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A circle is inscribed inside right triangle ABC shown above. What is

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A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png
T7909.png [ 6.71 KiB | Viewed 28077 times ]

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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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Area of triangle = 1/2 * 4 * 4 = 8

Radius of incircle = 2(Area)/(Sum of the sides of triangle)
= 2*8/(4 + 4 + 4*\(\sqrt{2}\)))
= \(2(2 - \sqrt{2})\)
= \(2\sqrt{2}(\sqrt{2} - 1)\)

Area of incircle = pi*(radius)^2 = \(8pi(\sqrt{2} - 1)^2\)

Answer: B
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 22 Oct 2016, 10:05
We can use Hero's formula as well to calculate area of triangle

A = sqrt{s(s-a)(s-b)(s-c)} where s={a+b+c}/2

However this will unnecessarily consume time. So we can use 1/2.b.h to calculate area

1/2.b.h = 1/2.4.4. (since being a right angled isosceles triangle)
Area = 8

radius of incircle = 2(Area of triangle) / Sum of sides
= 2.8/ 8 + 4 sq rt 2
= 2sq rt2 /1+ sq rt 2

Area of circle = Pi * R^2
= Pi * (2sq rt2 /1+ sq rt 2)^2
= 8 * Pi ( 3 - 2sq rt2)
Now look for options B or E

B can be re-written as 8 * Pi ( 2+1 - 2 sq rt 2)

Hence this matches our earlier reached out equation. Option B is correct choice.
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A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 30 Oct 2016, 11:22
Vyshak wrote:
Area of triangle = 1/2 * 4 * 4 = 8

Radius of incircle = 2(Area)/(Sum of the sides of triangle)
= 2*8/(4 + 4 + 4*\(\sqrt{2}\)))
= \(2(2 - \sqrt{2})\)
= \(2\sqrt{2}(\sqrt{2} - 1)\)

Area of incircle = pi*(radius)^2 = \(8pi(\sqrt{2} - 1)^2\)

Answer: B



Is the formula you are using applicable to all the circles inscribed in a right triangle? Or only for those inscribed in 45-45-90?

And do you have an explanation for the formula?

Thanks!
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A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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Bunuel wrote:
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A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png



Solution :


Given : 1. ABC is a right angled triangle
2. A circle is inscribed in the triangle
3. AB = 4 cm and angle ACB = 45 degrees

Find : Area of the circle.

Approach :

To find the area of the circle, we need to find the radius of the circle.

It is given that ABC is a right angled triangle and angle ACB is 45 degrees.

From the above information, we can infer that -

angle BAC = \(180 - ACB - ABC = 180 - 45 - 90 = 45\)

Therefore we can conclude that ABC is an isosceles right-angled triangle.

And AB will be equal to BC.


We know that area of the triangle =1/2 x base x height.........(i)

AND

We can also find the area of the triangle by using the formula = in-radius of the circle x semi-perimeter.......(ii)

Equating the above two formulas, we can find the in-radius of the circle.

Once we have the in-radius, we can find the area of the circle.

Working :

Since AB is equal to BC

We can write,

AB = BC = 4cm


Area of the triangle = 1/2 x AB x BC
Area of the triangle = 1/2 x 4 x 4
Therefore,
Area of the triangle = 8..................(i)


Also
Using Pythagorus Theorem, we can write,

\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 4^2 + 4^2\)
\(AC^2 = 16 + 16\)
AC = 4\(\sqrt{2}\)

Perimeter of the triangle of ABC = AB + BC + CA
= \(4 + 4+ 4\sqrt{2}\)
= \(4 (2 + \sqrt{2})\)

Semi - perimeter =\(\frac{Perimeter of Triangle}{2}\)
= \(4 (2 + \sqrt{2})/2\)
= \(2 (2 + \sqrt{2})\)


As discussed in the analysis,
area of the triangle can also be written as = in-radius x semi-perimeter
Area of the triangle = in-radius x \(2 (2 + \sqrt{2})\) ..........(ii)

Equating (i) and (ii), we get

in-radius x \(2 (2 + \sqrt{2})\) = 8

in - radius = \(4/(2+\sqrt{2})\)
= \(4(2-\sqrt{2})/(2-\sqrt{2})(2+\sqrt{2})\)
= \(4(2-\sqrt{2})/(4 -2)\)


Therefore,
in-radius = \(2(2- \sqrt{2})\)
= \(2\sqrt{2}(\sqrt{2} - 1)\)
Therefore, Area of the circle = pi x \((in-radius)^2\)
Area of the circle = pi x \([2\sqrt{2}(\sqrt{2} - 1)]^2\)
Area of the circle = pi x \(8(\sqrt{2}- 1)]^2\)


Hence, the correct answer is Option B


Thanks,
Saquib
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A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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Bunuel wrote:
Image
A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png


Hi all,
Are there we to know inradius concept.. Maybe NO..
So there has to be some other way, and a simpler one...

Two ways..

1) proper method

Let the circle touch the side AB at E..
So BE will be radius and = AB-BE
Since the ∆ is isosceles,45°at C, AB=BC=4 and hypotenuse AC is 4√2..
the circle will touch hypotenuse AC at mid point,say F, as it is isosceles..
So AF is 4√2/2=2√2. But AF will also be equal to AE..
a point outside circle will be equidistant from the tangents to it..


So now we have our radius BE as \(AB-BE = 4-2√2\)..
So Area \(=\pi(4-2√2)^2= \pi*(2√2)^2*(√2-1)^2=8\pi (√2-1)^2\)

B

2) choices..
Now what is the Area of triangle \(\frac{1}{2} *4*4=8\)...
Can the area of circle be greater than 8, NO
So D and E are straight OUT..
C is \(2π=2*3.14=6.28\), again very close to 8, out..
Now left A and B..
B is \(8*3.14*0.414^2=1.28*3.14~4\), close to half of ∆...OK
A is half that is 2, almost 1/4 of triangle.. it can be seen that circle is almost Half of ∆
B
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 21 Nov 2016, 02:12
bazu wrote:
Vyshak wrote:
Area of triangle = 1/2 * 4 * 4 = 8

Radius of incircle = 2(Area)/(Sum of the sides of triangle)
= 2*8/(4 + 4 + 4*\(\sqrt{2}\)))
= \(2(2 - \sqrt{2})\)
= \(2\sqrt{2}(\sqrt{2} - 1)\)

Area of incircle = pi*(radius)^2 = \(8pi(\sqrt{2} - 1)^2\)

Answer: B



Is the formula you are using applicable to all the circles inscribed in a right triangle? Or only for those inscribed in 45-45-90?

And do you have an explanation for the formula?




Thanks!



This is valid for all right triangles.
I think if you start with the concept that tangential distance from any external points is equal. Then you can derive the formula.
Hope it helps :)
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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bazu wrote:
Vyshak wrote:
Area of triangle = 1/2 * 4 * 4 = 8

Radius of incircle = 2(Area)/(Sum of the sides of triangle)
= 2*8/(4 + 4 + 4*\(\sqrt{2}\)))
= \(2(2 - \sqrt{2})\)
= \(2\sqrt{2}(\sqrt{2} - 1)\)

Area of incircle = pi*(radius)^2 = \(8pi(\sqrt{2} - 1)^2\)

Answer: B



Is the formula you are using applicable to all the circles inscribed in a right triangle? Or only for those inscribed in 45-45-90?

And do you have an explanation for the formula?

Thanks!


please clarify the operation as much as possible.
many thanks
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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hatemnag wrote:
bazu wrote:
Vyshak wrote:
Area of triangle = 1/2 * 4 * 4 = 8

Radius of incircle = 2(Area)/(Sum of the sides of triangle)
= 2*8/(4 + 4 + 4*\(\sqrt{2}\)))
= \(2(2 - \sqrt{2})\)
= \(2\sqrt{2}(\sqrt{2} - 1)\)

Area of incircle = pi*(radius)^2 = \(8pi(\sqrt{2} - 1)^2\)

Answer: B



Is the formula you are using applicable to all the circles inscribed in a right triangle? Or only for those inscribed in 45-45-90?

And do you have an explanation for the formula?

Thanks!


please clarify the operation as much as possible.
many thanks


Yes it is applicable to all circles inscribed in a triangle and is not restricted to just right triangle.
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 21 Mar 2017, 01:48
Bunuel wrote:
Image
A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png


Area of the triangle = r.s = 1/2 .b .h
s=(4+4+4√2)/2 = 4+2√2

1/2.b.h = 1/2 . 4 . 4 = 8

therefore

r. (4+2√2) = 8
r = 4/(2+√2)

therefore area = pi. r^2 = pi.{2(2-√2)}^2 = 8 pi(√2 - 1)^2
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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Bunuel wrote:
Image
A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
The attachment T7909.png is no longer available


For any right triange the shortcut r=(a+b-c)/2 holds true (see uploaded explanation).
So, in this problem, r=\((4+4-4\sqrt{2})/2\)=\(2\sqrt{2}(\sqrt{2}-1)\).
And S=\(8\pi(\sqrt{2}-1)^2\).
Hence, B.
Attachments

C313DC18-ACBC-4ACA-8912-793804313ACA.jpeg
C313DC18-ACBC-4ACA-8912-793804313ACA.jpeg [ 378.54 KiB | Viewed 2997 times ]


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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 12 Apr 2018, 00:55
Tulkin987 wrote:
Bunuel wrote:
Image
A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png


For any right triange the shortcut r=(a+b-c)/2 holds true (see uploaded explanation).
So, in this problem, r=\((4+4-4\sqrt{2})/2\)=\(2\sqrt{2}(\sqrt{2}-1)\).
And S=\(8\pi(\sqrt{2}-1)^2\).
Hence, B.


Superb approach !!! Thanks :)
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 12 Apr 2018, 21:23
spetznaz wrote:
Superb approach !!! Thanks :)


Happy to help!!! :-)
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 23 Apr 2018, 15:53
chetan2u wrote:
Bunuel wrote:
Image
A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png


Hi all,
Are there we to know inradius concept.. Maybe NO..
So there has to be some other way, and a simpler one...

Two ways..

1) proper method

Let the circle touch the side AB at E..
So BE will be radius and = AB-BE
Since the ∆ is isosceles,45°at C, AB=BC=4 and hypotenuse AC is 4√2..
the circle will touch hypotenuse AC at mid point,say F, as it is isosceles..
So AF is 4√2/2=2√2. But AF will also be equal to AE..
a point outside circle will be equidistant from the tangents to it..


So now we have our radius BE as \(AB-BE = 4-2√2\)..
So Area \(=π{4-2√2)^2= π*(2√2)^2*(√2-1)^2=8π(√2-1)^2\)

B

2) choices..
Now what is the Area of triangle \(\frac{1}{2} *4*4=8\)...
Can the area of circle be greater than 8, NO
So D and E are straight OUT..
C is \(2π=2*3.14=6.28\), again very close to 8, out..
Now left A and B..
B is \(8*3.14*0.414^2=1.28*3.14~4\), close to half of ∆...OK
A is half that is 2, almost 1/4 of triangle.. it can be seen that circle is almost Half of ∆
B


could you please help explain method 1? Can't see your calculation clearly.
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Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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New post 23 Apr 2018, 21:20
getitdoneright wrote:
chetan2u wrote:
Bunuel wrote:
Image
A circle is inscribed inside right triangle ABC shown above. What is the area of the circle, if the leg AB of the right triangle is 4?

A. \(4\pi(\sqrt{2}-1)^2\)

B. \(8\pi(\sqrt{2}-1)^2\)

C. \(2\pi\)

D. \(4\pi\)

E. \(8\pi(1-\sqrt{2})\)

Attachment:
T7909.png


Hi all,
Are there we to know inradius concept.. Maybe NO..
So there has to be some other way, and a simpler one...

Two ways..

1) proper method

Let the circle touch the side AB at E..
So BE will be radius and = AB-BE
Since the ∆ is isosceles,45°at C, AB=BC=4 and hypotenuse AC is 4√2..
the circle will touch hypotenuse AC at mid point,say F, as it is isosceles..
So AF is 4√2/2=2√2. But AF will also be equal to AE..
a point outside circle will be equidistant from the tangents to it..


So now we have our radius BE as \(AB-BE = 4-2√2\)..
So Area \(=π{4-2√2)^2= π*(2√2)^2*(√2-1)^2=8π(√2-1)^2\)

B

2) choices..
Now what is the Area of triangle \(\frac{1}{2} *4*4=8\)...
Can the area of circle be greater than 8, NO
So D and E are straight OUT..
C is \(2π=2*3.14=6.28\), again very close to 8, out..
Now left A and B..
B is \(8*3.14*0.414^2=1.28*3.14~4\), close to half of ∆...OK
A is half that is 2, almost 1/4 of triangle.. it can be seen that circle is almost Half of ∆
B


could you please help explain method 1? Can't see your calculation clearly.


Edited that post. Hope it's clear now.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A circle is inscribed inside right triangle ABC shown above. What is   [#permalink] 23 Apr 2018, 21:20
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