Can you find the area of the square? скачать в хорошем качестве

Can you find the area of the square?

Todays problem is an example of thinking of out the box. Its a problem that is interesting and involves a bit of patience. References: https://www.mathsisfun.com/pythagoras... https://www.mathsisfun.com/algebra/qu... Script: Hello World and welcome to another funza academy video. Before I continue with todays problem, we would like to give a big shout out a strong supporter of our channel - G Vincent Voltaire. Thank you so much for your motivating comments. Its comments such as these that keep us going. Ok, now coming to todays problem So here it is. ABCD is a square of side x E is a point on the 2 dimensional plane. The length of EA is 5, EB is 2 and EC is 4. Find the area of x? Please pause the video and give the problem a try. If you do not wish to solve the problem, please write down how you will approach the problem. And then check your approach with ours. Right welcome back. So we will try to get some order into this problem. We have got some known values such as 2, 4 and 5 and one unknown, x. Lets see if we can relate the 2 and come up with a value for x. Our friend that will help us do so is the Pythagoras theorem. But first we need some right angle triangles. So here is one, and here is another and finally here is another. Now we will introduce 2 more unknown values- a and b Its interesting, isnt it. We wish to find one unknown but now we are introducing more unknowns to find 1 unknown. Ah the wonders of life ! And that is the fun bit. Somehow by introducing more unknowns, we will find our unknown. Ok, now let us focus on our right angle triangles and find relationships between our unknowns. So first let us look at triangle EBF. Using Pythagoras theorem we can say: EB square = EF square + BF square And this is our first equation Now lets focus on the next triangle - ECG. We can say EC square = CG square + EG square And simplifying this further we get. But, we saw in our previous equation that a sq + b sq is 4. So lets put that 4 in here. Now we will solve for a and this is what we get. At this point, you may say, why are we solving for a when we want x. This is a good question. Actually we not solving for a. We are getting a in terms of x. This means that later we can replace a with x and that will help us arrive at x. We will see this shortly. Ok, so this is our second equation. Ok guys, now lets come to our 3rd triangle EAF. It is a right angle triangle. So we can say EA sq = EF sq + FA sq And simplifying this further we get. But, we saw in our previous equation that a sq + b sq is 4. So lets put that 4 in here. Now we will solve for b and this is what we get. Now we have both a and b in terms of x. Does that help us? Well it does and we can use it in this equation. Notice how this equation is in terms of a and b. But since we calculated a and b in terms of x, we can re-write this equation , in terms of x. And if we have an equation in x, we can solve for x. This equation looks a bit complex. Now lets see if we can do something about x square. x square is the area of the square. Let us simplify this but replacing x square with u. So this is what we get. And simplifying it further, this is what we get So here, my friends, we have a quadratic equation. And so solve it we will use a formula by Brahmagupta an ancient Indian mathematician and astronomer. So here is the formula on the left. We will replace the values of a, b and c with the values in our equation and solve for u. These are the 2 values we get. And solving further, we get the values of u as, which leaves us with But which of the 2 values is the correct one? Can you please pause the video and try to figure this out? Ok, we can find out answer by looking at one of our equations, this one. We know that we have 2 values for u, 9 and 32. But what is u. u is x Square And here is a in terms x. Can a be negative. It cannot. So 12 - x square cannot be negative, right iF X2 is 9, then it will not be negative but x2 is 32, we get -20 which is negative. This means that X 2 or u has the value of 9. So finally, we can say that the area of our square is approx 9.