Genady Posted August 1, 2023 Share Posted August 1, 2023 Link to comment Share on other sites More sharing options...
Carrock Posted August 1, 2023 Share Posted August 1, 2023 (edited) I can't see how to do this without at least implicit equations... Edited August 1, 2023 by Carrock no spoiler 1 Link to comment Share on other sites More sharing options...
Genady Posted August 1, 2023 Author Share Posted August 1, 2023 8 minutes ago, Carrock said: I can't see how to do this without at least implicit equations... Well, you equate an unknown to something known - this is an equation indeed, but you don't need to solve this equation to find the unknown. Anyway, this is rather semantics, and the fact is that your answer is correct. +1 I'd like to ask you to hide your solution using the Spoiler feature, so that others could try as well. Link to comment Share on other sites More sharing options...
TheVat Posted August 1, 2023 Share Posted August 1, 2023 (edited) Spoiler Bottom of blue triangle is two, so square is 16 sq m. Edited August 1, 2023 by TheVat nftgh Link to comment Share on other sites More sharing options...
Genady Posted August 1, 2023 Author Share Posted August 1, 2023 1 hour ago, TheVat said: Hide contents Bottom of blue triangle is two, so square is 16 sq m. No, it is not Link to comment Share on other sites More sharing options...
studiot Posted August 1, 2023 Share Posted August 1, 2023 (edited) Let KI intersect EF in M. The triangle ABJ is similar to JCK is sim to KEM is sim to FMI. That is the triangles along the diagonal all have the same angles and areas in proportion. I take it ABCD, CEFG and EHIF are squares. Edited August 1, 2023 by studiot Link to comment Share on other sites More sharing options...
TheVat Posted August 1, 2023 Share Posted August 1, 2023 Spoiler OK, I think I see it. 12? I can provide my steps, if needed. 1 Link to comment Share on other sites More sharing options...
Genady Posted August 1, 2023 Author Share Posted August 1, 2023 1 hour ago, studiot said: Let KI intersect EF in M. The triangle ABJ is similar to JCK is sim to KEM is sim to FMI. That is the triangles along the diagonal all have the same angles and areas in proportion. I take it ABCD, CEFG and EHIF are squares. Yes, all correct. The last three are equal squares. This square's area is the question. 45 minutes ago, TheVat said: Reveal hidden contents OK, I think I see it. 12? I can provide my steps, if needed. Right. +1. The steps are not needed, but would be interesting to see, as they might be different. Link to comment Share on other sites More sharing options...
studiot Posted August 1, 2023 Share Posted August 1, 2023 (edited) So AI is the diagonal of a rectangle 3 squares wide by 2 squares high and K must be the centre point of this rectangle. So by no equations I assume that I can use formulae you mean no equations need solving ? So I can say the area of this rectangle is 3 squares x 2 squares which makes 6 squares. Edited August 1, 2023 by studiot Link to comment Share on other sites More sharing options...
Genady Posted August 1, 2023 Author Share Posted August 1, 2023 14 minutes ago, studiot said: So AI is the diagonal of a rectangle 3 squares wide by 2 squares high and K must be the centre point of this rectangle. So by no equations I assume that I can use formulae you mean no equations need solving ? So I can say the area of this rectangle is 3 squares x 2 squares which makes 6 squares. "No equations" is a vague statement, I realize. Let's say, equations are not needed, but allowed. Don't take this part too seriously. We are asked to find the area of the square in m2. Please provide your answers using the "Spoiler" feature for others not to be affected. Link to comment Share on other sites More sharing options...
OldChemE Posted August 1, 2023 Share Posted August 1, 2023 Spoiler clearly 12 m2, since the line is 2/3 up the side of the middle box 1 Link to comment Share on other sites More sharing options...
Genady Posted August 1, 2023 Author Share Posted August 1, 2023 8 minutes ago, OldChemE said: Hide contents clearly 12 m2, since the line is 2/3 up the side of the middle box Right. +1 Link to comment Share on other sites More sharing options...
mistermack Posted August 2, 2023 Share Posted August 2, 2023 (edited) Spoiler All of the triangles have the same angles, so they all have sides of ratio 2 to 3. So the diagonal line cuts e-f 2/3 of the way up. The top section of the middle square takes 4 small triangles, and there are two more identical areas, below it so that adds up 12 small triangle, so 12 m2. Edited August 2, 2023 by mistermack 1 Link to comment Share on other sites More sharing options...
Genady Posted August 2, 2023 Author Share Posted August 2, 2023 10 minutes ago, mistermack said: Hide contents All of the triangles have the same angles, so they all have sides of ratio 2 to 3. So the diagonal line cuts e-f 2/3 of the way up. The top section of the middle square takes 4 small triangles, and there are two more identical areas, below it so that adds up 12 small triangle, so 12 m2. Indeed. +1 Link to comment Share on other sites More sharing options...
Genady Posted August 2, 2023 Author Share Posted August 2, 2023 After a couple of comments regarding "no equations" phrase in the title I've realized that it would be better to clarify, like: 13 hours ago, studiot said: no equations need solving Link to comment Share on other sites More sharing options...
studiot Posted August 2, 2023 Share Posted August 2, 2023 (edited) 12 Area units is correct Here is my derivation, avoiding equations one way or another. I use the property of rectilinear similar figures that the areas are in proportion to the squares of corresponding sides. So firstly complete the rectangle ANIM by adding 3 extra squares, making 6 in all. DH is a right bisector of two sides so passes through the centre of the rectangle at K AI is also a diagonal of the rectangle and more importantly bisects CE. Thus CK is equal to KE. Call this s. CK is one side of the triangle CKJ which we are given has having 1 square area unit. Triangle CKJ contains the same angles as triangle AMI since AN is parallel to MI. Hence triangle CKJ is similar to triangle AMI., with side CK corresponding to side MI Since the length of each side of a square is 2s and side MI occupies 3 squares its length is 6s. So the area of triangle AMI is (6)2 times the area of triangle CKJ or 36 area units. Triangle AMI is exactly half the rectangle ANIM. So the total area of the rectangle is 72 area units. The rectangle is comprised of 6 squares so each square has an area od 12 area units as required. Edited August 2, 2023 by studiot 2 Link to comment Share on other sites More sharing options...
Genady Posted August 2, 2023 Author Share Posted August 2, 2023 25 minutes ago, studiot said: 12 Area units is correct Here is my derivation, avoiding equations one way or another. I use the property of rectilinear similar figures that the areas are in proportion to the squares of corresponding sides. So firstly complete the rectangle ANIM by adding 3 extra squares, making 6 in all. DH is a right bisector of two sides so passes through the centre of the rectangle at K AI is also a diagonal of the rectangle and more importantly bisects CE. Thus CK is equal to KE. Call this s. CK is one side of the triangle CKJ which we are given has having 1 square area unit. Triangle CKJ contains the same angles as triangle AMI since AN is parallel to MI. Hence triangle CKJ is similar to triangle AMI., with side CK corresponding to side MI Since the length of each side of a square is 2s and side MI occupies 3 squares its length is 6s. So the area of triangle AMI is (6)2 times the area of triangle CKJ or 36 area units. Triangle AMI is exactly half the rectangle ANIM. So the total area of the rectangle is 72 area units. The rectangle is comprised of 6 squares so each square has an area od 12 area units as required. Very smooth! +1 Link to comment Share on other sites More sharing options...
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