Area of the circle inscribed in a right triangle: r = (a + b - c)/2 Where a and b are base and height or height and base, and c is the hypotenuse Once the hypotenuse was not given, there was the additional effort to use the pythagorean theorem to solve for the hypotenuse
Dude, I'm 15.. I solved it in about 5 second or so, I was pretty much sure I'm wrong.. but I clicked the video and skipped to the end and saw I got the same answer 😁
If we join the vertices of the triangle with centre of the circle we get three small triangles with each side as base and radius as height. Now we may equate the area of original triangle= Sum of areas of three new triangles. 1/2 ×27×36=1/2 × radius × (27+36+45) 27×36 = r ×108 Radius r = 9 This is BSP sarma a retired Bank employee from Hyderabad, I hope you will appreciate the same.
Thanks for the great comment! I appreciate a second approach to the question! I have to add more videos like this in the upcoming school year. This question is from an old Japanese Geometry math book. They have questions that are way more challenging!
I've just watched a video on similar problem. The inscribed triangle can be any triangle, not just right-angled triangle. As lengths of 3 sides of the triangle are given, its area can be found by Heron's formula. Then using your method of dividing the triangle into 3 small triangles each with their height as the radius of the circle. Equating the sum of areas of the 3 triangles with that of the whole triangle can solve value of radius accordingly. Simple method is vital in examination.
I always argue that if a triangle is right angled,it should be specifically be mentioned. Presumption should be avoided even if it is drawn and looks like it
Fair point! Would you say that it is enough to mention the triangle is a right triangle with legs measuring 36 and 27 units long? I believe that the information given in the original question establishes that we have a right triangle, with the right angle forming at the point where the legs meet.
I appreciate that! I am trying to make as much quality content as possible. This question came from a book called Sacred Mathematics Japanese Temple Geometry. That book is loaded with excellent questions.
Hi, since the triangle legs are not the same length, are you sure that the line you draw on 1:03 is perpendicular to the hypothenuse/tangent? Imagine a square triangle whose legs would measure 20 and 65 (ex). The hypothenuse wouldn’t be a tangent to the circle in the same point. So it’s perpendicular line wouldn’t be the same.
Yes, the concept of an inscribed circle is that it is tangent to the triangle at three locations. The radius drawn to those 3 locations forms a 90-degree angle
I am branching out and doing some challenge problems, this is definitely a stretch for a standard math class, but if you can do the challenge questions, the regular questions will seem easy!
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Area of the circle inscribed in a right triangle:
r = (a + b - c)/2
Where a and b are base and height or height and base, and c is the hypotenuse
Once the hypotenuse was not given, there was the additional effort to use the pythagorean theorem to solve for the hypotenuse
Thank you for the formula, what if you're given the value of the base only, how will someone solve that kind of question?
Dude, I'm 15.. I solved it in about 5 second or so, I was pretty much sure I'm wrong.. but I clicked the video and skipped to the end and saw I got the same answer 😁
If we join the vertices of the triangle with centre of the circle we get three small triangles with each side as base and radius as height.
Now we may equate the area of original triangle=
Sum of areas of three new triangles.
1/2 ×27×36=1/2 × radius × (27+36+45)
27×36 = r ×108
Radius r = 9
This is BSP sarma a retired Bank employee from Hyderabad,
I hope you will appreciate the same.
Thanks for the great comment! I appreciate a second approach to the question! I have to add more videos like this in the upcoming school year. This question is from an old Japanese Geometry math book. They have questions that are way more challenging!
I've just watched a video on similar problem. The inscribed triangle can be any triangle, not just right-angled triangle. As lengths of 3 sides of the triangle are given, its area can be found by Heron's formula. Then using your method of dividing the triangle into 3 small triangles each with their height as the radius of the circle. Equating the sum of areas of the 3 triangles with that of the whole triangle can solve value of radius accordingly. Simple method is vital in examination.
I always argue that if a triangle is right angled,it should be specifically be mentioned. Presumption should be avoided even if it is drawn and looks like it
Fair point!
Would you say that it is enough to mention the triangle is a right triangle with legs measuring 36 and 27 units long? I believe that the information given in the original question establishes that we have a right triangle, with the right angle forming at the point where the legs meet.
very helpfull video keep going !!!
Thanks, will do! My next big video will be on triangle proofs.
You’re doing good work man
I appreciate that! I am trying to make as much quality content as possible. This question came from a book called Sacred Mathematics Japanese Temple Geometry. That book is loaded with excellent questions.
Hi, since the triangle legs are not the same length, are you sure that the line you draw on 1:03 is perpendicular to the hypothenuse/tangent? Imagine a square triangle whose legs would measure 20 and 65 (ex). The hypothenuse wouldn’t be a tangent to the circle in the same point. So it’s perpendicular line wouldn’t be the same.
Yes, the concept of an inscribed circle is that it is tangent to the triangle at three locations. The radius drawn to those 3 locations forms a 90-degree angle
Great video
Glad you enjoyed it!
Great video, all though it was a bit confusing
I am branching out and doing some challenge problems, this is definitely a stretch for a standard math class, but if you can do the challenge questions, the regular questions will seem easy!
27-(45-36)/2=9
54 - 45 = 9
Bro... just divide 27 by 3
hypotenuse=45unit.
😂radius(r)=9 unitans