Not clear from the diagram whether 6mm is meant to be the whole diagonal of the big square or half the diagonal. Assuming 6mm is the blue bit, i.e. half the diagonal: The big square is two isosceles triangles, each with base 12mm (the full diagonal of the big square) and height 6mm (half the diagonal of the big square). The area of each triangle is therefore ½ × 12 × 6 so the area of the big square is double that, which is 12 × 6 = 72 mm² The small square is a quarter the area of the big square so its area is 72/4 = 18 mm²
It is not really an equilateral triangles. It has two sides that are the same not three, 45 degree verses 60 degree. But if you take the 6mm diagonal & mirror it using the vertical axis at center of large square, you end up with a triangle with two sides of 6mm. That is from the center to the two upper corners of the large square. The square area we are looking for is equivalent to area of this triangle: Area of triangle = 1/2 base × height = 1/2 ( 6 x6) = 1/2 x 36 = 18
You need less than that. Just knowing that the area of a triangle is half the base × the height is enough. Oh, and recognising that the small square is one quarter the area of the big square.
@@gavindeane3670 And there are 4 small squares in the big square, but only 2 triangles of base = 12, height = 6 in the big square. One big ⬛ two big triangles, 4 small squares.
The sine or cosine for 45° are the same, 0.70710678... Opposite / Hypotenuse= sin Adjacent / Hypotenuse = cos I was confused when I didn't notice my cheapo phone calculator was set to radians! Every time I restart the phone it opens with Rad, not Deg. Weird.
Solved in my head after listening to the problem, and IF I'm reading it right after that, the answer is that the small square has an area of 18 mm^2. If the diagonal from the center of the large square to one corner is 6mm, we can use that as the hypoteneuse of a right triangle. Using the Pythagorean theorem, a^2 + b^ 3= c^2, so 6^2=36. Half of that, 18, would then be the square of the length of each of the two sides of the small square, but since we want the area of that square, we can stop at 18.
Sides of the yellow square = a mm then Pythagoras: a² + a² = (6/2)² assuming the diagonal of the bigger square is 6mm so 2a² = 9 or a² = 9/2 mm² and THAT is also the area of the yellow square while Area = a . a = a²
The diagonal bisect the 90° angle of the larger square into 2 complementary angles of 45° and also creates 2 special right angles triangles The smaller Right-Angle-Triangle on the left formed by the 1/2 diagonal = 6mm ---> Hypotenuse lets call it side c one of the side of the smaller square --> lets call it side b one part of the side of the larger square --> lets call it side a Now we have Right-Angle-Triangle with sides a - b - c and angles 90° - 45 ° - x° Using the triangle theorem , the sum of the inner angles of a triangle is equal to 180° so x° = 180° - (90° + 45°) = 45° , so its a special Right-Angle-Triangle 45°- 45° - 90° with side ratios --> x : x : x√2 side c (hypotenuse) = 6 mm --> x√2 side b (side of the smaller square) = x side a (part of the side of the larger square) = x side b = c /√2 side b = 6 /√2 --> [6/√2] * [√2/√2] = 6√2 / 2 = 3√2 side of the smaller square = 3√2 Area (smaller square) = side * side Area(smaller square) = 3√2 * 3√2 = [3√2]² = 18 mm²☑ (common error [3√2]² = 11 mm² --> 3√2 ≠ 3 + √2 --> 3√2 = 3 * √2)
When you get to "side b = 6 /√2" you don't need to get rid of the irrational in the denominator, because we know that squaring it will produce 2. And we know that we'll be squaring it because we're looking for an area. So.. Area = (6 /√2)(6 /√2) =36/2 = 18 mm²
Let each side of the big square be X mm. The diagonal of the big square is given to be (2x6)= 12mm According to Pythagorean theorem, X^2+X^2=12^2 2X^2=144 X^2 = 144/2 = 72 The area of the small square in question = 1/2 X * 1/2 X = 1/4 X^2 Substituting X^2 = 72, 1/4 X^2 = 1/4*72=18 Therefore, the area of the small square is 18 square mm.
square --> 45° angles --> sides: 1/1/sqrt(2) here 6 relates to sqrt(2) i.e large square side = 6/sqrt(2) small square side = 3/sqrt(2) area small square = (3/sqrt(2))^2 = 9/2 = 4.5 mm^2
Let’s look at the triangle above the yellow square formed by the top edge of the yellow square, the 6mm length, and the part of the big square’s left edge above the yellow square. This forms a right triangle. Also, because the yellow square’s top right corner separates the big square’s diagonal exactly in half, this triangle is a 45-45-90 right triangle, where the two legs are of equal length. Let’s call this length s. Per the Pythagorean Theorem: x^2 + y^2 = z^2 As mentioned, in this case x = y = s and we know z = 6 mm. Substituting in: s^2 + s^2 = (6 mm)^2 2*s^2 = 36 mm^2 (note mm^2 means square millimeters, a measure of area) s^2 = 36 mm^2 / 2 = 18 mm^2 s = +/- sqrt(18 mm^2) = +/- sqrt(9 mm^2 *2) = +/- 3 * sqrt(2) mm Since we are dealing with a real geometric shape, the negative answer is discarded as nonsensical, thus s = 3 * sqrt(2) mm. This is the length of the top edge of the yellow square. And since this is a square, by definition, all its sides will have the same length s. The area of any square is: A = s^2 , where A is the area and s is the length of a side of the square. In our case, we can plug in our value for s: A = (3 * sqrt(2) mm)^2 A = 9 * 2 mm^2 A = 18 mm^2 The area of the yellow square is 18 square millimeters.
So here again, a square with diagonal = 12 has an area which is a multiple of the diagonal, but it's 6 times the diagonal (72 = 6×12), not 3 times as with the 18 sq. mm square. What if the diagonal was length = 1? Would the area of the d = 1 square be 1/2 d^2, meaning 0.5 mm^2?
When you know what the diagonal of a square measures, is the area of the square always 3 times the diagonal? Because this time a diag of 6mm crosses an area of 18mm^2. And 3 × 6 = 18.
18 Six is to the square root of 2, as x is to 1, when x is the side of the square. 6/(√2) : x/1 . . . So x = 6/√2 And the ⬛ area = (6/√2)^2 = 36/2 = 18 (ANSWER) CHECK: Does a base of 12, height of 6 triangle have twice this area, or 36? (1/2) 12 x 6 = 36✓
In geometry, one hash mark is used to identify a length "a" and two hash marks for length "b" (for example). I don't think there is a name for the hash mark or the segment of line they cross, it's just a convention for equating the length of one segment (or side of a polygon) with another segment. The segments do NOT have to be aligned, but these 2 are. In this case, the pair of hash marks mean that one 6mm diagonal has the same length as the other diagonal segment crossed with a hash mark.
![กี่หมื่นฟ้า | Your Sky Series EP.1 [1/4]](http://i.ytimg.com/vi/vLGjxFL6U_I/mqdefault.jpg)
Not clear from the diagram whether 6mm is meant to be the whole diagonal of the big square or half the diagonal. Assuming 6mm is the blue bit, i.e. half the diagonal:
The big square is two isosceles triangles, each with base 12mm (the full diagonal of the big square) and height 6mm (half the diagonal of the big square).
The area of each triangle is therefore
½ × 12 × 6
so the area of the big square is double that, which is
12 × 6 = 72 mm²
The small square is a quarter the area of the big square so its area is
72/4 = 18 mm²
It is not really an equilateral triangles.
It has two sides that are the same not three, 45 degree verses 60 degree.
But if you take the 6mm diagonal & mirror it using the vertical axis at center of large square, you end up with a triangle with two sides of 6mm.
That is from the center to the two upper corners of the large square.
The square area we are looking for is equivalent to area of this triangle:
Area of triangle = 1/2 base × height
= 1/2 ( 6 x6) = 1/2 x 36 = 18
@@chamberizerThanks! That was a careless error. I meant isosceles!
I will correct it.
@gavindeane3670
Nobody is perfect.
Pytagoras is all you need to solve this.
You need less than that. Just knowing that the area of a triangle is half the base × the height is enough. Oh, and recognising that the small square is one quarter the area of the big square.
@@gavindeane3670
And the triangle with base = 12 and height = 6 has half area of the big square.
@@gavindeane3670
And there are 4 small squares in the big square, but only 2 triangles of base = 12, height = 6 in the big square. One big ⬛ two big triangles, 4 small squares.
Since it is 45 degrees:
Multiply 6 × .707 to get length of a side.
Both sides are equal thus
(6 × .707) (6 x .707) = area = 18
Where do you get 0.707?
The sine or cosine for 45° are the same, 0.70710678...
Opposite / Hypotenuse= sin
Adjacent / Hypotenuse = cos
I was confused when I didn't notice my cheapo phone calculator was set to radians! Every time I restart the phone it opens with Rad, not Deg. Weird.
Solved in my head after listening to the problem, and IF I'm reading it right after that, the answer is that the small square has an area of 18 mm^2.
If the diagonal from the center of the large square to one corner is 6mm, we can use that as the hypoteneuse of a right triangle. Using the Pythagorean theorem, a^2 + b^ 3= c^2, so 6^2=36.
Half of that, 18, would then be the square of the length of each of the two sides of the small square, but since we want the area of that square, we can stop at 18.
Sides of the yellow square = a mm then Pythagoras: a² + a² = (6/2)² assuming the diagonal of the bigger square is 6mm so
2a² = 9 or a² = 9/2 mm² and THAT is also the area of the yellow square while Area = a . a = a²
The diagonal bisect the 90° angle of the larger square into
2 complementary angles of 45° and also creates 2 special right angles triangles
The smaller Right-Angle-Triangle on the left
formed by the 1/2 diagonal = 6mm ---> Hypotenuse lets call it side c
one of the side of the smaller square --> lets call it side b
one part of the side of the larger square --> lets call it side a
Now we have Right-Angle-Triangle with sides a - b - c
and angles 90° - 45 ° - x°
Using the triangle theorem , the sum of the inner angles of a triangle is equal to 180°
so x° = 180° - (90° + 45°) = 45° , so its a special Right-Angle-Triangle 45°- 45° - 90°
with side ratios --> x : x : x√2
side c (hypotenuse) = 6 mm --> x√2
side b (side of the smaller square) = x
side a (part of the side of the larger square) = x
side b = c /√2
side b = 6 /√2 --> [6/√2] * [√2/√2] = 6√2 / 2 = 3√2
side of the smaller square = 3√2
Area (smaller square) = side * side
Area(smaller square) = 3√2 * 3√2 = [3√2]² = 18 mm²☑
(common error [3√2]² = 11 mm² --> 3√2 ≠ 3 + √2 --> 3√2 = 3 * √2)
When you get to "side b = 6 /√2" you don't need to get rid of the irrational in the denominator, because we know that squaring it will produce 2. And we know that we'll be squaring it because we're looking for an area. So..
Area = (6 /√2)(6 /√2)
=36/2
= 18 mm²
Let each side of the big square be X mm.
The diagonal of the big square is given to be (2x6)= 12mm
According to Pythagorean theorem,
X^2+X^2=12^2
2X^2=144
X^2 = 144/2 = 72
The area of the small square in question =
1/2 X * 1/2 X = 1/4 X^2
Substituting X^2 = 72,
1/4 X^2 = 1/4*72=18
Therefore, the area of the small square is
18 square mm.
a² + b² = c²
Here a = b as mentioned
So 2a² = 36
Or a² = 18
Area = a² = 18
square --> 45° angles
--> sides: 1/1/sqrt(2)
here 6 relates to sqrt(2)
i.e large square side = 6/sqrt(2)
small square side = 3/sqrt(2)
area small square
= (3/sqrt(2))^2
= 9/2
= 4.5 mm^2
Let’s look at the triangle above the yellow square formed by the top edge of the yellow square, the 6mm length, and the part of the big square’s left edge above the yellow square. This forms a right triangle. Also, because the yellow square’s top right corner separates the big square’s diagonal exactly in half, this triangle is a 45-45-90 right triangle, where the two legs are of equal length. Let’s call this length s. Per the Pythagorean Theorem:
x^2 + y^2 = z^2
As mentioned, in this case x = y = s and we know z = 6 mm. Substituting in:
s^2 + s^2 = (6 mm)^2
2*s^2 = 36 mm^2 (note mm^2 means square millimeters, a measure of area)
s^2 = 36 mm^2 / 2 = 18 mm^2
s = +/- sqrt(18 mm^2) = +/- sqrt(9 mm^2 *2) = +/- 3 * sqrt(2) mm
Since we are dealing with a real geometric shape, the negative answer is discarded as nonsensical, thus s = 3 * sqrt(2) mm. This is the length of the top edge of the yellow square. And since this is a square, by definition, all its sides will have the same length s.
The area of any square is:
A = s^2 , where A is the area and s is the length of a side of the square. In our case, we can plug in our value for s:
A = (3 * sqrt(2) mm)^2
A = 9 * 2 mm^2
A = 18 mm^2
The area of the yellow square is 18 square millimeters.
Shorter form videos are much more helpful.
Area of a square with known diagonal is 1/2 d^2 so d of big square is 12 so area is 72. Little square is 1/4 of 72 or 18.
So here again, a square with diagonal = 12 has an area which is a multiple of the diagonal, but it's 6 times the diagonal (72 = 6×12), not 3 times as with the 18 sq. mm square.
What if the diagonal was length = 1?
Would the area of the d = 1 square be 1/2 d^2, meaning 0.5 mm^2?
@@btpcmsag Yes. The area of a square with a diagonal of length 1 mm is 0.5 mm².
a = (6/√2)²
a = 36/2
a = 18 mm²
When you know what the diagonal of a square measures, is the area of the square always 3 times the diagonal? Because this time a diag of 6mm crosses an area of 18mm^2.
And 3 × 6 = 18.
18
Six is to the square root of 2, as x is to 1, when x is the side of the square.
6/(√2) : x/1 . . . So x = 6/√2
And the ⬛ area = (6/√2)^2 = 36/2 = 18 (ANSWER)
CHECK:
Does a base of 12, height of 6 triangle have twice this area, or 36? (1/2) 12 x 6 = 36✓
A1=(6+6)*6/2=12*3=
=36mm^2
A1=2A//:2 A=(A1)/2
A=18mm^2
I got right! Yay❤
Niggle point/question here: if distance from a corner to center of square is 6mm, what is purpose of the two black hash marks across the diagonal?
In geometry, one hash mark is used to identify a length "a" and two hash marks for length "b" (for example). I don't think there is a name for the hash mark or the segment of line they cross, it's just a convention for equating the length of one segment (or side of a polygon) with another segment. The segments do NOT have to be aligned, but these 2 are.
In this case, the pair of hash marks mean that one 6mm diagonal has the same length as the other diagonal segment crossed with a hash mark.
@@btpcmsag Thanks. I did not know that.
4:00 triangle
got 18 5 seconds small triangle side is 2X^2 = 36 so side = sr 18.
area of small square is sr 18 X sr 18 so area = 18
thanks for the fun
maybe put your units on the answer?
I found that the small square is 4.5
18 mm squared
18 sq mm
Inside this, not inside of this.