Can you find area of the Blue Square? | (Justify your answer) | #math #maths | #geometry

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Can you find area of the Blue Square? | (Justify your answer) | #math #maths | #geometry

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[zupitoxyt]

Here's another exciting and challenging question ahead.
Thanks professor

[harikatragadda]

By Similarity of ∆CPB and ∆MDC, PC = 4/2=2 and BC² = 20= Area

[andrepiotrowski]

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.

[FSIec-xx]

Use similar triangles to figure out CM=4+4/(2^2)=5 and Blue Square Area BP×CM=4×5=20

[alster]

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

[georgebliss]

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.

[marioalb]

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 √ )

[unknownidentity]

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

[imetroangola]

*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.*

[newme]

If 20 people like this comment I'll tell you the answer:)

[shashijoshi]

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!

[Waldlaeufer]

s^2 = 4^2 + 2^2
s^2 = square area = 16 + 4 = 20 square units

[marioalb]

Right triangle BCP
h/b = 1/2 = h/4 --> h= 2 cm
Pytagorean theorem:
s² = b²+h² = 4²+2² = 20cm² (Solved √)

[marioalb]

A = 4 A₁ + A₂ = 4 (½b.h) + (s₂)²
A = 4 (½*4*2) + (½4)² = 16 + 4
A = 20 cm² ( Solved √ )

[marioalb]

A = s² = (4/cos[atan(1/2)] )²
A = 20 cm² ( Solved √ )

[santiagoarosam]

Razón entre catetos =1/2---> PC=4/2=2---> BC=√20---> Área ABCD =20 u².
Gracias y saludos

[imetroangola]

*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.

[zawatsky]

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.

[MrPaulc]

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.

[PrithwirajSen-njqq]

Sir,
May I write
x^2=x√5
> x =√5(dividing both sides by x)