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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink]
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incenter of a circle= Area/semi-perimeter

Area= \(\frac{1}{2}*6*8\)=24
Semi-perimeter= \(\frac{6+8+10}{2}\)=12

Incenter= 24/12=2





Bunuel wrote:
XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the hypotenuse. A circle with center O and radius x has been inscribed in the triangle. What is the value of x?

A. 1.8
B. 2
C. 2.4
D. 3
E. 4


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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink]
nick1816 wrote:
incenter of a circle= Area/semi-perimeter

Area= \(\frac{1}{2}*6*8\)=24
Semi-perimeter= \(\frac{6+8+10}{2}\)=12

Incenter= 24/12=2





Bunuel wrote:
XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the hypotenuse. A circle with center O and radius x has been inscribed in the triangle. What is the value of x?

A. 1.8
B. 2
C. 2.4
D. 3
E. 4


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I don't think this formula is required on the GMAT. Is there any way to derive this solution for the given problem?
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XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink]
Incircle (r)= Area of Traingle/Semi Perimeter
.


r = 6*8/2*12
r = 2

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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink]
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don't know if this is a common rule we need to know for the GMAT, but it is helpful in this 1 Case.

In a Right Triangle, the Inscribed Triangle's Radius (Inradius) is Equal to = (semi-perimeter) - Hypotenuse

x = (6 + 8 + 10) / 2 - 10
x = 12 - 10 = 2

Answer choice: B (if you know the obscure Rule, takes less than 30 seconds)
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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink]
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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink]
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