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How to find the radius of this smaller circle?
Find legs of a right triangle if radius of circumscribed circle is 15 and radius of inscribed circle is 6What is the radius of the circle?Prove that $O$ is the center of circle inscribed in triangle $O_1,O_2,O_3$ iff circles $k_1,k_2,k_3$ have same radiusHow to find the radius of small inscribed circle?Finding the radius of a circle inside of a triangleFind the radius of the circle.Find length of triangle as well as radius of inscribed circleFind the length of the sides of an equilateral triangle inscribed in a circle of radius $6$cm.What is the radius of a circle that has a 8 15 17 inscribed within?Proving radius of circle $dfractriangleatan^2dfracA2$
$begingroup$
The question says, "A circle is inscribed in a triangle whose sides are $40$ cm, $40$ cm and $48$ cm respectively. A smaller circle is touching two equal sides of the triangle and the first circle. Find the radius of smaller circle."
I can find the radius of the inscribed circle fairly easily by assuming the radius as $r$, and using the Heron's Formula: $$frac12 * r * (40 + 40 + 48) = sqrtleft(frac40 + 40 + 482right) left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-48right)$$
Which evaluates to give : $r = 12$, so The inscribed circle has a radius of $12$ cm.
But The smaller circle is only in touch with the other circle, and I can't get anything to work like constructions or etc. Trigonometry doesn't work too (maybe I'm doing it wrong, I'm a Grade 11 student anyway).
The most I can do is to find the area which is not occupied by the circle, but occupied by the triangle simply by subtracting the areas of both. [Which is $768 - pi*(12)^2$ cm].
And this question was on a small scholarship paper I've attended, and it also had some more questions like it (I came to solve most of them).
geometry
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
$endgroup$
add a comment |
$begingroup$
The question says, "A circle is inscribed in a triangle whose sides are $40$ cm, $40$ cm and $48$ cm respectively. A smaller circle is touching two equal sides of the triangle and the first circle. Find the radius of smaller circle."
I can find the radius of the inscribed circle fairly easily by assuming the radius as $r$, and using the Heron's Formula: $$frac12 * r * (40 + 40 + 48) = sqrtleft(frac40 + 40 + 482right) left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-48right)$$
Which evaluates to give : $r = 12$, so The inscribed circle has a radius of $12$ cm.
But The smaller circle is only in touch with the other circle, and I can't get anything to work like constructions or etc. Trigonometry doesn't work too (maybe I'm doing it wrong, I'm a Grade 11 student anyway).
The most I can do is to find the area which is not occupied by the circle, but occupied by the triangle simply by subtracting the areas of both. [Which is $768 - pi*(12)^2$ cm].
And this question was on a small scholarship paper I've attended, and it also had some more questions like it (I came to solve most of them).
geometry
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
$endgroup$
1
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
May 16 at 3:55
$begingroup$
@Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all
$endgroup$
– Soumalya Pramanik
May 16 at 4:05
add a comment |
$begingroup$
The question says, "A circle is inscribed in a triangle whose sides are $40$ cm, $40$ cm and $48$ cm respectively. A smaller circle is touching two equal sides of the triangle and the first circle. Find the radius of smaller circle."
I can find the radius of the inscribed circle fairly easily by assuming the radius as $r$, and using the Heron's Formula: $$frac12 * r * (40 + 40 + 48) = sqrtleft(frac40 + 40 + 482right) left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-48right)$$
Which evaluates to give : $r = 12$, so The inscribed circle has a radius of $12$ cm.
But The smaller circle is only in touch with the other circle, and I can't get anything to work like constructions or etc. Trigonometry doesn't work too (maybe I'm doing it wrong, I'm a Grade 11 student anyway).
The most I can do is to find the area which is not occupied by the circle, but occupied by the triangle simply by subtracting the areas of both. [Which is $768 - pi*(12)^2$ cm].
And this question was on a small scholarship paper I've attended, and it also had some more questions like it (I came to solve most of them).
geometry
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
$endgroup$
The question says, "A circle is inscribed in a triangle whose sides are $40$ cm, $40$ cm and $48$ cm respectively. A smaller circle is touching two equal sides of the triangle and the first circle. Find the radius of smaller circle."
I can find the radius of the inscribed circle fairly easily by assuming the radius as $r$, and using the Heron's Formula: $$frac12 * r * (40 + 40 + 48) = sqrtleft(frac40 + 40 + 482right) left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-40right)left(frac40 + 40 + 482-48right)$$
Which evaluates to give : $r = 12$, so The inscribed circle has a radius of $12$ cm.
But The smaller circle is only in touch with the other circle, and I can't get anything to work like constructions or etc. Trigonometry doesn't work too (maybe I'm doing it wrong, I'm a Grade 11 student anyway).
The most I can do is to find the area which is not occupied by the circle, but occupied by the triangle simply by subtracting the areas of both. [Which is $768 - pi*(12)^2$ cm].
And this question was on a small scholarship paper I've attended, and it also had some more questions like it (I came to solve most of them).
geometry
geometry
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
asked May 16 at 3:48
Soumalya PramanikSoumalya Pramanik
310112
310112
1
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
May 16 at 3:55
$begingroup$
@Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all
$endgroup$
– Soumalya Pramanik
May 16 at 4:05
add a comment |
1
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
May 16 at 3:55
$begingroup$
@Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all
$endgroup$
– Soumalya Pramanik
May 16 at 4:05
1
1
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
May 16 at 3:55
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
May 16 at 3:55
$begingroup$
@Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all
$endgroup$
– Soumalya Pramanik
May 16 at 4:05
$begingroup$
@Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all
$endgroup$
– Soumalya Pramanik
May 16 at 4:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
See the figure below. One unit on the paper is six units in your problem. $AB=48,AC=40,BC=40$. Circle $D$ has radius $12$ as you say. $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$. $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $frac 14$ the size of the large one. That says the radius of the small circle is $frac 14 cdot 12=3$
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
$endgroup$
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
2
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
$endgroup$
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC=40$ and $BC=48.$
Also, let $(I,12)$ be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively, $(O,x)$ be the little circle, which touched to $AC$ in the point $F$.
Thus, $$AE=frac40+40-482=16$$ and since $Delta AIEsim Delta AOF,$ we obtain:
$$fracAFAE=fracOFIE$$ or
$$fracAF16=fracx12,$$ which gives
$$AF=frac43x,$$
$$FE=16-frac43x$$ and by the Pythagoras's theorem we obtain:
$$FE=2sqrtIEcdot OF$$ or
$$16-frac43x=2sqrt12x.$$
Can you end it now?
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See the figure below. One unit on the paper is six units in your problem. $AB=48,AC=40,BC=40$. Circle $D$ has radius $12$ as you say. $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$. $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $frac 14$ the size of the large one. That says the radius of the small circle is $frac 14 cdot 12=3$
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
$endgroup$
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
2
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
add a comment |
$begingroup$
See the figure below. One unit on the paper is six units in your problem. $AB=48,AC=40,BC=40$. Circle $D$ has radius $12$ as you say. $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$. $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $frac 14$ the size of the large one. That says the radius of the small circle is $frac 14 cdot 12=3$
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
$endgroup$
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
2
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
add a comment |
$begingroup$
See the figure below. One unit on the paper is six units in your problem. $AB=48,AC=40,BC=40$. Circle $D$ has radius $12$ as you say. $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$. $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $frac 14$ the size of the large one. That says the radius of the small circle is $frac 14 cdot 12=3$
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
$endgroup$
See the figure below. One unit on the paper is six units in your problem. $AB=48,AC=40,BC=40$. Circle $D$ has radius $12$ as you say. $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$. $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $frac 14$ the size of the large one. That says the radius of the small circle is $frac 14 cdot 12=3$
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
answered May 16 at 4:05
Ross MillikanRoss Millikan
306k24202379
306k24202379
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
2
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
add a comment |
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
2
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
$begingroup$
How can we say that the small triangle is $frac14$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth.
$endgroup$
– Soumalya Pramanik
May 16 at 4:10
2
2
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct.
$endgroup$
– Ross Millikan
May 16 at 4:13
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good!
$endgroup$
– Soumalya Pramanik
May 16 at 4:16
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
$EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$
$endgroup$
– Ross Millikan
May 16 at 4:18
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
May 16 at 4:26
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
$endgroup$
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
$endgroup$
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
$endgroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
answered May 16 at 4:13
CY AriesCY Aries
19.8k12046
19.8k12046
add a comment |
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC=40$ and $BC=48.$
Also, let $(I,12)$ be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively, $(O,x)$ be the little circle, which touched to $AC$ in the point $F$.
Thus, $$AE=frac40+40-482=16$$ and since $Delta AIEsim Delta AOF,$ we obtain:
$$fracAFAE=fracOFIE$$ or
$$fracAF16=fracx12,$$ which gives
$$AF=frac43x,$$
$$FE=16-frac43x$$ and by the Pythagoras's theorem we obtain:
$$FE=2sqrtIEcdot OF$$ or
$$16-frac43x=2sqrt12x.$$
Can you end it now?
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
$endgroup$
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC=40$ and $BC=48.$
Also, let $(I,12)$ be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively, $(O,x)$ be the little circle, which touched to $AC$ in the point $F$.
Thus, $$AE=frac40+40-482=16$$ and since $Delta AIEsim Delta AOF,$ we obtain:
$$fracAFAE=fracOFIE$$ or
$$fracAF16=fracx12,$$ which gives
$$AF=frac43x,$$
$$FE=16-frac43x$$ and by the Pythagoras's theorem we obtain:
$$FE=2sqrtIEcdot OF$$ or
$$16-frac43x=2sqrt12x.$$
Can you end it now?
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
$endgroup$
add a comment |
$begingroup$
Let in $Delta ABC$ we have $AB=AC=40$ and $BC=48.$
Also, let $(I,12)$ be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively, $(O,x)$ be the little circle, which touched to $AC$ in the point $F$.
Thus, $$AE=frac40+40-482=16$$ and since $Delta AIEsim Delta AOF,$ we obtain:
$$fracAFAE=fracOFIE$$ or
$$fracAF16=fracx12,$$ which gives
$$AF=frac43x,$$
$$FE=16-frac43x$$ and by the Pythagoras's theorem we obtain:
$$FE=2sqrtIEcdot OF$$ or
$$16-frac43x=2sqrt12x.$$
Can you end it now?
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
$endgroup$
Let in $Delta ABC$ we have $AB=AC=40$ and $BC=48.$
Also, let $(I,12)$ be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively, $(O,x)$ be the little circle, which touched to $AC$ in the point $F$.
Thus, $$AE=frac40+40-482=16$$ and since $Delta AIEsim Delta AOF,$ we obtain:
$$fracAFAE=fracOFIE$$ or
$$fracAF16=fracx12,$$ which gives
$$AF=frac43x,$$
$$FE=16-frac43x$$ and by the Pythagoras's theorem we obtain:
$$FE=2sqrtIEcdot OF$$ or
$$16-frac43x=2sqrt12x.$$
Can you end it now?
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
answered May 16 at 4:09
Michael RozenbergMichael Rozenberg
113k1899202
113k1899202
add a comment |
add a comment |
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$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
May 16 at 3:55
$begingroup$
@Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all
$endgroup$
– Soumalya Pramanik
May 16 at 4:05