# Geometry - no equations needed

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I can't see how to do this without at least implicit equations...

Edited by Carrock
no spoiler
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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.

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Spoiler

Bottom of blue triangle is two, so square is 16 sq m.

Edited by TheVat
nftgh
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1 hour ago, TheVat said:
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Bottom of blue triangle is two, so square is 16 sq m.

No, it is not

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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 by studiot
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OK, I think I see it.  12?  I can provide my steps, if needed.

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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:
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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.

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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 by studiot
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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.

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clearly 12 m2, since the line is 2/3 up the side of the middle box

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8 minutes ago, OldChemE said:
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clearly 12 m2, since the line is 2/3 up the side of the middle box

Right. +1

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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 by mistermack
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10 minutes ago, mistermack said:
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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

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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

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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 by studiot
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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

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