X=-3 represents a square whose top right corner is the bottom left corner of the green rectangle and whose bottom left corner is the point B. Don't reject it just because it's 'out of the box'
121 The width of the rectangle = 3 (15/5) Draw a line from the edge of the rectangle to the square to form a right triangle . Let the length of the square = n, then the bases of the triangle are , n-5, and n-3. Hence, 10^2 = (n-5)^2 + (n-3)^2 =n^2 + 25-10n + n^2 +9-6n =2n^2 + 34- 16n 50 = n^2 + 17-8n 0 = n^2 - 33-8n 0 = (n-11) (n+3) 11=n 121= n^2
Green rectangle: Aɢ = hw 15 = h(5) h = 15/5 = 3 Draw CA, where C is the point on the top side of the blue square where BC and CA are perpendicular. If s is the side length of the blue square, then by observation, BC = s-5 and CA = s-3. Triangle ∆ABC: BC² + CA² = AB² (s-5)² + (s-3)² = 10³ s² - 10s + 25 + s² - 6s + 9 = 100 2s² -16s - 66 = 0 s² - 8s - 33 = 0 (s-11)(s+3) = 0 s = 11 | s = -3 ❌ s > 0 Blue square: Aʙ = s² = 11² = 121 cm²
green rectangle is of 5 cm long and 3 cm high Side of blue square be a This implies ( a - 5)^2 + ( a - 3)^2 = 10 ^2 2 a ^2 - 16 a + 34 = 100 a^2 - 8 a + 64 = 130 a = 8 + √ (130) Hereby a^2 = 194 + 16 √ (130)
Solution: a = horizontal side of the green rectangle = 5, b = vertical side of the green rectangle = 15/5 = 3, x = side of the blue square. Pythagoras: (x-5) ²+(x-3)² = 10² ⟹ x²-10x+25+x²-6x+9 = 100 |-100 ⟹ 2x²-16x-66 = 0 |/2 ⟹ x²-8x-33 = 0 |p-q-formula ⟹ x1/2 = 4±√(16+33) = 4±7 ⟹ x1 = 4+7 = 11 and x2 = 4-7 = -3 [not allowed in geometry] ⟹ Blue square area = 11² = 121[cm²]
First ly name the points of intersection as A. B. C. D. E And F. Extend EF to meet BC at G. Now the four sides of the square are AB. BC. CD. and. DA. Let that be =X cms. Take the green figure Area is 25 cm^2. One side is 3 cm. Hence the other side Namely ED and FH =3cm. Now the triangle BFG you have BG as =X-3and FG as =X-5. So in the triangle Rt angle FB^2= (x-3)^2+(x-5)^2. 100=2x^2-16x+44. So 2x^2-16x=56. By dividing by 2 .We have. X^2-8X-28=0. So in that Quadratic equation you get two values. Take the positive one as length is positive. That when multiples by X itself becomes the area.
I made one stupid mistake (x-3)²+(x-5)²=10² x²-6x+9=x²-10x+25=100 By adding I made the mistake 2x²-16x+36=100 instead of 2x²-16x+34=100 Buthey, this happens to all of us. We're all human beings. One of the best math channels on youtube, clear explanation. You can't say that of some other math channels
Let's find the area: . .. ... .... ..... The height of the green rectangle can be calculated as follows: A = (base b)*(height h) ⇒ h = A/b = (15cm²)/(5cm) = 3cm Now let's add point C on the right side of the blue square such that ABC is a right triangle. In this case we can apply the Pythagorean theorem and with s being the side length of the square we obtain: AB² = AC² + BC² AB² = (s − b)² + (s − h)² (10cm)² = (s − 5cm)² + (s − 3cm)² 100cm² = s² − (10cm)*s + 25cm² + s² − (6cm)*s + 9cm² 0 = 2*s² − (16cm)*s − 66cm² 0 = s² − (8cm)*s − 33cm² s = 4cm ± √[(4cm)² + 33cm²] = 4cm ± √(16cm² + 33cm²) = 4cm ± √(49cm²) = 4cm ± 7cm Since the side length is a positive quantity, the only useful solution is: s = 4cm + 7cm = 11cm Now we are able to calculate the area of the blue square: A = s² = (11cm)² = 121cm² Best regards from Germany
My way of solution ▶ Agreen= 15 cm² x= 5 cm ⇒ 15= 5y y= 3 cm [AB]= 10 cm The length [AB] is the hypothenuse of a right triangle: the length of the blue square a the base of this rectangular = a-5 the height of this triangular= a-3 By applyling the Pythagorean theorem we get: 10²= (a-3)²+(a-5)² 100= a²-6a+9+a²-10a+25 2a²-16a-66=0 a²-8a-33=0 Δ= 196 √Δ= 14 ⇒ a₁= (8+14)/2 a₁= 11 cm a₂= (8-14)/2 a₂= -3 < 0 ❌ ⇒ a= 11 cm Asquare= 11² Asquare= 121 cm²
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X=-3 represents a square whose top right corner is the bottom left corner of the green rectangle and whose bottom left corner is the point B. Don't reject it just because it's 'out of the box'
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121
The width of the rectangle = 3 (15/5)
Draw a line from the edge of the rectangle to the square to
form a right triangle .
Let the length of the square = n,
then the bases of the triangle are , n-5, and n-3.
Hence, 10^2 = (n-5)^2 + (n-3)^2
=n^2 + 25-10n + n^2 +9-6n
=2n^2 + 34- 16n
50 = n^2 + 17-8n
0 = n^2 - 33-8n
0 = (n-11) (n+3)
11=n
121= n^2
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A₁ = 15 cm² = b.h = 5h
h = 3 cm
Pytagorean theorem:
(s-3)²+(s-5)²=10²
(s²-6s+9)+(s²-10s+25)=100
2s² - 16s + 34 = 100
s² - 8s - 33 = 0
s = 11 cm
s² = 121 cm² ( Solved √ )
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Knew the answer by looking. A right triangle with a hypotenuse of 10 gives it away.
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I agree with you! It is indeed a 6-8-10 triangle in this problem😊!
Green rectangle:
Aɢ = hw
15 = h(5)
h = 15/5 = 3
Draw CA, where C is the point on the top side of the blue square where BC and CA are perpendicular. If s is the side length of the blue square, then by observation, BC = s-5 and CA = s-3.
Triangle ∆ABC:
BC² + CA² = AB²
(s-5)² + (s-3)² = 10³
s² - 10s + 25 + s² - 6s + 9 = 100
2s² -16s - 66 = 0
s² - 8s - 33 = 0
(s-11)(s+3) = 0
s = 11 | s = -3 ❌ s > 0
Blue square:
Aʙ = s² = 11² = 121 cm²
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(l-3)^2+(l-5)^2=100...l=11
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(a - 5)² + (a - 3)² = 10²
a² - 10a + 25 + a² - 6a + 9 = 100
2a² - 16a + 34 = 100
a² - 8a + 17 = 50
a² - 8a = 33
a² - 8a + 16 = 33 + 16 = 49
(a - 4)² = 49
a - 4 = 7 (negative solution invalid)
a = 11
A(square) = 11² = 121 [cm²]
green rectangle is of 5 cm long and 3 cm high
Side of blue square be a
This implies
( a - 5)^2 + ( a - 3)^2 = 10 ^2
2 a ^2 - 16 a + 34 = 100
a^2 - 8 a + 64 = 130
a = 8 + √ (130)
Hereby a^2 = 194 + 16 √ (130)
Solution:
a = horizontal side of the green rectangle = 5,
b = vertical side of the green rectangle = 15/5 = 3,
x = side of the blue square.
Pythagoras:
(x-5) ²+(x-3)² = 10² ⟹
x²-10x+25+x²-6x+9 = 100 |-100 ⟹
2x²-16x-66 = 0 |/2 ⟹
x²-8x-33 = 0 |p-q-formula ⟹
x1/2 = 4±√(16+33) = 4±7 ⟹
x1 = 4+7 = 11 and x2 = 4-7 = -3 [not allowed in geometry] ⟹
Blue square area = 11² = 121[cm²]
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(x - 3)^2 + (x - 5)^2 = 10^2
2x^2 - 16x - 66 = 0
x^2 - 8x - 33 = 0
x = (4 + ✓49) cm = 11 cm
area = (11 cm)^2 = 121 cm^2
First ly name the points of intersection as A. B. C. D. E And F. Extend EF to meet BC at G. Now the four sides of the square are AB. BC. CD. and. DA. Let that be =X cms. Take the green figure Area is 25 cm^2. One side is 3 cm. Hence the other side Namely ED and FH =3cm. Now the triangle BFG you have BG as =X-3and FG as =X-5. So in the triangle Rt angle FB^2= (x-3)^2+(x-5)^2.
100=2x^2-16x+44.
So 2x^2-16x=56. By dividing by 2 .We have. X^2-8X-28=0. So in that Quadratic equation you get two values. Take the positive one as length is positive. That when multiples by X itself becomes the area.
S=121 cm²
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(a-5)²+[a-(15/5)]²=10²→ a=11→ a²=121 = Área del cuadrado azul.
Gracias y saludos.
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Fine, that's simple.
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I made one stupid mistake
(x-3)²+(x-5)²=10²
x²-6x+9=x²-10x+25=100
By adding I made the mistake
2x²-16x+36=100 instead of 2x²-16x+34=100
Buthey, this happens to all of us. We're all human beings.
One of the best math channels on youtube, clear explanation. You can't say that of some other math channels
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15/5=3 5+x=3+y y=x+2
x²+(x+2)²=10² 2x²+4x+4=100 2x²+4x-96=0 x²+2x-48=0
(x+8)(x-6)=0 x>0 , x=6 y=8 5+6=3+8=11
Blue Square area = 11*11 = 121cm²
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I used s instead of x, but yeah, got 121.
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(X-5)^2 + (X-3)^2 = 100
X = 11
S = 121
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Let x is the side of square
So (x-3)^2+(x-5)^2=10^2
So x=11cm
Area of the blue square=11^2=121 cm^2.❤❤❤
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The area is 121 cm squared
Another geometry problems that is *easier* than it looks!!!
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Ar. square=11^2=121 sq.unit.
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Not 25 but 15. So the other side is 3. Cms
Solution:
A = Lenght × Width
15 = 5 × W
W = 3 cm
Applying Pythagorean Theorem, we'll have:
(a - 5)² + (a - 3)² = 10²
a² - 10a + 25 + a² - 6a + 9 = 100
2a² - 16a + 34 = 100
2a² - 16a - 66 = 0 (÷2)
a² - 8a - 33 = 0
a = [8 ± √(64 + 132)]/2
a = [8 ± 14]/2
a = 11
A = a²
A = (11)²
A = 121 cm²
=========
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Lets close the Rigth Triangle creating Point C.
02) BC = X
03) AC = Y
04) AB = 10
05) 100 = X^2 + Y^2
06) Square Side = 5 + Y
07) Square Side = 3 + X
08) 5 + Y = 3 + X ; Y = X - 2
09) 100 = X^2 + (X - 2)^2 ; 100 = X^2 + X^2 - 4X + 4 ; 2X^2 - 4X + 4 = 100 ; 2X^2 - 4X - 96 = 0 ; X^2 - 2X - 48 = 0
10) Two Solutions : X = - 6 or X = 8
11) So, X = 8 and Y = 6
12) Square Side = 5 + 6 = 3 + 8 = 11
13) Area = 11^2 = 121
Therefore,
OUR ANSWER :
Blue Square Area equal 121 Square Centimeters.
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Let's find the area:
.
..
...
....
.....
The height of the green rectangle can be calculated as follows:
A = (base b)*(height h) ⇒ h = A/b = (15cm²)/(5cm) = 3cm
Now let's add point C on the right side of the blue square such that ABC is a right triangle. In this case we can apply the Pythagorean theorem and with s being the side length of the square we obtain:
AB² = AC² + BC²
AB² = (s − b)² + (s − h)²
(10cm)² = (s − 5cm)² + (s − 3cm)²
100cm² = s² − (10cm)*s + 25cm² + s² − (6cm)*s + 9cm²
0 = 2*s² − (16cm)*s − 66cm²
0 = s² − (8cm)*s − 33cm²
s = 4cm ± √[(4cm)² + 33cm²] = 4cm ± √(16cm² + 33cm²) = 4cm ± √(49cm²) = 4cm ± 7cm
Since the side length is a positive quantity, the only useful solution is:
s = 4cm + 7cm = 11cm
Now we are able to calculate the area of the blue square:
A = s² = (11cm)² = 121cm²
Best regards from Germany
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121
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@ 7:52 is a classic example of throwing the baby out with the bathwater by the overzealous physicists in search of dark matter . ...just sayin! 🙂
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121.
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My way of solution ▶
Agreen= 15 cm²
x= 5 cm
⇒
15= 5y
y= 3 cm
[AB]= 10 cm
The length [AB] is the hypothenuse of a right triangle:
the length of the blue square a
the base of this rectangular = a-5
the height of this triangular= a-3
By applyling the Pythagorean theorem we get:
10²= (a-3)²+(a-5)²
100= a²-6a+9+a²-10a+25
2a²-16a-66=0
a²-8a-33=0
Δ= 196
√Δ= 14
⇒
a₁= (8+14)/2
a₁= 11 cm
a₂= (8-14)/2
a₂= -3 < 0 ❌
⇒
a= 11 cm
Asquare= 11²
Asquare= 121 cm²
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121
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121.
121
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