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

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Math Expert
Joined: 02 Sep 2009
Posts: 43916
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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Question Stats:

38% (02:34) correct 62% (02:30) wrong based on 196 sessions

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

[Reveal] Spoiler:
Attachment:

T7909.png [ 6.71 KiB | Viewed 18587 times ]
[Reveal] Spoiler: OA

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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
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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 = $$8pi(\sqrt{2} - 1)^2$$

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.
Intern
Joined: 17 Aug 2016
Posts: 49
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!
Intern
Joined: 16 Oct 2016
Posts: 13
GMAT 1: 460 Q30 V23
Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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30 Oct 2016, 11:49
I totally lost when I saw the formula for the incircle!

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 806
A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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19 Nov 2016, 12:55
1
KUDOS
Expert's post
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. $$8\pi(1-\sqrt{2})$$

[Reveal] Spoiler:
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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Math Expert
Joined: 02 Aug 2009
Posts: 5663
Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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20 Nov 2016, 05:35
Expert's post
2
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BOOKMARKED
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. $$8\pi(1-\sqrt{2})$$

[Reveal] Spoiler:
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 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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Intern
Joined: 10 Nov 2016
Posts: 16
GMAT 1: 700 Q50 V35
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: 113
Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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22 Nov 2016, 07:54
1
KUDOS
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
SC Moderator
Joined: 13 Apr 2015
Posts: 1590
Location: India
Concentration: Strategy, General Management
WE: Analyst (Retail)
Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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

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.
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1277
Location: Malaysia
A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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21 Mar 2017, 00:13
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. $$8\pi(1-\sqrt{2})$$

[Reveal] Spoiler:
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 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

Dear Bunuel, Does inradius concept test in GMAT?
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Intern
Joined: 12 Dec 2016
Posts: 12
Re: A circle is inscribed inside right triangle ABC shown above. What is [#permalink]

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21 Mar 2017, 00:29
I did this by ballparking. Just assumed that since the circle is approx. half of the triangle, it should be somewhere around 4, and the closest value was answer B. Can anyone confirm my approach?
Senior Manager
Joined: 05 Jan 2017
Posts: 434
Location: India
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. $$8\pi(1-\sqrt{2})$$

[Reveal] Spoiler:
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
Re: A circle is inscribed inside right triangle ABC shown above. What is   [#permalink] 21 Mar 2017, 00:48
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