@@egillandersson1780 Same for me - except that it took much longer than 20 seconds! I must be getting slow in my old age. It took me a long time to realise that the side of the square was equal to twice the length of half of it!
I had one of those rare 'outside the box' moments with this question, where the Universe delivered an insight to my dull mathematical mind. Up to 3:11 where Mr. Premath finds MC=x√5 my process was the same. But then I extended CM and BA, to intersect off to the left at a point K. By showing △CDM ≅ △KAM by ASA, we see that area □ABCD = area △BCK Then calculating the area of △BCK two different ways and setting them equal reveals the value of x, as follows: (½)(2⋅CM)(PB) = (½)(2⋅AB)(BC) (½)(2x√5)(4) = (½)(4x)(2x) etc.
I think if we denote the side length of square as X by similar triangles we get X/4=cm/X From Pythagoras theorem cm=sqrt((X/2)^2+x^2)=X*√5/2 => X/4=√5/2 X=2√5 so area is 20 you guys are totally right
Call the square's sides 2x. The area will be 4x^2 BPC is similar to MDC. (2x)/4 = (MC)/(2x) 4*(MC) = 4x^2 Therefore, MC = x^2 Sides are now: x. 2x, x^2 (PC), 4, 2x (2x)/(x) = (4)/(PC) 4x = (2x)*(PC), so PC = 2. Ref BPC: 4^2 + 2^2 = 4x^2 16 + 4 = 4x^2 20 = 4x^2 which is the square's area. I used 2x as the square's sides rather than plain x. This was becaise there was an early warning that half sides were involved and I wanted to avoid too many fractions.
Using similar triangle approaches, we already know that PC should be half of 4 = 2. Since triangle CDM shows that DM = x and CD = 2x. The area of the square with length 2x is 4x^2. From Pythagoras, 4x^2 = 4^2 + 2^2 =20 (QED)
There is a faster way to solve this. Since we know MC= x√5 we can immediately substitute it in the Similarity Formula to get x= √5 (x√5)/2x= 2x/4 √5/2= x/2 x= √5 Since we know that the square's area is 4x² 4(√5)² 4(5) A= 20 units²
*Another solution:* Draw the segment BM. The triangle ∆BMC is isosceles, since BM = MC.Let h be the height drawn from vertex M in ∆BMC, then we can find the area of this triangle as follows: h × BC/2 = BP×MC/2 *h × BC= 4MC (1).* By Pythagoras in ∆GCD, we have: MC² = MD² + DC² = (DC/2)² + DC² MC² = DC²(1/4 + 1) = DC²/4 × 5 MC = √5DC/2, being DC = BC, then MC = √5BC/2. Substituting in (1): h × BC= 4 √5BC/2 h = 2√5 = √20. Now, h= AB=DC=AD=BC. Thus, the area of the square is given by h², i.e,(√20)² = *20.*
A very good solution. Another alternate approach will be: After determining that MC = x. Sqrt5, Join points M and B. Then the area of triangle MBC can be calculated in 2 ways as 1/2.MC.4 , and 1/2.2x.2x Equating these 2 values, we will get 4.MC = 4.xsq = area of the square ABCD!
Let's find the area: . .. ... .... ..... First of all we apply the Pythagorean theorem to the right triangle CDM. With s being the side length of the square we obtain: CM² = CD² + DM² = CD² + (AD/2)² = s² + (s/2)² = s² + s²/4 = 4s²/4 + s²/4 = 5s²/4 ⇒ CM = √(5s²/4) = (√5/2)s Since ∠BPC=∠CDM=90° and ∠CBP=∠DCM, we know that the triangles BCP and CDM are similar. So we can conclude: BC/BP = CM/CD s/4 = (√5/2)s/s s/4 = √5/2 ⇒ s = 4*√5/2 = 2√5 Now we are able to calculate the area of the blue square: A(ABCD) = s² = (2√5)² = 20 Best regards from Germany
Разрезаем по линии МС, поворачиваем ▲CDM на 180° вокруг точки М, складываем MD с АМ. В результате из малого треугольника и трапеции получается большой треугольник, подобный малому (вдвое больше), с площадью, равной площади квадрата. РВ является его высотой. Точку, в которую перешла т. С, обозначим за Е. ▲ВСЕ прямоугольный, потому площадь также равна произведению катетов. Сторону квадрата за х, тогда х*2х=2х²=4*СЕ/2=2СЕ⇔х²=СЕ. Но СЕ²=х²+(2х)², как гипотенуза, получаем уравнение четвёртой степени: х⁴=х²+4х²=5х². Отрицательные и комплексные корни искать не нужно, потому смело сокращаем: х²=5, откуда х=√5. Можно решать и через пропорцию, заметив, что катеты 1 к 2, гипотенуза кратна √5, а высота, соответственно, должна быть кратна 1/√5. 4х/√5=4, откуда х=√5.
*Solution:* In ∆MCD: Let ∠DCM = α. Hence, tg α= MD/DC = MD/2MD= 1/2. In ∆BCP: tg (90 - α) = 4/PC →1/tg α = 4/PC 2 = 4/PC → PC = 2. By Pythagoras, we have: BC² = 4² + 2² = 20, this tells us that the area of the square is 20.
We use an orthonormal center D and first axis (DC) and be c the side length of the square. We have C(c; 0) and M(0; c/2). VectorCM(c/2; -c) is colinear to VectorU(-2; 1). The equation of (CM) is: (1).(x - c) - (-2).(y) = 0 or x + 2.y - c = 0 The distance from B(c; c) to (CM) is abs(c + 2.c - c)/sqrt((2^2) + (1^2)) = (2.c)/(sqrt(5)), and this distance is also 4, so we have (2.c)/(sqrt(5)) = 4, so c = 2.sqrt(5) and the area of the square is c^2 = 4.5 = 20. (Sorry, I prefer analytic geometry each time it is possible!)
Let AM = DM = x. By the Segment Addition Postulate, AD = AM + DM = x + x = 2x. By definition of squares, BC = CD = 2x. Use the Pythagorean Theorem on △CDM. a² + b² = c² x² + (2x)² = (CM)² x² + 4x² = (CM)² 5x² = (CM)² CM = √(5x²) = x√5 The root cannot be negative because x must be positive. Let α & β be the measures of complementary angles. Let m∠CBP = α. Then, m∠BCP = β by definition of acute angles in a right triangle. Because ∠BCD is a right angle by definition of squares, m∠DCM = α & m∠CMD = β. So, △BPC ~ △CDM by SAS. Define proportions. BP/CD = BC/CM 4/(2x) = (2x)/(x√5) 2x(2x) = 4(x√5) 4x² = 4x√5 4x² - 4x√5 = 0 4x(x - √5) = 0 4x = 0 or x - √5 = 0 x = 0 x = √5 But if x = 0, the diagram would not exist (It would also contradict the statement made earlier about x). So, x = √5. The side length of square ABCD is 2x = 2(√5) = 2√5 u. Find the area of square ABCD. A = s² = (2√5)² = 2² * (√5)² = 4 * 5 = 20 So, the area of the blue square is 20 square units.
Potremmo risolvere una equazione che mette in relazione la superficie del quadrato calcolata utilizzando la formula L*L ovvero considerando L=2x S=4x² e la somma delle superfici dei 3 triangoli DCM ABM MCB, inoltre considerando che MC si può ricavare col teorema di pitagora applicato al triangolo BCM possiamo sviluppare il nostro sistema risolutivo: La superficie del triangolo MBA = S1 = 2x*x/2 = x² La superficie del triangolo DCM = S2 = 2x*x/2 = x² La superficie del triangolo CBM = S3 = MC*4/2 = 2*MC dove MC è la base del triangolo MCB di altezza 4 con pitagora determino il lato MC MC=√(x²+(2x)²) = x√5 Quindi: S=S1+S2+S3 S=x²+x²+2*x√5 S=2x²+(2√5)x Risolvo il sistema: 2x²+(2√5)x=4x² -2x²+(2√5)x=0 Questa è una quadratica pertanto: x1,2=(-b+/-√(b²-4ac))/2a le soluzioni sono x1=0 x2=√5 scartiamo x1 pertanto la soluzione è x=√5 sostituendo a S= 4x² otteniamo S=4*(√5)²=20 La superficie del quadrato è 20
Solution: Since M is the midpoint of AD, let's label DM = AM = a Therefore AB = BC = CD = DA = 2a Applying Pythagorean Theorem in ∆ CDM CD² + DM² = CM² (2a)² + (a)² = CM² CM² = 5a² CM = a√5 ∆ CDM and ∆ BCP are similar by AA, and, like that, we have proportions Hence: BC/BP = CM/CD 2a/4 = a√5/2a a/2 = √5/2 a = 2√5/2 a = √5 If a = √5, the blue square side length is 2√5 Blue Square Area = (2√5)² Blue Square Area = 20 Square Units ✅
A right Triangle with one side equal 4, dictates that the other leg is 3 and the hypotunuse is 5. The square has all equal sides of 5 . The area is teh product of the sidea 5 X 5 = 25
MY RESOLUTION PROPOSAL : 01) Square Side = 2X lin un 02) AM = MD = X lin un 03) (MCD) Angle Tangent = 2X / X = 2 04) (CMD) Angle Tangent = X / 2X = 1 / 2 05) Angle (CBP) = Angle (DCM) 06) Angle (CMD) = Angle (BCM) 07) Angle (BCM) is a Complemantary Angle with Angle (DCM). DEFINITION : " Two angles are Complementary if the Sum of the Angles is equal to 90º (Ninety Degrees). " 08) Wich means that Triangle [CDM] is similar to Triangle [CBP] 09) tan(CBP) = 1 / 2 ; tan(CBP) = PC / 4 10) 1 / 2 = PC / 4 ; PC = 2 lin un 11) BC^2 = 4^2 + 2^2 ; BC^2 = (16 + 4) ; BC^2 = 20 Therefore, MY BEST ANSWER IS : Blue Square Area equal 20 Square Units.
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By Similarity of ∆CPB and ∆MDC, PC = 4/2=2 and BC² = 20= Area
Same for me : solvable this way in less then 20 sec
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@@egillandersson1780 Same for me - except that it took much longer than 20 seconds! I must be getting slow in my old age. It took me a long time to realise that the side of the square was equal to twice the length of half of it!
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Way too complicated! MDC and BPC are similar, the ratio of the catheti is 1:2, so PC is 2. 4²+2² = 20 = (BC)² , which is the area we were searching.
So much better! 😊
Similar triangles DMC & PCB.
X / 2X = PC / 4.
PC = 4X / 2X = 2.
CB^2 = 4^2 + PC^2.
CB^2 = 16 + 4 = 20.
CB^2 = area of square.
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Thanks to you sir!
And to the other solutions offered in the posts
I had one of those rare 'outside the box' moments with this question, where the Universe delivered an insight to my dull mathematical mind.
Up to 3:11 where Mr. Premath finds MC=x√5 my process was the same. But then I extended CM and BA, to intersect off to the left at a point K.
By showing △CDM ≅ △KAM by ASA, we see that area □ABCD = area △BCK
Then calculating the area of △BCK two different ways and setting them equal reveals the value of x, as follows:
(½)(2⋅CM)(PB) = (½)(2⋅AB)(BC)
(½)(2x√5)(4) = (½)(4x)(2x)
etc.
Use similar triangles to figure out CM=4+4/(2^2)=5 and Blue Square Area BP×CM=4×5=20
Yep - I did that in my head in about 30s
I think if we denote the side length of square as X by similar triangles we get
X/4=cm/X
From Pythagoras theorem cm=sqrt((X/2)^2+x^2)=X*√5/2
=> X/4=√5/2
X=2√5 so area is 20 you guys are totally right
It's not necessary to consider the side of the square,
SquareArea=2×ΔBCMarea=BP×CM
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Call the square's sides 2x. The area will be 4x^2
BPC is similar to MDC.
(2x)/4 = (MC)/(2x)
4*(MC) = 4x^2
Therefore, MC = x^2
Sides are now:
x. 2x, x^2
(PC), 4, 2x
(2x)/(x) = (4)/(PC)
4x = (2x)*(PC), so PC = 2.
Ref BPC: 4^2 + 2^2 = 4x^2
16 + 4 = 4x^2
20 = 4x^2 which is the square's area.
I used 2x as the square's sides rather than plain x. This was becaise there was an early warning that half sides were involved and I wanted to avoid too many fractions.
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Exercício interessante
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Using similar triangle approaches, we already know that PC should be half of 4 = 2. Since triangle CDM shows that DM = x and CD = 2x.
The area of the square with length 2x is 4x^2.
From Pythagoras, 4x^2 = 4^2 + 2^2 =20 (QED)
There is a faster way to solve this.
Since we know MC= x√5 we can immediately substitute it in the Similarity Formula to get x= √5
(x√5)/2x= 2x/4
√5/2= x/2
x= √5
Since we know that the square's area is 4x²
4(√5)²
4(5)
A= 20 units²
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Similarity of right triangles:
4/s = s / √[s²+(½s)²]
s² = 4 √[5/4 s²] = 4 √5/2 s
s² = (2√5)² = 20 cm² ( Solved √ )
Let 2X - side of the square,
Then, for similar triangles:
2X / X = 4 / ((4X^2-16)^.5)
X = 5^0.5, S = 20
*Another solution:*
Draw the segment BM. The triangle ∆BMC is isosceles, since BM = MC.Let h be the height drawn from vertex M in ∆BMC, then we can find the area of this triangle as follows:
h × BC/2 = BP×MC/2
*h × BC= 4MC (1).*
By Pythagoras in ∆GCD, we have:
MC² = MD² + DC² = (DC/2)² + DC²
MC² = DC²(1/4 + 1) = DC²/4 × 5
MC = √5DC/2, being DC = BC, then MC = √5BC/2. Substituting in (1):
h × BC= 4 √5BC/2
h = 2√5 = √20. Now, h= AB=DC=AD=BC. Thus, the area of the square is given by h², i.e,(√20)² = *20.*
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s^2 = 4^2 + 2^2
s^2 = square area = 16 + 4 = 20 square units
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A = s² = (4/cos[atan(1/2)] )²
A = 20 cm² ( Solved √ )
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A very good solution.
Another alternate approach will be:
After determining that MC = x. Sqrt5,
Join points M and B.
Then the area of triangle MBC can be calculated in 2 ways as
1/2.MC.4 , and 1/2.2x.2x
Equating these 2 values, we will get 4.MC = 4.xsq = area of the square ABCD!
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Let's find the area:
.
..
...
....
.....
First of all we apply the Pythagorean theorem to the right triangle CDM. With s being the side length of the square we obtain:
CM² = CD² + DM² = CD² + (AD/2)² = s² + (s/2)² = s² + s²/4 = 4s²/4 + s²/4 = 5s²/4 ⇒ CM = √(5s²/4) = (√5/2)s
Since ∠BPC=∠CDM=90° and ∠CBP=∠DCM, we know that the triangles BCP and CDM are similar. So we can conclude:
BC/BP = CM/CD
s/4 = (√5/2)s/s
s/4 = √5/2
⇒ s = 4*√5/2 = 2√5
Now we are able to calculate the area of the blue square:
A(ABCD) = s² = (2√5)² = 20
Best regards from Germany
Still keen eye on you 🤨
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Разрезаем по линии МС, поворачиваем ▲CDM на 180° вокруг точки М, складываем MD с АМ. В результате из малого треугольника и трапеции получается большой треугольник, подобный малому (вдвое больше), с площадью, равной площади квадрата. РВ является его высотой. Точку, в которую перешла т. С, обозначим за Е. ▲ВСЕ прямоугольный, потому площадь также равна произведению катетов. Сторону квадрата за х, тогда х*2х=2х²=4*СЕ/2=2СЕ⇔х²=СЕ. Но СЕ²=х²+(2х)², как гипотенуза, получаем уравнение четвёртой степени: х⁴=х²+4х²=5х². Отрицательные и комплексные корни искать не нужно, потому смело сокращаем: х²=5, откуда х=√5. Можно решать и через пропорцию, заметив, что катеты 1 к 2, гипотенуза кратна √5, а высота, соответственно, должна быть кратна 1/√5. 4х/√5=4, откуда х=√5.
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√(5l^2/4-16)+√(l^2-16)=√5l/2...l^2=20
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A = 4 A₁ + A₂ = 4 (½b.h) + (s₂)²
A = 4 (½*4*2) + (½4)² = 16 + 4
A = 20 cm² ( Solved √ )
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S=20
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*Solution:*
In ∆MCD:
Let ∠DCM = α. Hence,
tg α= MD/DC = MD/2MD= 1/2.
In ∆BCP:
tg (90 - α) = 4/PC →1/tg α = 4/PC
2 = 4/PC → PC = 2. By Pythagoras, we have:
BC² = 4² + 2² = 20, this tells us that the area of the square is 20.
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If x^2 = x√5, why not just divide both sides by x to get √5?
We have two similar right triangles
If the square side length = 2x
(√5)x/2x = 2x/4
4x^2 = 4(√5)x
2x = 2√5
Square area = (2√5)^2 = 20
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@ Thank YOU 🙂
Right triangle BCP
h/b = 1/2 = h/4 --> h= 2 cm
Pytagorean theorem:
s² = b²+h² = 4²+2² = 20cm² (Solved √)
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20
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Razón entre catetos =1/2---> PC=4/2=2---> BC=√20---> Área ABCD =20 u².
Gracias y saludos
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MDC∞BPC MD : DC = 1 : 2 PC/4=1/2 PC=2
BC=√[2²+4²]=√20
Blue Square area = √20*√20 = 20
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We use an orthonormal center D and first axis (DC) and be c the side length of the square.
We have C(c; 0) and M(0; c/2). VectorCM(c/2; -c) is colinear to VectorU(-2; 1). The equation of (CM) is: (1).(x - c) - (-2).(y) = 0 or x + 2.y - c = 0
The distance from B(c; c) to (CM) is abs(c + 2.c - c)/sqrt((2^2) + (1^2)) = (2.c)/(sqrt(5)), and this distance is also 4, so we have (2.c)/(sqrt(5)) = 4,
so c = 2.sqrt(5) and the area of the square is c^2 = 4.5 = 20.
(Sorry, I prefer analytic geometry each time it is possible!)
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Sir,
May I write
x^2=x√5
> x =√5(dividing both sides by x)
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x.2X/2+x.2X/2+4*(5^0.5)x/2=2x.2x
x^2+x^2+2(5^0.5)x=4x^2
5^0.5=x
4x^2=4*5=20
a = lado del cuadrado
tg A = 1/2
También cos A = 4/a
Entonces a = 4 sec A
a^2 = 16 (sec A)^2 = 16 [1+(tgA)^2]
a^2=16 [1 + 1/4] = 20
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Let AM = DM = x. By the Segment Addition Postulate, AD = AM + DM = x + x = 2x.
By definition of squares, BC = CD = 2x. Use the Pythagorean Theorem on △CDM.
a² + b² = c²
x² + (2x)² = (CM)²
x² + 4x² = (CM)²
5x² = (CM)²
CM = √(5x²)
= x√5
The root cannot be negative because x must be positive.
Let α & β be the measures of complementary angles.
Let m∠CBP = α.
Then, m∠BCP = β by definition of acute angles in a right triangle.
Because ∠BCD is a right angle by definition of squares, m∠DCM = α & m∠CMD = β.
So, △BPC ~ △CDM by SAS. Define proportions.
BP/CD = BC/CM
4/(2x) = (2x)/(x√5)
2x(2x) = 4(x√5)
4x² = 4x√5
4x² - 4x√5 = 0
4x(x - √5) = 0
4x = 0 or x - √5 = 0
x = 0 x = √5
But if x = 0, the diagram would not exist (It would also contradict the statement made earlier about x). So, x = √5.
The side length of square ABCD is 2x = 2(√5) = 2√5 u.
Find the area of square ABCD.
A = s²
= (2√5)²
= 2² * (√5)²
= 4 * 5
= 20
So, the area of the blue square is 20 square units.
@ 6:47 , 🤔 ... the world needs more ideas and solutions , not fewer! Rejection is a miserable creature. 😊
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شكرا لكم على المجهودات
يمكن استعمال
S(ABCD) = 2S(ABM) + S(BMC)
........
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25
√5
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🎉Los triángulos PBC y DCM son semejantes, luego los ángulos PCB y CMD son iguales, cuya tangente es 2 A partir de eso, la solució es obvia😂
Potremmo risolvere una equazione che mette in relazione la superficie del quadrato calcolata utilizzando la formula L*L ovvero considerando L=2x S=4x² e la somma delle superfici dei 3 triangoli DCM ABM MCB, inoltre considerando che MC si può ricavare col teorema di pitagora applicato al triangolo BCM possiamo sviluppare il nostro sistema risolutivo:
La superficie del triangolo MBA = S1 = 2x*x/2 = x²
La superficie del triangolo DCM = S2 = 2x*x/2 = x²
La superficie del triangolo CBM = S3 = MC*4/2 = 2*MC dove MC è la base del triangolo MCB di altezza 4
con pitagora determino il lato MC
MC=√(x²+(2x)²) = x√5
Quindi:
S=S1+S2+S3
S=x²+x²+2*x√5
S=2x²+(2√5)x
Risolvo il sistema:
2x²+(2√5)x=4x²
-2x²+(2√5)x=0
Questa è una quadratica pertanto:
x1,2=(-b+/-√(b²-4ac))/2a
le soluzioni sono x1=0 x2=√5
scartiamo x1 pertanto la soluzione è x=√5
sostituendo a S= 4x² otteniamo
S=4*(√5)²=20
La superficie del quadrato è 20
Solution:
Since M is the midpoint of AD, let's label DM = AM = a
Therefore AB = BC = CD = DA = 2a
Applying Pythagorean Theorem in ∆ CDM
CD² + DM² = CM²
(2a)² + (a)² = CM²
CM² = 5a²
CM = a√5
∆ CDM and ∆ BCP are similar by AA, and, like that, we have proportions
Hence:
BC/BP = CM/CD
2a/4 = a√5/2a
a/2 = √5/2
a = 2√5/2
a = √5
If a = √5, the blue square side length is 2√5
Blue Square Area = (2√5)²
Blue Square Area = 20 Square Units ✅
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😂😂median formula pm=4 sind cd since pm90⁰to cd so md=1/2pm so md 2 by applying Pythagarous theorm to triangle pmd we get 20
A right Triangle with one side equal 4, dictates that the other leg is 3 and the hypotunuse is 5. The square has all equal sides of 5 . The area is teh product of the sidea 5 X 5 = 25
Who was your math teacher😂?
It’s a 1, 2, sqrt(5) right triangle.
more magic from the math wizard
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MY RESOLUTION PROPOSAL :
01) Square Side = 2X lin un
02) AM = MD = X lin un
03) (MCD) Angle Tangent = 2X / X = 2
04) (CMD) Angle Tangent = X / 2X = 1 / 2
05) Angle (CBP) = Angle (DCM)
06) Angle (CMD) = Angle (BCM)
07) Angle (BCM) is a Complemantary Angle with Angle (DCM). DEFINITION : " Two angles are Complementary if the Sum of the Angles is equal to 90º (Ninety Degrees). "
08) Wich means that Triangle [CDM] is similar to Triangle [CBP]
09) tan(CBP) = 1 / 2 ; tan(CBP) = PC / 4
10) 1 / 2 = PC / 4 ; PC = 2 lin un
11) BC^2 = 4^2 + 2^2 ; BC^2 = (16 + 4) ; BC^2 = 20
Therefore,
MY BEST ANSWER IS :
Blue Square Area equal 20 Square Units.
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