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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 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
2 hours ago
$begingroup$
@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
2 hours ago
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 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
asked 2 hours ago
Soumalya PramanikSoumalya Pramanik
304
304
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Soumalya Pramanik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
2 hours ago
$begingroup$
@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
2 hours ago
add a comment |
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
2 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
2 hours ago
$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
2 hours ago
$begingroup$
@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
2 hours ago
$begingroup$
@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
2 hours ago
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 2 hours ago
Ross MillikanRoss Millikan
304k24201376
$endgroup$
$begingroup$
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
2 hours ago
1
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered 2 hours ago
CY AriesCY Aries
18.9k11844
$endgroup$
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 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
$endgroup$
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 2 hours ago
Ross MillikanRoss Millikan
304k24201376
$endgroup$
$begingroup$
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
2 hours ago
1
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
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 2 hours ago
Ross MillikanRoss Millikan
304k24201376
$endgroup$
$begingroup$
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
2 hours ago
1
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
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 2 hours ago
Ross MillikanRoss Millikan
304k24201376
$endgroup$
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 2 hours ago
Ross MillikanRoss Millikan
304k24201376
answered 2 hours ago
Ross MillikanRoss Millikan
304k24201376
answered 2 hours ago
Ross MillikanRoss Millikan
304k24201376
answered 2 hours ago
Ross MillikanRoss Millikan
304k24201376
304k24201376
$begingroup$
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
2 hours ago
1
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
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
2 hours ago
1
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
1
1
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
$EC=32$ 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
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
$begingroup$
Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha
$endgroup$
– Soumalya Pramanik
2 hours ago
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered 2 hours ago
CY AriesCY Aries
18.9k11844
$endgroup$
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered 2 hours ago
CY AriesCY Aries
18.9k11844
$endgroup$
add a comment |
$begingroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered 2 hours ago
CY AriesCY Aries
18.9k11844
$endgroup$
$$fracR12=frac2x48=frac40-(24+x)40 implies fracR12=frac2x+2(16-x)48+2(40)$$
answered 2 hours ago
CY AriesCY Aries
18.9k11844
answered 2 hours ago
CY AriesCY Aries
18.9k11844
answered 2 hours ago
CY AriesCY Aries
18.9k11844
answered 2 hours ago
CY AriesCY Aries
18.9k11844
18.9k11844
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 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
$endgroup$
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 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
$endgroup$
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 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
$endgroup$
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 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
answered 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
answered 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
answered 2 hours ago
Michael RozenbergMichael Rozenberg
112k1898202
112k1898202
add a comment |
add a comment |
Soumalya Pramanik is a new contributor. Be nice, and check out our Code of Conduct.
Soumalya Pramanik is a new contributor. Be nice, and check out our Code of Conduct.
Soumalya Pramanik is a new contributor. Be nice, and check out our Code of Conduct.
Soumalya Pramanik is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Drawing the common tangent of the two circles other than the two equal sides may be useful.
$endgroup$
– Yuta
2 hours ago
$begingroup$
@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
2 hours ago