I just got a straight edge and compass yesterday, it's amazing how much more intuitive these problems are with these tools, even if you don't fully construct the solution.
On the second problem, I worked with similar triangles. Comparing the two legs of two similar right triangles, the following ratio is valid: (7-r)/r = r/(3-r). This becomes 21-10r+r^2 = r^2 and simplifies to 21 = 10r, or r= 2.1. Done.
Whoa! Someone wrote a math puzzle involving a right triangle with integral sides, and for once it ISN'T a 3-4-5 triangle. I didn't know that was possible! 😊
Problem 1 can be solved without invoking Pythagoras by using the characteristic of right triangles wherein the radius of the inscribed circle is [product of legs]/perimeter. From the information given and derived @ 3:56: 12 = (a+12) (56)/[2*(a + 56)]; hence a = 21.
to the 2nd prob. let's generalize. 3 - > x, 7-> y. based on the similar triangles, the solution is r = xy/(x+y) = 1/(1/x+1/y), so the diameter of the circle is the HARMONIC MEAN of the two parallel sides of the trapezoid.
There's a better way for solving equation at the end of the first problem: 1. We start with (a + 12)^2 + 56^2 = (a + 44)^2 2. Subtract (a + 12)^2 from both sides => 56^2 = (a + 44)^2 - (a + 12)^2 3. Use the formula A^2 - B^2 = (A - B)(A + B) => 56^2 = 32(2a + 56) 4. 56 factored to primes is 2^3*7, so 56^2 = 2^6*7^2. 32 is 2^5. => 2^6 * 7^2 = 2^5 * (2a + 56) 5. Extract 2 from the parenthesis => 2^6 * 7^2 = 2^6 * (a + 28) 6. Divide both sides by 2^6 => a + 28 = 49 7. a = 21
@RAG981 yes. The video skimmed over the numeric calculations, and that's why it looks simpler. Do therm without a calculator and see for yourself. [edit] "skimmed" not "skinned". God, I hate typing on the phone! :D
a = 21 This is the easiest one of the two. The base = 56 ( 12 + 44 ) tangent circle theorem The height = 12 + a tangent circle theorem The hypotenuse = 44 + a (given) Pythagorean Theorem: (44 + a)^2 = (12 + a)^2 + 56^2 a^2 + 88a + 1936 = a^2 + 24a + 144 + 3136 88a - 24a = 3280 - 1936 64a = 1344 a = 1344/64 a =21
As far as I was able to observe, the setup of Problem 2 did not say anything about units. The introduction of centimeters at 9:20 does not seem to be called for.
9:20 - I'm going to have to take off points for units appearing out of nowhere. Centimeters are as good a unit as any but that was not stated in the original problem.
The second problem is probs the first problem where i found a mich easier to solution than the one in the video. You create a rectangle as per the video with the sides being 3 and r. Then you half it to create two triangles and solve for r with cos(45⁰) * 3 ~2.1
The radius of the circle = 2.1 Answer Draw a triangle on top of line 3 cm (APT) to form a triangle on top of the quadrilateral to form triangle ABC. Triangle ATP is similar to triangle ABC since both have angle A, and both are a right triangle (90 degrees) Draw a straight line through the circle from T to the base of triangle ABC to form another triangle. The base = 4 cm (7cm -3cm) Since APT is similar to ABC and the height of the circle is 2r (two radii), then 2r/4 = ?/3 ? = the height of triangle APT 2r/4 = ?/3 2r/ 4 *3= ? 6r/4 =? 3r/2 = ? 1.5 r =? Hence, the height of triangle APT in terms of r = 1.5 r Hence, the height of triangle ABC is 2r + 1.5 r = 3.5 r Hence, triangle ABC has a base of 7 and a height of 3.5 r So, if we get the length of the hypotenuse in terms of r or any number, we could find r or the circle's radius. Finding the hypotenuse of triangle ABC: Draw a square within the triangle ABC starting at the circle's center from the two-point of tangency The length of the square = r According to two tangent circle theorem, from the point of tangency to line C or (7 -r) equals the point of tangency on the hypotenuse of triangle ABC to line C. hence, part of the height of AC (or the hypotenuse of triangle ABC) is 7- r Also, according to the theorem from the point of tangency of the vertical line (above the square) to A is equal to the distance from the point of tangency of the hypotenuse to A or 3.5 r - r or 2.5 r Hence, the length of the hypotenuse = 2.5 r + 7- r = 7 + 1.5 r To find r, employed the Pythagorean Theorem by using 7, 3.5 r, and 7 + 1.5 r 7^2 + (3.5 r)^2 = (7 + 1.5 r)^2 49 + 12.25 r^2 = 49 + 2.25 r^2 + 21 r 12.25 r^2 = 2.25 r^2 + 21 r (49 cancels each other) 12.25 r^2 - 2.25 r^2 + 21r 10 r^2 = 21 r 10 r = 21 (divide both sides by r) r = 21/10 r = 2.1 Answer
Question 1's numbers are also delightfully well-designed for non-calculator solutions. 56^2-44^2 = 100 * 12 by difference of two squares, then adding 144=12^2 gives a total of 112*12 on the RHS. Dividing by 64 (16|112 [easily seen from 100=25*4 => 112=28*4=16*7] and 4|12) gives a=7*3=21 without working out any of the two-digit squares.
If you have learned trigonometry, both problems becomes quite straightforward. Q1. Given tan x, find tan (pi/4 - x) Q2. 3-r = r tan x, 7 - r = r cot x, solve
For the 2nd problem, I equated the areas of a rectangle and smaller trapezoid constituting the larger trapezoid and the area of the larger trapezoid... and I got the radius as 2, not 2.1 How???
Every side of the quadrilateral is equal to the radius of the circle. One of the angles shares the same right angle as the right triangle and the other 2 opposite angles are formed by points of tangency, making them right. It’s pretty easy to tell it’s a square.
yeah. stating its cm is wrong. the result is 2.1 UNITS - since what type of units was naver stated, there is no type in the answer. it could be meters. or mm. or feet. we dont know, and the math itself doesnt really care either
1 The lengths of the two tangents drawn from an external point to a circle are equal. This we learned as Crow's Beak Theorem. Yes, this IS a theorem. 2 Too complicated and easy to make a error. Why not use the formula? We learned in 6th grade I think. r=(b*B)/(b+B); here b=3 and B=7 so r=21/10=2.1 And to expand: Area is A=b*B=21, perimeter P=2(b+B)=20. And also r=2A/P Formulas that we learned in 6th grade.
1 Hmm. Doesn't look like anyone else calls the tangents to a circle theorem the 'crow's beak theorem'. I've never heard it called that before, and some googling fails to provide any results for it at all. Maybe it was just your maths teacher that dubbed it so? 2 I'd hardly call a bit of Pythagoras too complicated.
@@Grizzly01-vr4pn I agree. This isn’t a common enough situation to memorize a formula for it. We’re far better off deriving the answer using the Pythagorean theorem and a little mathematical know-how.
Crow's Beak Theorem. Yes, maybe it was just my maths teacher that called so, but that was in late 80s, so I do not remember if it have other name. Googe have no result. The theorem I find in my old notebook where I writed all sorts of formulae, constants, theorems, notebook that I have for 30 years or so. However, the "The lengths of the two tangents drawn from an external point to a circle are equal" seems to be theorem 10.2, as I find in some english sites. It is not about aplying Pythagora Theorem, it is about the errors that can appear. Yes, is kinda redundant to memorize a formula rarely used but that formula was teached and who knows it have a nice shortcut. And at an exam, is useful to minimize the errors and gain time.
I just got a straight edge and compass yesterday, it's amazing how much more intuitive these problems are with these tools, even if you don't fully construct the solution.
Lmao that's wild
On the second problem, I worked with similar triangles. Comparing the two legs of two similar right triangles, the following ratio is valid: (7-r)/r = r/(3-r). This becomes 21-10r+r^2 = r^2 and simplifies to 21 = 10r, or r= 2.1. Done.
Whoa! Someone wrote a math puzzle involving a right triangle with integral sides, and for once it ISN'T a 3-4-5 triangle. I didn't know that was possible! 😊
Gracias por la dedicación y trabajo, excelente página!!!
From. Bogotá D.C. COLOMBIA
Problem 1 can be solved without invoking Pythagoras by using the characteristic of right triangles wherein the radius of the inscribed circle is [product of legs]/perimeter. From the information given and derived @ 3:56: 12 = (a+12) (56)/[2*(a + 56)]; hence a = 21.
Wow, great ❤
About the first exercise: because >, we can get the right answer this way:
• consider triangle (of the picture) is ABC (with A being the top angle and C the 90° angle)
• consider O as the center of the circle
then:
angle ABO = arctan(12/44) = 15.25°
angle ABC = 2·ABO = 2·15.25° = 30.5°
angle BAC = 180° - 90° - 30.5° = 59.5°
angle BAO = BAC/2 = 59.5°/2 = 29.75°
=>
tan(29.75°) = 12/x
x·tan(29.75°) = 12
x = 12/tan(29.75°)
x ≈ 20.99
■ final result: x = 21 units of length
🙂 AND THANK YOU FOR THE VIDEO!!!
fun fact: Hey, this is Presh Talwalkar
Pythagoras
(a+44)^2=(a+12)^2+56^2
aa+88a+121*16=aa+24a+144 +49*16*4
64a= 16(49*4+9-121)
()=(196-121+9)=(75*9=84=12*7=6*14
64a=16*6*14
a=21
alternatively
Area=rs
Area=(a+12)(56)/2=(a+12)28
s=(a+12 + 56 + a+44)/2=(2a+112)/2=a+56
rs=12(a+56)=
Area=28(a+12)
12a+12.56= 28a+12.28
12.56-12.28= 28a-12a=16a
12.28=16a
4.3.4.7=16.a
3.7=a=21
to the 2nd prob. let's generalize. 3 - > x, 7-> y.
based on the similar triangles, the solution is r = xy/(x+y) = 1/(1/x+1/y),
so the diameter of the circle is the HARMONIC MEAN of the two parallel sides of the trapezoid.
There's a better way for solving equation at the end of the first problem:
1. We start with (a + 12)^2 + 56^2 = (a + 44)^2
2. Subtract (a + 12)^2 from both sides => 56^2 = (a + 44)^2 - (a + 12)^2
3. Use the formula A^2 - B^2 = (A - B)(A + B) => 56^2 = 32(2a + 56)
4. 56 factored to primes is 2^3*7, so 56^2 = 2^6*7^2. 32 is 2^5. => 2^6 * 7^2 = 2^5 * (2a + 56)
5. Extract 2 from the parenthesis => 2^6 * 7^2 = 2^6 * (a + 28)
6. Divide both sides by 2^6 => a + 28 = 49
7. a = 21
Oh dear, you really think that's better?
@RAG981 yes. The video skimmed over the numeric calculations, and that's why it looks simpler. Do therm without a calculator and see for yourself.
[edit] "skimmed" not "skinned". God, I hate typing on the phone! :D
a = 21
This is the easiest one of the two.
The base = 56 ( 12 + 44 ) tangent circle theorem
The height = 12 + a tangent circle theorem
The hypotenuse = 44 + a (given)
Pythagorean Theorem:
(44 + a)^2 = (12 + a)^2 + 56^2
a^2 + 88a + 1936 = a^2 + 24a + 144 + 3136
88a - 24a = 3280 - 1936
64a = 1344
a = 1344/64
a =21
Thanks for taking notes while watching the video.
As far as I was able to observe, the setup of Problem 2 did not say anything about units.
The introduction of centimeters at 9:20 does not seem to be called for.
9:20 - I'm going to have to take off points for units appearing out of nowhere. Centimeters are as good a unit as any but that was not stated in the original problem.
Incorrect. Units were stated in the original problem. See 0:05. They only appeared for about a second-but they were definitely there.
For the second problem, you have similar triangles so r/(7-r) = (3-r)/r. Solving for r gives the same answer as in the video.
The second problem is probs the first problem where i found a mich easier to solution than the one in the video.
You create a rectangle as per the video with the sides being 3 and r. Then you half it to create two triangles and solve for r with cos(45⁰) * 3 ~2.1
Wow great ...thanks a lot
Great Problem
As a student of 9th I can confidently say that " *I CAN'T SOLVE THIS PROBLEM* "
1:50 I always thought of AOC as more of a square, but I'll play along.
That's what I was going to say
Well "put." Well "put."
Your videos are a treasure. Thank you for your talent and creativity.🦐💿😵
Where did the unit of measurement come from (cm)? It wasn't listed in the problem.
Indeed... and why does nobody never formulate such problems e.g. with parsecs? 😊 or at least with kms?
The radius of the circle = 2.1 Answer
Draw a triangle on top of line 3 cm (APT) to form a triangle on top of the quadrilateral to form
triangle ABC.
Triangle ATP is similar to triangle ABC since both have angle A, and both are a right triangle (90 degrees)
Draw a straight line through the circle from T to the base of triangle ABC to form another triangle.
The base = 4 cm (7cm -3cm)
Since APT is similar to ABC and the height of the circle is 2r (two radii), then
2r/4 = ?/3 ? = the height of triangle APT
2r/4 = ?/3
2r/ 4 *3= ?
6r/4 =?
3r/2 = ?
1.5 r =?
Hence, the height of triangle APT in terms of r = 1.5 r
Hence, the height of triangle ABC is 2r + 1.5 r = 3.5 r
Hence, triangle ABC has a base of 7 and a height of 3.5 r
So, if we get the length of the hypotenuse in terms of r or any number, we could find r or the circle's radius.
Finding the hypotenuse of triangle ABC:
Draw a square within the triangle ABC starting at the circle's center from the two-point of tangency
The length of the square = r
According to two tangent circle theorem, from the point of tangency to line C or (7 -r)
equals the point of tangency on the hypotenuse of triangle ABC to line C.
hence, part of the height of AC (or the hypotenuse of triangle ABC) is 7- r
Also, according to the theorem from the point of tangency of the vertical line (above the square)
to A is equal to the distance from the point of tangency of the hypotenuse to A or 3.5 r - r
or 2.5 r
Hence, the length of the hypotenuse = 2.5 r + 7- r = 7 + 1.5 r
To find r, employed the Pythagorean Theorem by using 7, 3.5 r, and 7 + 1.5 r
7^2 + (3.5 r)^2 = (7 + 1.5 r)^2
49 + 12.25 r^2 = 49 + 2.25 r^2 + 21 r
12.25 r^2 = 2.25 r^2 + 21 r (49 cancels each other)
12.25 r^2 - 2.25 r^2 + 21r
10 r^2 = 21 r
10 r = 21 (divide both sides by r)
r = 21/10
r = 2.1 Answer
These two were fun.
Question 1's numbers are also delightfully well-designed for non-calculator solutions. 56^2-44^2 = 100 * 12 by difference of two squares, then adding 144=12^2 gives a total of 112*12 on the RHS. Dividing by 64 (16|112 [easily seen from 100=25*4 => 112=28*4=16*7] and 4|12) gives a=7*3=21 without working out any of the two-digit squares.
Two nice but rather routine and certainly doable problems - no big deal - kinda like doing 10 pushups for the brain.
If you have learned trigonometry, both problems becomes quite straightforward.
Q1. Given tan x, find tan (pi/4 - x)
Q2. 3-r = r tan x, 7 - r = r cot x, solve
(2R)^2 + (7 cm - 3 cm)^2 = (7 cm - R + 3 cm - R)^2
4R^2 = 100 cm^2 - 16 cm^2 - 40R cm + 4R^2
84 cm^2 - 40R cm = 0
R = (84/40) cm = 2.1 cm
First time when I solved all the problems of presh in a single video on my own
Simple and elegant solution.
1ST PROBLEM
(a + 44)^2 = (a +12)^2 + (12 + 44)^2
(a + 44)^2 = (a + 12)^2 + 56^2
a = 21
2ND PROBLEM
(7 + 3 - 2R)^2 = (2R)^2 + 16
(10 - 2R)^2 = 4R^2 + 16
R = 21 / 10
2ed proplem , when use formula : rs = √abcd its give other results,, can please explain it!!!!!
with which tool or software you have generated this video??
56²+(12+a)²=(44+a)²
a=21
when did 2nd prob become "cm"?
See 0:05.
First problem should be abother: a=21 and b=44, find r. 🤷♀️
For the 2nd problem, I equated the areas of a rectangle and smaller trapezoid constituting the larger trapezoid and the area of the larger trapezoid... and I got the radius as 2, not 2.1 How???
Tried it on Desmos as a small diagram, got a poor 2.2 visually
I actually know what an inscribed angle is let's go
Bro these prblems get easier and easier...
You’re just getting smarter
1. x=a/4; 12/4=3; 44/4=11
(x+11)²=(x+3)²+(11+3)²
x²+22x+121=x²+6x+9+196
16x=84; a=4x=21 😁
2. (7-3)/2=2; 2r/2=r; (10-2r)/2=5-r
2²+r²=(5-r)²; 4+r²=25-10r+r²
10r=21; r=2.1 😁
And this is the answer.
He spoke and said.
This or that?
Without thinking too much... 22 ?
2.1
So the answer for the second question is always xy/x+y
Yes is a formula we learned in 6th grade if I remember correctly.
you camt say that is a square js by seeing it, prove it
Every side of the quadrilateral is equal to the radius of the circle. One of the angles shares the same right angle as the right triangle and the other 2 opposite angles are formed by points of tangency, making them right. It’s pretty easy to tell it’s a square.
Centimeters? Problem 2 only gave the top as 3 and the bottom as 7...
yeah. stating its cm is wrong. the result is 2.1 UNITS - since what type of units was naver stated, there is no type in the answer. it could be meters. or mm. or feet. we dont know, and the math itself doesnt really care either
See 0:04. The original problem stated cm.
@@verkuilb Thanks
You didn’t prove that 21 matches to solution. -1 point in Olympiad
1 The lengths of the two tangents drawn from an external point to a circle are equal. This we learned as Crow's Beak Theorem. Yes, this IS a theorem.
2 Too complicated and easy to make a error. Why not use the formula? We learned in 6th grade I think.
r=(b*B)/(b+B); here b=3 and B=7 so r=21/10=2.1
And to expand: Area is A=b*B=21, perimeter P=2(b+B)=20. And also r=2A/P
Formulas that we learned in 6th grade.
1 Hmm. Doesn't look like anyone else calls the tangents to a circle theorem the 'crow's beak theorem'.
I've never heard it called that before, and some googling fails to provide any results for it at all. Maybe it was just your maths teacher that dubbed it so?
2 I'd hardly call a bit of Pythagoras too complicated.
@@Grizzly01-vr4pn I agree. This isn’t a common enough situation to memorize a formula for it. We’re far better off deriving the answer using the Pythagorean theorem and a little mathematical know-how.
And googling “Crow’s Beak Theorem” results in…nothing related to geometry.
Crow's Beak Theorem. Yes, maybe it was just my maths teacher that called so, but that was in late 80s, so I do not remember if it have other name. Googe have no result. The theorem I find in my old notebook where I writed all sorts of formulae, constants, theorems, notebook that I have for 30 years or so. However, the "The lengths of the two tangents drawn from an external point to a circle are equal" seems to be theorem 10.2, as I find in some english sites.
It is not about aplying Pythagora Theorem, it is about the errors that can appear. Yes, is kinda redundant to memorize a formula rarely used but that formula was teached and who knows it have a nice shortcut. And at an exam, is useful to minimize the errors and gain time.
Only 132 views in 3 minutes? Bro fell off.
3rd ?
Am I really this Early