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XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the
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Bunuel
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XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  25 Nov 2019, 00:26
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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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nick1816
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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  31 Jul 2020, 08:20
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1. the radius of the circle to the point of tangency is perpendicular to the tangent line.

2. two tangents drawn from an external point to a circle are of equal length. [You can prove it by congruence of triangles]

Attachment:
Untitled.png
Untitled.png [ 5.52 KiB | Viewed 4638 times ]


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

r = 2










catinabox wrote:
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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Archit3110
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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  25 Nov 2019, 02:38
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Radius of Incircle of Right angled triangle is given r=(a+b−c)/2
a=6,b=8 and c=10
we get ; 4/2 ; 2
IMO B
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


Are You Up For the Challenge: 700 Level Questions
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nick1816
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Joined: 19 Oct 2018
Posts: 2266
Location: India
Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  25 Nov 2019, 02:47
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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


Are You Up For the Challenge: 700 Level Questions
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catinabox
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Joined: 27 Nov 2019
Posts: 78
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GMAT 1: 720 Q49 V40
GMAT 2: 770 Q50 V44
Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  31 Jul 2020, 07:31
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


Are You Up For the Challenge: 700 Level Questions


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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Hemanthdasu13
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XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  31 Jul 2020, 09:01
Incircle (r)= Area of Traingle/Semi Perimeter
.


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

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Fdambro294
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Re: XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the [#permalink] New post  15 Aug 2020, 20:54
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]
15 Aug 2020, 20:54
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