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The sum is 11 and the product is 24, so a equals 3. b equals 8. Therefore, 1/a + 1/b = 1/3 + 1/8 = 11/24 The sum is 11 and the product is 24. Answer 11/24

Funny one (tricky too)!!! We can also move as follows: (a+b)/ab=11/24 Bifurcating these we will get (a/ab)+(b/ab)=11/24 Resulting as(1/b)+(1/a)=11/24 But however thanks to you is minimum appreciation.

2.(10-R) = 10 / cos 45° 20 - 2R = 10√2 R = 10 - 5√2 = 2,929 cm ( Solved √ )

[2.(10-R)]² = 2. 10² 4 . (10²-20R + R²) = 200 R² - 20R + 50 = 0 R = 2,929 cm ( Solved √ )

Thank you for these videos! 🌟

this is the best way to solve this problem we should always try to use basic methods which we never forgot in future. Good problem and nice solution.

Área ABC =10*10/2=50 =2*(10-r)r+r²→ 10-5√2 =2,92893... ud. Gracias y un saludo.

The radius is 5[2-sqrt(2)]. I checked that the radius is more than zero. This should indicate that I have done a sanity check!!!

I find it highly counter-intuitive. In the notation first is the radius and then the angle. And drawing the point you always 'draw' a line at the correct angle and THEN measure the 'distance' radius.

Really helpful illustrations, thanks for this video!

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So good and clear. Thank you a lot for this!

This is challenge that could be used to fine tune youe ability to practice Law of Cosines and Law of Sines.

Wonderfully explained, thank you so much!!

Done it on my own by finding third side applying cos rules followed by heron's formula. Honour for you 🙏

The first method was the fastest!!!

The answer is 48sin(110°). I am kind if glad that there are ways of calculating the area of certain shapes depending on given information. I am planning on mixing and matching problems of subtract other areas type with this type so that I can shapoe my geometry even further!!!

1st: if you calculate all angles with the Law of Cosines you can check the result because all the angles should add up to 180 2nd: you can simplefy the Law of Sines step by calculating the 'triangle factor' (distance over sin(angle)) of the known angle, and calculate the other angles with that number

Good lesson mam. Thanks a lot

You are best teacher I've ever known! Thank you!

the circle is called the incircle and the radius, inradius

I solved this by inspection and conjecture. 27 = 9 x 3. Assume, therefore, the long side is 9 sqrt(2). 15 = 3 x 5. Assume, therefore, the short side is 5 sqrt (2). Area of the rectangle is thus 9 x 5 x sqrt(2) x sqrt(2) = 90. Area of blue = 90 - 27 -15 - 12 = 36.

Thank you, I appreciate your content.

Your explanation was so systematic, clear and easy to understand. Thanks!

OE^2+EC^2-2*OE*EC*cos(135°)=OC^2=R^2, (COS THEOREM), OE=EF/√2

Love It! Thank you for helping me solve my homework! Life just became much simpler!

Nicely done.

It is so useful! Thank you!

Beautiful question! 👍

In competitive exams we have to make use of formulae, we dont have much time.

We have a formula to find out x Product of sides/sum of sides 24x16/(24+16)=48/5

Is 250 is closer to 200 or 300???

24(2π/3-√3/2)-72(π/3-√3/2)=-8π+24 √3

S=9*231^0.5/4 r=231^0.5/6

(a+3)^2+a^2=45 2a^2+6a-36=0 a=(-6+-18)/4=3;-6 a=3 S=9

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Love this!

thank you for your explanation♥♥♥♥

a=10, b=9

ar. of inscribed square92.16 sq unit.ans

The way the task was posed must always apply to variable radii of the circle. The radius could therefore be zero, the triangle collapses, the line BC also becomes zero and the former triangle only consists of two lines that are parallel to each other and both have a length of 5. The circumference is therefore 2 * 5 = 10. This solution approach of mine is based solely on the assumption that the task could only be posed in this way if this solution approach works, without discussing the geometric basis for why the task could be posed in this way. The solution in the video is similar to mine, but it contains the complete explanation of why the perimeter must always be 10, and thus also explains why this problem could be posed in exactly this way.

Can also be solved using 'Special Case' argument. Point F can be anywhere in the circle between points D and E. Suppose point B is very very close to point D. In this case line BC is almost the same length as AD. Which gives the clue as perimeter of triangle is 2x5 = 10.

BD=BF; CE=CF ---> ABF+ACF=AD+AE=5+5=10=AB+BC+CA

Third method is very insightful. Thank you for making this video, I appreciate that you approached it from different strategies.

Excellent explanation

Very helpful!