle in right angled can be determined by

(a+b-c)/2

where a, b & c are sides of right angled ∆:

; side of triangle 8: 8: 8 sqrt 2 as its a 45:45:90 ∆ x:x:x√2
radius of circle 8-4 sqrt 2.
OPTION C

debjit1990 · Answer

Ans: C
angles are:90-45-45
so ab=bc=8
ac=8√2
r=(a+b-c)/2
=(8+8-8√2)/2=8−4√2

monikakumar · Answer

45 45 90 triangle
so 1:1:root2

a=8, b=8, c =8root2
inradius = a+b-c/2
8+8-8root2/2
8(2-root2)/2
4(2-root2)
8-4root2

Ans C

v12345 · Answer

Radius of inscribed circle in isosceles right triangle =  \frac{side}{2}(2-\sqrt{2}) = 4(2-\sqrt{2})  =  8 - 4\sqrt{2}

Hence, OA is (C).

Gowtham91 · Answer

The given triangle is an isosceles right triangle. And hence AB=BC=8. And AC= 8\sqrt{2}

Formula for inradius of a circle inscribed in a right triangle when the sides are known =  \frac{a+b-c}{2}   , where a and b are sides of the triangle and c is the hypotenuse.

Threfore we have inradius =  \frac{8+8-8\sqrt{2}}{2}  =  8-4\sqrt{2}  -  Option C

bumpbot · Answer

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The circle above is inscribed inside right triangle ABC. What is the

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