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

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

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19 Oct 2016, 22:31
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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 [ 6.71 KiB | Viewed 40724 times ]

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

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20 Nov 2016, 05:35
2
4
Bunuel wrote:

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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19 Oct 2016, 22:54
1
4
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 = $$8pi(\sqrt{2} - 1)^2$$

##### General Discussion
Manager
Joined: 11 Jul 2016
Posts: 80
Re: A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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22 Oct 2016, 09:05
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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Joined: 17 Aug 2016
Posts: 48
A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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30 Oct 2016, 10:22
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 = $$8pi(\sqrt{2} - 1)^2$$

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!
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2203
A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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19 Nov 2016, 12:55
1
Bunuel wrote:

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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Posts: 21
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Re: A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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21 Nov 2016, 01:12
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 = $$8pi(\sqrt{2} - 1)^2$$

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
Manager
Joined: 20 Apr 2014
Posts: 92
Re: A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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22 Nov 2016, 07:54
1
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 = $$8pi(\sqrt{2} - 1)^2$$

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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Joined: 13 Apr 2015
Posts: 1689
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
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Re: A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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22 Nov 2016, 08:13
1
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 = $$8pi(\sqrt{2} - 1)^2$$

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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21 Mar 2017, 00:48
Bunuel wrote:

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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11 Apr 2018, 04:31
1
Bunuel wrote:

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 [ 378.54 KiB | Viewed 12330 times ]

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

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11 Apr 2018, 23:55
Tulkin987 wrote:
Bunuel wrote:

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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12 Apr 2018, 20:23
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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23 Apr 2018, 14:53
chetan2u wrote:
Bunuel wrote:

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

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Joined: 02 Sep 2009
Posts: 50619
Re: A circle is inscribed inside right triangle ABC shown above. What is  [#permalink]

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23 Apr 2018, 20:20
getitdoneright wrote:
chetan2u wrote:
Bunuel wrote:

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

Edited that post. Hope it's clear now.
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Re: A circle is inscribed inside right triangle ABC shown above. What is &nbs [#permalink] 23 Apr 2018, 20:20
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