Can you find the Green shaded Square area? | (Triangle) | #math #maths | #geometry

Added: 2026-05-28

[00:00:00]Welcome to Primath. In this video, we have got these two blue squares along with this uh green shaded square put together as you can see in this given diagram. Such that the area of this uh smaller blue square is 1 cm square whereas the area of this uh larger blue square is 2 cm square. And now our task is to calculate the area of this uh green shaded square. Please don't forget to give a thumbs up and subscribe. And please keep in mind that this figure may not be 100% true to the scale. Let's go ahead and get started.

[00:00:45]And here's our very first step since we are dealing with all these uh squares.

[00:00:50]So therefore all these squares have the same side length. If I label this side length as lowerase A, then this is going to be lowerase A, lowerase A and lowerase A. And likewise, if I label this uh larger blue square length as lowerase B, then this side length is going to be lowerase B, lowerase P, and lowerase P across the board. And likewise, let's focus on this uh green shaded square. I'm going to label this side length as lowerase X. Then this is going to be lowerase x lowerase x and lowerase x across the board. And now let's recall the area of a square formula. Area equ= to s² where s represents the side length of the square. And now let's focus on this uh smaller blue square. Its side length is lowerase a and its area has been given to us as 1. So therefore we can write down a² = 1. And now I'm going to undo this square by taking a square root on both sides. So therefore lowerase a value turns out to be equal to 1 cm. So therefore our lowerase a value turns out to be 1 across the board. And now let's focus on this uh other uh blue square as well. It side length is lowerase b and the area has been given to us as two. So therefore we can write down b² = to 2. I'm going to undo this square by taking a square root on both sides. So therefore uh our lowerase B value turns out to be square<unk> of 2 cm. So thus the side length B of this uh blue square turns out to be square roo<unk> of 2 across the board. And finally let's focus on this uh green shaded square as well and its side length is x across the board.

[00:03:02]So therefore the green square area is going to become x².

[00:03:08]And now our task is to find the value of x². And now let's bear in mind that this uh side a c is equal to this side c.

[00:03:22]These sides lengths are congrent. And now let's focus on this uh tiny right triangle a b c. And we are going to apply the Pythagorean theorem on this triangle. And here's our pyagorean theorem. A square + B square= to C square. And in our case, our hypotenuse is X. Whereas our two other legs are 1 and this uh line segment B C. Let's go ahead and fill in the blanks in this Pythagorean formula. So we got uh b c² + uh 1 square = x². Let's simplify. So we are going to have a b c² + 1 square is same as 1 = x². Now I'm going to move this one to the other side. So therefore b c² is going to be x² -1 and now I'm going to undo this square by taking a square root on both sides.

[00:04:28]So therefore our this b c side length turns out to be square root of x² minus one. So thus our this uh side B C length turns out to be square root of X square minus one. And now we are going to focus on this other right triangle C D E. We are going to apply the Pythagorean theorem on this triangle as well. And here's our Pythagorean theorem once again. A square + B square= to C square. And uh in this case we have our hypotenuse x whereas our two other legs are square root of 2 and this uh side uh cd. Let's go ahead and fill in the blanks. So this is going to give us uh c d² + uh square<unk> of 2² = x².

[00:05:26]Let's simplify. we're going to get uh c d² and that is going to give us simply 2 = x². I'm going to move this to to the right hand side. So therefore c d² turns out to be x² - 2. I'm going to undo this square by taking the square root on both sides. So therefore our C d side length turns out to be the square roo<unk> of x² - 2. So therefore our this uh cd length turns out to be square root of x² - 2.

[00:06:10]And now let's assume that alpha and beta are our two complimentary angles. In other words, the sum of these two angles alpha plus beta must be equal to 90°.

[00:06:24]If I label this angle as alpha, we know this is our 90° angle. So therefore, this angle has got to be our angle beta.

[00:06:35]And furthermore, we know this angle is 90°. This angle is beta. So therefore, this angle has got to be our angle alpha. And finally we know this angle is 90° this angle is alpha. So therefore this angle has got to be our angle beta.

[00:06:53]So thus we conclude that these two right triangles this right triangle A B C and this uh other right triangle C D E are our congruent triangles according to angle side angle congruency theorem.

[00:07:10]Since these two side lengths are equal in length and since these two triangles are congrent. So therefore the side opposite to this angle beta we have square root of x square minus 2. And likewise uh the side opposite to this angle beta is 1. So therefore I can write down the square root of x² - 2 is going to be equal to 1. And now I am going to undo this square root by taking a square on both sides. And here we can see this square and square root undo each other. So we are ended up with x² - 2 = 1. And now I'm going to move this -2 on the other side. So therefore we can write x² = 2 + 1 or simply x² value turns out to be 3. So thus our x square value turns out to be 3. And now we can see that this uh green square area is being represented by x square.

[00:08:22]So therefore I'm going to substitute this uh x square value three over here.

[00:08:27]So therefore we conclude that our this uh green square area turns out to be 3 cm square. So thus after all the calculations and manipulations the area of this uh green shaded square turns out to be 3 cm square. In other words the area of this uh green shaded square is going to be 3 cm square. And that's our final answer. Thanks for watching and please don't forget to subscribe to my channel for more exciting videos. Bye.