Can you find area of the big Blue Square? | (Circle) | |

[00:00:00]Welcome to PreMath. In this video, we have got this yellow shaded circle with the center O and this white square PECF fully inscribed in a bigger blue square ABCD, as you can see in this given diagram.

[00:00:19]Such that the area of this yellow circle has been given to us as pi cm squared.

[00:00:25]And moreover, this segment DE equals to the segment AC.

[00:00:33]And likewise, this segment FC equal to this segment uh BF.

[00:00:41]And furthermore, these points M and N are the points of tangency.

[00:00:48]And now, our task is to calculate the area of this bigger blue square ABCD.

[00:00:57]Please don't forget to give a thumbs-up and subscribe.

[00:01:00]And please keep in mind that this figure may not be 100% true to the scale.

[00:01:05]Let's go ahead and get started, and here's our very first step. We know the area of this yellow circle has been given to us as pi cm squared. And we are going to calculate the radius of this circle. I'm going to label the radius as lowercase r. Now, we are going to find the value of lowercase r. And now, let's recall the area of a circle formula area is always equal to pi r squared, where lowercase r is the radius. And the area has been given to us as pi.

[00:01:38]So, therefore, we can write the area of this circle as pi equals to pi times lowercase r squared.

[00:01:48]And now, we are going to divide both sides by pi to isolate r squared. This pi and pi is gone. So, therefore, r squared value turns out to be equal to 1.

[00:02:01]And now I'm going to undo this square by taking square root on both sides. So, therefore, our lower case r radius value turns out to be positive 1 cm. So, thus the radius of this yellow circle turns out to be equal to 1 cm.

[00:02:21]And here's our next step. Let's focus on this uh square PECF.

[00:02:27]I'm going to label its side lengths as lower case a across the board.

[00:02:35]And since this segment DE equal to this segment EC, if this segment is lower case a, then this is going to be lower case a as well. And likewise, this segment BF is going to be lower case a as well since this segment equal to this segment EC. So, thus we can see the side length of this big blue square is going to be a plus a is going to be equal to two times lower case a. And now our task is to find the value of this lower case a before we could calculate the area of this big blue square ABCD.

[00:03:22]And now let's make an observation. We can see that these segments are equal.

[00:03:29]Likewise, these segments are equal as well. So, therefore, we conclude that this point P is at the center of this big blue square ABCD.

[00:03:42]And in this next step, I have connected these vertices A and C as you can see such that all these points uh are collinear on this diagonal.

[00:03:59]And now let's focus on this right triangle PEC.

[00:04:03]And we are going to use the Pythagorean theorem on this triangle. And here's our Pythagorean theorem a squared plus b squared equal to c squared. And in our case this PC is the hypotenuse. Whereas these two segments lower case a and lower case a are our two other legs.

[00:04:27]Let's go ahead and fill in the blanks in this Pythagorean formula. So we got a squared plus a squared is going to be equal to our hypotenuse PC squared. I'm going to undo this square by taking a square root on both sides.

[00:04:48]So therefore our PC segment length is going to be a times square root of two.

[00:04:57]So that's our this PC segment length turns out to be a times square root of two.

[00:05:04]And now let's make an observation. We can see that this uh segment AP is going to be equal to this segment PC. As you can see in this given equation.

[00:05:19]And we know our PC length is a times square root of two. So I'm going to substitute that value a times square root of two over here. So therefore we can write our AP length is going to be a times square root of two. I'm going to label this one as our equation number one.

[00:05:43]And here in this next step I'm going to connect this center O with these two points of tangency M and N as you can see in this next step. And now let's make an observation. We can see that this uh ON is the radius of this circle and we know our radius is one. So this radius is going to be one and likewise this OM is radius as well.

[00:06:12]So this is going to be one unit as well.

[00:06:15]And this OP is radius as well. So this is going to be 1 cm as well.

[00:06:22]And now let's recall the circular theorem. According to this theorem, the angle between the radius and the tangent line will always be exactly 90°. So no wonder these angles are going to be 90° each since these are our radii and these are our tangent lines. So therefore we can see that this ANOM is the square.

[00:06:51]So therefore we conclude that this segment AM is going to be one and likewise this segment AN is going to be one as well.

[00:07:01]And now let's focus on this right triangle AMO and we are going to apply the Pythagorean theorem on this triangle.

[00:07:09]And here's our Pythagorean theorem once again, a² + b² = c². And in our case our hypotenuse is this segment uh AO length.

[00:07:20]Whereas our two other legs are one and one.

[00:07:25]Let's go ahead and fill in the blanks in this Pythagorean formula. So that is going to be equal to 1² + 1² is going to be equal to AO² I'm going to undo this square by taking a square root on both sides. So, therefore, our this AO segment length turns out to be square root of two.

[00:07:56]So, that's our this AO segment length turns out to be square root of two.

[00:08:02]And now let's make an observation. We can see that this whole AP length is going to be equal to the sum of these two individual lengths, so AO plus OP, as you can see in this given equation. And we know our AO segment length is square root of two. So, I'm going to substitute that value over here. And likewise, this OP length is one. So, I'm going to substitute one over here.

[00:08:36]However, from this equation one, we can see our AP length is A times square root of two. So, therefore, I'm going to substitute that value A times square root of two over here.

[00:08:50]So, therefore, after the substitution, this is going to become A times square root of two is going to be equal to square root of two plus one. And now I am going to divide both sides by square root of two to isolate A. This two square root of two and square root of two is gone. So, therefore, A value turns out to be square root of two plus one divided by square root of two.

[00:09:24]And now we are going to rationalize this denominator by multiplying and dividing by square root of two at the very same time.

[00:09:35]And now we are going to multiply these numerators and likewise we are going to multiply these denominators as well. So, therefore, if we multiply these numerators square root of two times square root of two is going to give us simply two plus one times square root of two is going to give us square root of two divided by square root of two times square root of two is going to be simply two.

[00:10:08]So, this our lower case A value turns out to be square root of two plus two all over two.

[00:10:15]And now we know that our this big blue square side length is two times A as you can see in this equation.

[00:10:26]And our A value is square root of two plus two all over two.

[00:10:30]So, let's go ahead and fill in the blanks. So, this is going to be two times our A value is square root of two plus two all over two. And here we can see this two and two cancel each other out.

[00:10:48]So, therefore, our this big blue square side length turns out to be square root of two plus two.

[00:11:00]So, this this big blue square side length turns out to be square root of two plus two. And here's our final step. Now we are going to calculate the area of this big blue square ABCD.

[00:11:17]And now let's recall the area of a square formula. Area of a square is always equal to S squared where S represents the side length of the square.

[00:11:26]And in our case, the side length of the square is square root of two plus two.

[00:11:31]So, therefore, the area of this big blue square is going to be simply square root of two plus two whole square. And now, let's recall this famous identity a plus b whole square could be written as a square plus b square plus two times ab.

[00:11:53]And we are going to apply this identity on the right hand side of this equation.

[00:11:57]So, therefore, this is going to be sum two plus four plus four times square root of two. And if we simplify this thing, that is going to give us six plus four times square root of two centimeter square. So, thus after all the calculations and manipulations, the big blue square area turns out to be six plus four times square root of two centimeter square.

[00:12:31]And that is going to be approximately equal to 11.66 centimeter square as well.

[00:12:37]And that's our final answer.

[00:12:39]Thanks for watching and please don't forget to subscribe to my channel for more exciting videos. Bye.