There is more practical solution. Consider triangle DOC. This triangle is a right triangle at the point of O. So you can easily write that OE^2 = a x b. OE=7 then ab=7^2= 49
As the question looks like this, I can just assume a=b ( the question didn't have any clue for where point E is, you can make a rectangle ABCD instead of a trapezoid, then a=b=7, ab=49)
@@JLvatron I used similar logic; based on the fact that the upper angles are not specified I assume that the product ab must be constant regardless of the angle of the tangent, and if the tangent is horizontal the trapezoid becomes a rectangle in which the two sides and the height OE are all equal to 7.
@@JLvatron I prefer to call it a deduction, not an assumption. An assumption can be wrong, a deduction can't. We can deduce the lengths of a and b are irrelevant, as are the angles ACD and BDC simply by the fact that none are specified, and there is no other information in the question which would constrain them to specific values.
Another option to solve. As shown, AD=DE and EC=EB. Draw a line OE that is the radius of the semicircle, 7. This provides 2 kite shapes, AOED and OBCE. These are similar kites in that their angles are all the same, which means their sides are ratios of one another. You can then write AB/7=7/BC and cross multiply to get the product of the sides = 49.
Very nice job as always! From a pedagogical point of view, you might consider not labeling the original problem with a and b. It could create confusion for some students later in the solution when a, b, and c are used with the Pythagorean theorem to describe different but related lengths (i.e., b = a-b; but the b's are not the same).
I started off intending to find a and b. Thanks to your site and its wonderful problems I had a "wait a minute" moment...thought like you for that moment, and went looking for ab instead. As you said, quick and easy. Thanks!
Reflect the trapezoid ABCD across the x-axis. It will be the isosceles tangential trapezoid , it has an incircle. If the base are x and y, then the inradius is given by r = [sqrt(xy)]/2 , x=2a, y=2b 7 = [sqrt(2a*2b)]/2 ∴ ab = 7^2 = 49
This is one of the few I am able to solve. I merely turned the trapezoid into a regtangle. Angles A & B are both right angles. Draw a vertical line from center point o. Reposition the angled line so that it forms 2 right angles at the top of the arc. A will reduce to 7 and B will increase to 7. 7x7=49.
If we extend side b upward while shrinking side a downward, the point of tangency becomes a rolling fulcrum which will reach the center between a and b when a and b are equal, arriving at complete bilateral symmetry. Since the radius is constant, the symmetry is locked at 7 units vertical and 14 units horizontal. This seems intuitive to me, like a data sufficiency solution that I can't quite logically justify. Thank you for your excellent tutorials. If anybody needs me, I'll be over at the subscribe button.
Angles ADC and BCD are supplementary angles (BAD=ABD=90°). So their bisectors pass through the center of the semicircle O giving an angle DOC of 90°. Applying the second Euclid's theorem in right triangle DOC (whiere OE is the height relative to the hypotenuse) we have OE^2=DE*EC=a*b --> ab=49.
Thank you for an interesting question with different approaches. Your explanation was as usual very easy to understand. With respect, it’s worth adding that 4ab =(2r)² = 4r² Ηence, ab = r²
Wisdom has shown me: if you have asked for the product ab, then there must be a discrete, rather than general solution. I just redrew the tangent across the top of the circle, with both a and b by observation equal to 7. 5 second solve.
The same method I used to solve it. But I didn't square 14 in my head, I broke it down to 2^2 * 7^2, which I find to be easier to do in my head. It also really helped when dividing 196 by 4 at the end.
if the proportions of a and b are unknown, then we can assume that they are equal, and then a = 7 and b = 7, because this will be the radius. 7 × 7 = 49.
I always give your problems a few seconds of thought about the solution and then just sit back and watch you do it. Just lazy I guess. Keep up the good work
as a becomes longer, b becomes shorter and vice versa. therefore we can just take the average length of both which would be 7 as the radius of the circle shows. therefore the product of them is 7*7=49.
You are really amazing and the questions too . I am a student of class 10 .in starting I find the questions easy but when try solving it, then it start becoming hard for me to do. After some time, I quit the question and start watching your videos and the question not so hard.
I did it in my head in 5 seconds, but wasn't certain at first if I had it right.. I wondered if I oversimplified, but I came up with the same answer as the tutorial. Rather than a sloping line, I imagined CD as a horizontal line that forms a tangent with the top of the semicircle which is parallel with AB. This would make both AD and CB 7 units long (the same as the radius) and would give the answer as 49. As no angles are given, DC can be tilted like a seesaw where one length increases and the other decreases. But they must always produce 49.
axb = sq 7= 49. Just connect DO and CO, we notice that DO and CO are the bisector of the angle AOE, and BOE. We also notice that DA=DE and BC=CE. Also because E is the tangent point of the line DC to the circle so OE is peperdicular to CD. Because the angles AOE+BOE=180 degrees so the value of the angle DOC = 90 degrees. The triangle DOC is a right triangle in which OE is the height from the right angle to the hypotenuse, thus sq OE = DExEC = axb = sq AO, so axb= sq 7= 49
I also just did the 7x7=49 method. Though it's not obvious that the product of the two sides remain the same I assumed it did by the wording of the question.
We have OD is the bisector of angle AOE OC is the bisector of angle BOE => DOC = 90 degrees. (because AOB = 180) => DOC is right triangle => ab = height^2 = OE^2 = 7^2 = 49.
Method 2 consider triangle DOC, easily prove that it is a right angle triangle. so we have DO^2+OC^2=DC^2 (7^2+a^2)+(7^2+b^2)=(a+b)^2 expending, 2x7^2=2ab ab=49
A quick way to answer this question, since we don’t have any information on DC inclination then we can assume no matter the configuration ab is constant. We can then take the limit case were DC is parallel to AB, we have in this case a=7 and b=7 thus ab=49
I got the answer intuitively, then came back a few days later to see how it is worked out formally. Okay, that proves it formally. Nice. The intuitive answer, though, comes from assuming that there is a unique solution for ab no matter what angles ADE and BCE are. We can assume this because the problem statement asserts it while giving us no angles or lengths other than the radius. Therefore we can look for a degenerate case using the one piece of information that we do have: the radius. The degenerate case is where the angles ADE and BCE are right angles, ergo a and b are both equal to the radius since ABCD is a rectangle and the radius of the inscribed semicircle is 7. Ergo when a=7 and b=7, we know that ab is 49, and if this is not provably true for all angles this problem would be unsolvable in the general case.
Another way: Connect OD, OC, OE. Triangle DOC is 90 degrees. OE perpendicular to DC. We now have two similar triangles: OED & CEO. DE/OE= OE/EC --> OExOE = DExEC, where DE= a, EC=b, and OE=7. 7x7=axb.
Since the line CED is a tangent to the semicircle, and the point E is chosen arbitrarily, the tangent can be moved so that it becomes parallel to the line AB. Thus, a=b=7. therefore ab=49.
49? You can get a right triangle by drawing an horizon line from segment b to a. This line will be 14, being equal to the diameter. The hypotenuse will be a+b using the two tangents from a point to the circle are equal theorem. The third side of the triangle will be a-b. The Pythagoras theorem is used to get ab.
*Method 1:* Draw lines OD and OC from the center. This gives Right Triangle OAD with an altitude a and base 7 Similar to Right Triangle OBC with an altitude 7 and base b. a/7 = 7/b ab=49 *Method 2:* Draw a perpendicular OE from the center to the Hypotenuse of the Right Triangle DOC thus splitting it into a and b. We now have two Similar Right Triangles again, one with height b and base 7, and another with height 7 and base b. Hence ab= 49.
OAD and OED are congruent Right Triangles with θ at the vertex O. OEC and OBC are congruent Right Triangles with θ at the vertex C. Hence, OAD and OBC are similar. Image here : ruclips.net/video/YjrgzA9gFlI/видео.html
Or in short: a*b = AO^2 During the Thales theorem, the angle at E (between EA and EB) is always 90°, so the proportion of a/b is unimportant, a*b will always be the same. Imagine a=b (that would make ABCD a rectangle) a and b would both be 7, so ab is 7^2 = 49.
@@PreMath Am I right in thinking if a and b were the same height they would be 7 so 7*7 = 49 is the answer - as side a gains height, side b loses height - but ab is always the same?
It may not be obvious, but if you were to construct the diagonals AC and BD, they would cross at a point directly below E, at a height of a∙b/(a + b) above the line segment AB. That has nothing to do with the problem, but it's an interesting property of the construction.
Here's what I did: If there is a single answer, it is not constrained by what a or b are in particular, so just make the line from D to C parallel to that from A to B. Then its height is the radius 7, which is also the lengths of AD and BC, so the answer is 7*7 or 49.
I also got to the answer of 49 but in a more simplistic way. Imagining the line DC rocking over arc so a and be are equal in length. Both would then be 7 and the answer would be 7 * 7 = 49
I've an idea. If I m not wrong, then the product ab is independent of the inclination. If this is the case, then the semicircle may be assumed to be enclosed in a rectangle of side 7 & 14. So that ab=7.7=49.
I think the answer is 49 because if you see that 7 × 7 is 49 and if you put a and b where b is perpendicular to the edge (b is 0), a would technically be valued at infinity, so it should be a product of a and b since there is no clue otherwise?
Nice! I did take another approach. I see that someone else commented the same solution. There is a point where a and b can be the same length and in doing so, they would both be equal to 7. Simply putting in a=7 and b=7 into the question, we have 7*7= 49. :-) Thank you for your geometry puzzles. I had not been working my brain on these, I am not sure if I would have come up with that solution so quickly or at all. You solution it a much better proof of how this works.
I was curious if the estimate I did in my head was accurate: Assume the same nominclature. And that a=b Then a=b=r r^2=ab 49=ab If a>b or b>a But area of trapezoid is always the same, then Then ab=49?
Hi again!!! do not know why but as soon as I saw it, I said 7 * 7 = 49. Immediately after I realized that for every trapezoid of this type it holds that α * b = r ^ 2. Correct?
Alternatively: angles ODE + OCE = 90 degrees (because AD parallel to BC); therefore, angle DOC is 90 degrees, so triangle DOC is a right-angled triangle. If you then draw a new triangle, DFC, where F lies 7 units from E along line drawn perpendicular to DC, it's clear that points DOCF all lie on a circle, with DC as its diameter (because DFC and DOC are both right-angled triangles). Then, the product of DE and EC is the same as the product of OE and EF. But OE and EF are both 7; thus their product is 49 - and therefore the product of DE (a) and EC (b) is also 49.
Love the solution but I was able to do it in my head by first doubling it upside down to get A and B at 14X14 and than the product would be half AxB =49.Cheer
@@PreMath I use to get in so much trouble in school for seeing an answer and writing it down , teachers would subtract from the score if working was missing. 50 years later I understand why but it was so frustrating then.
AD = DE = a; BC = EC = b. Draw a perpendicular from C to AD ( call that point F ) In the triangle DCF (DE + EC)sq = CFsq + FDsq. (a + b)sq = 14sq + (a - b)sq Solve the equation to get 4ab = 14sq so ab = 49
I think we can do it simpler. DO bisects angle AOE, CO bisects angle EOB; so, angle DOC is half of the open angle AOB, and is itself a right angle. Now, right triangles DOC, DEO, and OEC are all similar by AA, their sides are proportional, and, inparticular, DE/OE=OE/EC; but DE=a, EC=b, and OR=7 is the radius of the circle; so, a/7=7/b; cross-multiplying, it becomes ab=49.
I did it a little differently, I identified angle ADO as alpha, angle BCO as beta. We know that angle ODE must also be alpha since line OD bisects the lines AD and DE. Call angle BCO delta, and angle BOC gamma. We know angle OCE is delta and angle COE is alpha because CO bisects lines BC and CE. Furthermore alpha+beta=90 and gamma+delta=90. And the angles between OA and OB are 2*beta+2*gamma = 180 so beta+gamma =90. Thus we find that beta=delta and alpha=gamma. This shows that the angle DOC is beta+alpha=90, so triangle DOC is a right triangle. By pythagorean applied to triangle AOD, OD is sqrt(49+a^2) and by pythagorean applied to triangle OBC, OC is sqrt(49+b^2). We now apply pythagorean to triangle DOC, which gives 49+a^2+49+b^2=(a+b)^2. Thus 2ab=98 and ab=49.
before watching :- join point of 'b' to point of 'a' to make rectangle since tangents from a common point are equal hypote..=a+b altitude ..=a-b base = 14 diameter 14²+ {a-b}²={a+b}² 4ab=196 ab=49 I am in 10th standard from:- INDIA
Nice and interesting. However, I lack the reasoning WHY ab = r^2. I feel that most viewers that are interested in maths would instinctively have guessed this (after some thought/reasoning and dimensional analysis), so that would have been fundamentally interesting to explain. Maybe with some kind of geometry/area visualization/simulation
Maybe a simpler way: take ADO and BOC, 2 right triangles. Since ADC + DCB = 180 degrees and DO and CO are halving 2 kites, ADO + OCB = 90 degrees, therefore ADO and BOC are SIMILAR right triangles. Since, 7/a = b/7 and from that follows ab = 49.
Easier method: Draw a line OD and a line OC. On the left we get two congruent triangles AOD and EOD On the right we get two congruent triangles BOC and EOC. Let the angle BOC= alpha. Then by congruent triangles angle EOC is also alpha. Call angle AOD= beta. By congruent triangles angle EOD is also beta. Now note that all these four angles at O cover the diameter AOB and thus total 180 degrees. 2*alpha +2*beta=180, therefore beta=90-alpha. Since angle AOD is thus 90-alpha, angle ADO has to be alpha. Consequently triangles AOD and BCO are similar. Thus AD/AO=BO/BC. Hence a/7=7/b. Thus ab=49.
Yo prolongue la base bc para formar un rectángulo como ec = bc= b y además AD = de = a la hipotenusa sería a+b los catetos son 14 y a-b entonces (a-b)^2 +14^2 = (a+b)^2 Ese desarrollo da 4ab=196 a*b=49
Here is how I have solved it: First, using Two-Tangent theorem, I have come to the fact that AD = DE and CE = CB. Since that, ab = DE × CE. Let's construct OE, OE = 7 and perpendicular to CD and also height of triangle DOC. Right triangles OEC and OBC are equal since EC = CB and OC is hypotenuse of both. Let's consider angle /ECO as x. /BCO is also x. Since /DCB is 2x, /ADC is equal to 180° - 2x. Right triangles ADO and EDO are equal since AD = DE and DO is hypotenuse of both. Then /ADO and /EDO are equal 90° - x each. In triangle DOC /CDO = 90° - x and /OCD = x. Then /DOC = 180° - x - (90° - x) = 90° and triangle DOC is right. In right triangle height squared is equal to segments of hypotenuse it divides to: OE^2 = DE × EC = AD × BC = 49. So, the answer is 49.
The real interest of this problem is that the answer is the same no matter where you put the tangent point on the circle - which is quite surprising. But your demonstration does not even mention this fact, much less use it to motivate the proof! Which makes your demonstration an almost indecipherable hash of early letters in the alphabet.
I solved it like this: a and b are variables, so they can be equal. If they are equal lime CD is horizontal and a=b=7. 7*7=49. This is no proof that other values for a and b will result in 49.
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](/img/1.gif)
That was beautifully explained, I enjoyed following along with the logic process.
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There is more practical solution. Consider triangle DOC. This triangle is a right triangle at the point of O. So you can easily write that OE^2 = a x b.
OE=7 then ab=7^2= 49
@@NT-ds8ic i saw the thumbnail and got answer by your way. So I clicked it to figure out what's the way different
As the question looks like this, I can just assume a=b ( the question didn't have any clue for where point E is, you can make a rectangle ABCD instead of a trapezoid, then a=b=7, ab=49)
Many options to solve this problem.
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This makes sense, but I just couldn't prove it on paper.
@@JLvatron I used similar logic; based on the fact that the upper angles are not specified I assume that the product ab must be constant regardless of the angle of the tangent, and if the tangent is horizontal the trapezoid becomes a rectangle in which the two sides and the height OE are all equal to 7.
Unfortunately it is just an assumption.
PreMath’s solution is the proof that it doesn’t rely on the angles.
@@JLvatron I prefer to call it a deduction, not an assumption. An assumption can be wrong, a deduction can't. We can deduce the lengths of a and b are irrelevant, as are the angles ACD and BDC simply by the fact that none are specified, and there is no other information in the question which would constrain them to specific values.
Another solution
Join OC & OD & OE
m
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Another option to solve. As shown, AD=DE and EC=EB. Draw a line OE that is the radius of the semicircle, 7. This provides 2 kite shapes, AOED and OBCE. These are similar kites in that their angles are all the same, which means their sides are ratios of one another. You can then write AB/7=7/BC and cross multiply to get the product of the sides = 49.
Many options to solve this problem.
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You are awesome Jay 😀
It works as a general proof. (when you say kite you mean trapezoid)
That was my solution, I considered also the solution in the video, but this one is absolutely faster.
I solved it the same way.
Very nice job as always! From a pedagogical point of view, you might consider not labeling the original problem with a and b. It could create confusion for some students later in the solution when a, b, and c are used with the Pythagorean theorem to describe different but related lengths (i.e., b = a-b; but the b's are not the same).
I started off intending to find a and b. Thanks to your site and its wonderful problems I had a "wait a minute" moment...thought like you for that moment, and went looking for ab instead. As you said, quick and easy. Thanks!
Reflect the trapezoid ABCD across the x-axis.
It will be the isosceles tangential trapezoid , it has an incircle.
If the base are x and y, then the inradius is given by
r = [sqrt(xy)]/2 , x=2a, y=2b
7 = [sqrt(2a*2b)]/2
∴ ab = 7^2 = 49
Many options to solve this problem.
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How do you mean isosceles?
This is one of the few I am able to solve. I merely turned the trapezoid into a regtangle. Angles A & B are both right angles. Draw a vertical line from center point o. Reposition the angled line so that it forms 2 right angles at the top of the arc. A will reduce to 7 and B will increase to 7. 7x7=49.
Just another display of awesomeness form the maestro, brilliant thanks again
If we extend side b upward while shrinking side a downward, the point of tangency becomes a rolling fulcrum which will
reach the center between a and b when a and b are equal, arriving at complete bilateral symmetry. Since the radius is
constant, the symmetry is locked at 7 units vertical and 14 units horizontal. This seems intuitive to me, like a data
sufficiency solution that I can't quite logically justify. Thank you for your excellent tutorials. If anybody needs me, I'll be
over at the subscribe button.
You are very welcome.
Excellent!
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Hello dear,
Not ordinary approach to the solution, almostly without geometry construction for a and b to be figured out, great 👍
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Angles ADC and BCD are supplementary angles (BAD=ABD=90°). So their bisectors pass through the center of the semicircle O giving an angle DOC of 90°. Applying the second Euclid's theorem in right triangle DOC (whiere OE is the height relative to the hypotenuse) we have OE^2=DE*EC=a*b --> ab=49.
Thank you for an interesting question with different approaches. Your explanation was as usual very easy to understand. With respect, it’s worth adding that
4ab =(2r)² = 4r²
Ηence, ab = r²
exellent
Wisdom has shown me: if you have asked for the product ab, then there must be a discrete, rather than general solution. I just redrew the tangent across the top of the circle, with both a and b by observation equal to 7. 5 second solve.
Yes, exactly how I did this.
Ditto.
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You are awesome Arnie 😀
The same method I used to solve it. But I didn't square 14 in my head, I broke it down to 2^2 * 7^2, which I find to be easier to do in my head. It also really helped when dividing 196 by 4 at the end.
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You are awesome Thorin 😀
if the proportions of a and b are unknown, then we can assume that they are equal, and then a = 7 and b = 7, because this will be the radius. 7 × 7 = 49.
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You are awesome Jakkima😀
@@PreMath . thank you! math is the best gymnastics for the mind)
I got that in a few seconds and I'm no Feynman. It just made sense.
I always give your problems a few seconds of thought about the solution and then just sit back and watch you do it. Just lazy I guess. Keep up the good work
Dear Bob, we are all lifelong learners! Keep persevering.
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as a becomes longer, b becomes shorter and vice versa. therefore we can just take the average length of both which would be 7 as the radius of the circle shows. therefore the product of them is 7*7=49.
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I like the way the teacher quotes a specific theorem for every new step in the proof.
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You are really amazing and the questions too .
I am a student of class 10 .in starting I find the questions easy but when try solving it, then it start becoming hard for me to do. After some time, I quit the question and start watching your videos and the question not so hard.
All the best Aditya dear
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@@PreMath I thought you are also from India. But thank you for the reply 😇😇
My strike rate started getting better by your teaching (blessings)
Wonderful!
You are awesome. Keep it up 👍
I did it in my head in 5 seconds, but wasn't certain at first if I had it right.. I wondered if I oversimplified, but I came up with the same answer as the tutorial. Rather than a sloping line, I imagined CD as a horizontal line that forms a tangent with the top of the semicircle which is parallel with AB. This would make both AD and CB 7 units long (the same as the radius) and would give the answer as 49. As no angles are given, DC can be tilted like a seesaw where one length increases and the other decreases. But they must always produce 49.
axb = sq 7= 49.
Just connect DO and CO, we notice that DO and CO are the bisector of the angle AOE, and BOE.
We also notice that DA=DE and BC=CE. Also because E is the tangent point of the line DC to the circle so OE is peperdicular to CD.
Because the angles AOE+BOE=180 degrees so the value of the angle DOC = 90 degrees.
The triangle DOC is a right triangle in which OE is the height from the right angle to the hypotenuse, thus sq OE = DExEC = axb = sq AO,
so axb= sq 7= 49
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I also just did the 7x7=49 method. Though it's not obvious that the product of the two sides remain the same I assumed it did by the wording of the question.
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We have
OD is the bisector of angle AOE
OC is the bisector of angle BOE
=> DOC = 90 degrees. (because AOB = 180)
=> DOC is right triangle => ab = height^2 = OE^2 = 7^2 = 49.
Excellent Tran!
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Very interesting and nice question! Thank you so much, Professor!
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Method 2
consider triangle DOC, easily prove that it is a right angle triangle.
so we have
DO^2+OC^2=DC^2
(7^2+a^2)+(7^2+b^2)=(a+b)^2
expending,
2x7^2=2ab
ab=49
A quick way to answer this question, since we don’t have any information on DC inclination then we can assume no matter the configuration ab is constant. We can then take the limit case were DC is parallel to AB, we have in this case a=7 and b=7 thus ab=49
I got the answer intuitively, then came back a few days later to see how it is worked out formally. Okay, that proves it formally. Nice. The intuitive answer, though, comes from assuming that there is a unique solution for ab no matter what angles ADE and BCE are. We can assume this because the problem statement asserts it while giving us no angles or lengths other than the radius. Therefore we can look for a degenerate case using the one piece of information that we do have: the radius. The degenerate case is where the angles ADE and BCE are right angles, ergo a and b are both equal to the radius since ABCD is a rectangle and the radius of the inscribed semicircle is 7. Ergo when a=7 and b=7, we know that ab is 49, and if this is not provably true for all angles this problem would be unsolvable in the general case.
Excellent we can conclude that the product a.b is constant and equal to (r)2. Where (r) is the Radius of the circle
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Angle COD is a right angle. OE is the height in that triangle. There is a theorm that OE^2 is ab. [EC*ED].
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Another way: Connect OD, OC, OE. Triangle DOC is 90 degrees. OE perpendicular to DC. We now have two similar triangles: OED & CEO. DE/OE= OE/EC --> OExOE = DExEC, where DE= a, EC=b, and OE=7. 7x7=axb.
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Since the line CED is a tangent to the semicircle, and the point E is chosen arbitrarily, the tangent can be moved so that it becomes parallel to the line AB. Thus, a=b=7. therefore ab=49.
49? You can get a right triangle by drawing an horizon line from segment b to a. This line will be 14, being equal to the diameter. The hypotenuse will be a+b using the two tangents from a point to the circle are equal theorem. The third side of the triangle will be a-b. The Pythagoras theorem is used to get ab.
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very well done bro, thanks for sharing this trapezoid problem
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Superb creation sir very short and technical explanation 🙏❤️
*Method 1:* Draw lines OD and OC from the center. This gives Right Triangle OAD with an altitude a and base 7 Similar to Right Triangle OBC with an altitude 7 and base b.
a/7 = 7/b
ab=49
*Method 2:* Draw a perpendicular OE from the center to the Hypotenuse of the Right Triangle DOC thus splitting it into a and b. We now have two Similar Right Triangles again, one with height b and base 7, and another with height 7 and base b.
Hence ab= 49.
Many options to solve this problem.
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How to prove triangle OAD similar to OBC
How did you get triangles OAD and OBC similar?
OAD and OED are congruent Right Triangles with θ at the vertex O.
OEC and OBC are congruent Right Triangles with θ at the vertex C.
Hence, OAD and OBC are similar.
Image here : ruclips.net/video/YjrgzA9gFlI/видео.html
CE=CB= b and DE=DA= a ( external tangents prop)
Thus CD = a+ b
By Pythagorean theorem,
(a+b)squared = 14•14 + (a-b)squared
4ab = 14•14
ab =49
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Or in short: a*b = AO^2
During the Thales theorem, the angle at E (between EA and EB) is always 90°, so the proportion of a/b is unimportant, a*b will always be the same. Imagine a=b (that would make ABCD a rectangle) a and b would both be 7, so ab is 7^2 = 49.
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Nice quetion and excellent solution, thank you teacher 🙏.
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@@PreMath Am I right in thinking if a and b were the same height they would be 7 so 7*7 = 49 is the answer - as side a gains height, side b loses height - but ab is always the same?
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You are absolutely correct! Great observation. Keep rocking 👍
@@PreMath Now I'm wondering what the maximum height of "a" can be - "14" wouldn't work.
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It can't be more than the radius!
I could solve it by another way and it give me the same solution
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Yes, there are many many ways to solve this problem.
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That's tan and inverse tan btw (a=r/b so r*a*r/a=r^2). E=. , might work if you can figure out x and y. x=r*cos(arctan(b)), y=r*sin(arctan(b))
It may not be obvious, but if you were to construct the diagonals AC and BD, they would cross at a point directly below E, at a height of a∙b/(a + b) above the line segment AB. That has nothing to do with the problem, but it's an interesting property of the construction.
Here's what I did: If there is a single answer, it is not constrained by what a or b are in particular, so just make the line from D to C parallel to that from A to B. Then its height is the radius 7, which is also the lengths of AD and BC, so the answer is 7*7 or 49.
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I also got to the answer of 49 but in a more simplistic way. Imagining the line DC rocking over arc so a and be are equal in length. Both would then be 7 and the answer would be 7 * 7 = 49
I've an idea. If I m not wrong, then the product ab is independent of the inclination. If this is the case, then the semicircle may be assumed to be enclosed in a rectangle of side 7 & 14. So that ab=7.7=49.
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I think the answer is 49 because if you see that 7 × 7 is 49 and if you put a and b where b is perpendicular to the edge (b is 0), a would technically be valued at infinity, so it should be a product of a and b since there is no clue otherwise?
Nice! I did take another approach. I see that someone else commented the same solution. There is a point where a and b can be the same length and in doing so, they would both be equal to 7. Simply putting in a=7 and b=7 into the question, we have 7*7= 49. :-) Thank you for your geometry puzzles. I had not been working my brain on these, I am not sure if I would have come up with that solution so quickly or at all. You solution it a much better proof of how this works.
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It was a very interesting and creative problem. I liked it. It is considered both easy and hard.
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I was curious if the estimate I did in my head was accurate:
Assume the same nominclature.
And that a=b
Then a=b=r
r^2=ab
49=ab
If a>b or b>a
But area of trapezoid is always the same, then
Then ab=49?
If we take the exceptional case where a and b are equal to the radius 7 so that we get ab are equal to 7x7=49
悔しいけど、解らんかったよ。PreMathさん、説明が上手ですね。
Thanks for video. Good luck!!!!!
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Interesting problem .
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Hi again!!! do not know why but as soon as I saw it, I said 7 * 7 = 49. Immediately after I realized that for every trapezoid of this type it holds that α * b = r ^ 2. Correct?
Yes!
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Alternatively: angles ODE + OCE = 90 degrees (because AD parallel to BC); therefore, angle DOC is 90 degrees, so triangle DOC is a right-angled triangle. If you then draw a new triangle, DFC, where F lies 7 units from E along line drawn perpendicular to DC, it's clear that points DOCF all lie on a circle, with DC as its diameter (because DFC and DOC are both right-angled triangles). Then, the product of DE and EC is the same as the product of OE and EF. But OE and EF are both 7; thus their product is 49 - and therefore the product of DE (a) and EC (b) is also 49.
Missed an initial step: angles ADE + ECB = 180. But ODE = half of ADE (by symmetry), and OCE = half of BCE (ditto). Thus ODE + OCE = half of (ADE + BCE) = half of 180 = 90. Then, as above.
DE = a and CE= b
(a+b)^2=(a-b)^2+14^2
4ab = 14*14
ab = 49
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Love the solution but I was able to do it in my head by first doubling it upside down to get A and B at 14X14 and than the product would be half AxB =49.Cheer
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@@PreMath I'm old now but at school it would not be a pass without showing work . Cheer
Draw trapezoid as a rectangle. A = 7 and B = 7. AB = 49. Takes about 5 seconds.
that's what I did
can't see any point doing all the rest
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@@PreMath I use to get in so much trouble in school for seeing an answer and writing it down , teachers would subtract from the score if working was missing. 50 years later I understand why but it was so frustrating then.
Actually it is easy to prove angle AOD = angle DOE & angle EOC = angle COB. Hence, angle DOC=90°. BO^2+OC^2=BC^2 => (a^2+7^2)+(7^2+b^2)=(a+b) ^2 => 98=2ab => ab=49
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If you are confident that the problem makes sense, you can choose the trapezoid to make it easy: the rectangle gives a=b=7 and so ab=49
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AD = DE = a; BC = EC = b. Draw a perpendicular from C to AD ( call that point F )
In the triangle DCF (DE + EC)sq = CFsq + FDsq. (a + b)sq = 14sq + (a - b)sq
Solve the equation to get 4ab = 14sq so ab = 49
Solved it using ur method!
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I think we can do it simpler. DO bisects angle AOE, CO bisects angle EOB; so, angle DOC is half of the open angle AOB, and is itself a right angle.
Now, right triangles DOC, DEO, and OEC are all similar by AA, their sides are proportional, and, inparticular, DE/OE=OE/EC; but DE=a, EC=b, and OR=7 is the radius of the circle; so, a/7=7/b; cross-multiplying, it becomes ab=49.
I did it a little differently, I identified angle ADO as alpha, angle BCO as beta. We know that angle ODE must also be alpha since line OD bisects the lines AD and DE. Call angle BCO delta, and angle BOC gamma. We know angle OCE is delta and angle COE is alpha because CO bisects lines BC and CE. Furthermore alpha+beta=90 and gamma+delta=90. And the angles between OA and OB are 2*beta+2*gamma = 180 so beta+gamma =90. Thus we find that beta=delta and alpha=gamma. This shows that the angle DOC is beta+alpha=90, so triangle DOC is a right triangle. By pythagorean applied to triangle AOD, OD is sqrt(49+a^2) and by pythagorean applied to triangle OBC, OC is sqrt(49+b^2). We now apply pythagorean to triangle DOC, which gives 49+a^2+49+b^2=(a+b)^2. Thus 2ab=98 and ab=49.
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Good Problem and solved nicely
before watching :-
join point of 'b' to point of 'a' to make rectangle
since tangents from a common point are equal
hypote..=a+b
altitude ..=a-b
base = 14 diameter
14²+ {a-b}²={a+b}²
4ab=196
ab=49
I am in 10th standard
from:- INDIA
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because question not define EOB's angle, so set EOB is 90, then a=7 b=7, a*b=49
Very Interesting Question!!!
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whatt about a and b. is any ab same value as the angled line tilts? lets see
Nice and interesting.
However, I lack the reasoning WHY ab = r^2.
I feel that most viewers that are interested in maths would instinctively have guessed this (after some thought/reasoning and dimensional analysis), so that would have been fundamentally interesting to explain.
Maybe with some kind of geometry/area visualization/simulation
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Line OE , ADEO is similar to BOEC, then AD/OE=OE/BC, AD.BC=OE.OE, an=7^2=49
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Maybe a simpler way: take ADO and BOC, 2 right triangles. Since ADC + DCB = 180 degrees and DO and CO are halving 2 kites, ADO + OCB = 90 degrees, therefore ADO and BOC are SIMILAR right triangles. Since, 7/a = b/7 and from that follows ab = 49.
This is also the method that I followed,after pondering for an hour.
Angle D + angle C = 180. Thus, angle ADO + angle BCO = 90. Thus, angle AOD + angle BOC = 90. Thus, angle DOC = 90. Thus, in the triangle DOC OE²=DE*EC --> 7²=a*b. a*b=49
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Awesome question
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Easier method: Draw a line OD and a line OC. On the left we get two congruent triangles AOD and EOD On the right we get two congruent triangles BOC and EOC. Let the angle BOC= alpha. Then by congruent triangles angle EOC is also alpha. Call angle AOD= beta. By congruent triangles angle EOD is also beta. Now note that all these four angles at O cover the diameter AOB and thus total 180 degrees. 2*alpha +2*beta=180, therefore beta=90-alpha. Since angle AOD is thus 90-alpha, angle ADO has to be alpha. Consequently triangles AOD and BCO are similar. Thus AD/AO=BO/BC. Hence a/7=7/b. Thus ab=49.
Whoops, I see I already did it a year ago. Looks like I solved it the same way as last time.😀
AD=DE,BC=CE. Draw line OE=7. OEC=90deg. ADC+AOE=180deg, AOE+BOE=180deg => ADE=BOE,AOE=BCE. So kite ADEO is similar to kite BCEO => b/7 = 7/a => ab=49
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Radius OE created similar quadrilaterals. a/7 = 7/b. ab = 49
Cool
Yo prolongue la base bc para formar un rectángulo como ec = bc= b y además AD = de = a la hipotenusa sería a+b los catetos son 14 y a-b entonces (a-b)^2 +14^2 = (a+b)^2
Ese desarrollo da 4ab=196 a*b=49
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Here is how I have solved it:
First, using Two-Tangent theorem, I have come to the fact that AD = DE and CE = CB. Since that, ab = DE × CE. Let's construct OE, OE = 7 and perpendicular to CD and also height of triangle DOC. Right triangles OEC and OBC are equal since EC = CB and OC is hypotenuse of both. Let's consider angle /ECO as x. /BCO is also x. Since /DCB is 2x, /ADC is equal to 180° - 2x. Right triangles ADO and EDO are equal since AD = DE and DO is hypotenuse of both. Then /ADO and /EDO are equal 90° - x each. In triangle DOC /CDO = 90° - x and /OCD = x. Then /DOC = 180° - x - (90° - x) = 90° and triangle DOC is right. In right triangle height squared is equal to segments of hypotenuse it divides to: OE^2 = DE × EC = AD × BC = 49.
So, the answer is 49.
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The real interest of this problem is that the answer is the same no matter where you put the tangent point on the circle - which is quite surprising. But your demonstration does not even mention this fact, much less use it to motivate the proof! Which makes your demonstration an almost indecipherable hash of early letters in the alphabet.
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I solved it like this: a and b are variables, so they can be equal. If they are equal lime CD is horizontal and a=b=7. 7*7=49.
This is no proof that other values for a and b will result in 49.
Can you get values for a and b ?
Amazing! I couldn't solve this, I didn't think about a triangle to Pythag.
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Pythagorean theorem
(a+b)^2=(a-b)^2+14^2
4ab=14^2
ab=14^2/4=49 🤗
Right triangle DFC
Very good.
can we generalize that to ab = r^2
This means the sqare of radius is the answer for that type of q
DE.EC = R^2
ab= R^2 = 7.7 = 49
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Ans : a x b = 98 units