Question

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Answer 1

Answer:

Step-by-step explanation:

Let the length of smaller square side be x . Note that Line AB of triangle is just equal to length of diagonal of small square . So

$AB= \sqrt{ {x}^{2} + {x}^{2} } = 2 \sqrt{x}$

Now using Pythogoras Theorem length of Side of big square will be

$= 2 \sqrt{x}$

And length of diagonal of big square will be

$= 2 \sqrt{x} ( \sqrt{2} ) = 2 \sqrt{2} x$

Length of green line say CP is equal to semi diagonal of big square so

$CP = \frac{2 \sqrt{2}x }{2} = x\sqrt{2}$

Now observe in ∆CDP

Angle COD is right angle ,

PD is length of square = x

$CP = x\sqrt{2}$

So use Pythogoras Theorem

$CD^2 = x^2+( x\sqrt{2} )^2\\ \\ CD=\sqrt{3x^2}=x\sqrt3$

Now finally

Area of triangle ABC = 1/2 × base × height

$4 = \frac{1}{2} \times AB \times CD \\ \\ \implies \: 4 = \frac{1}{2} \times \sqrt{2} x \times √3x \\ \\ \implies \: {x}^{2} = \frac{4√2}{ \sqrt{3} } \\ \\ \implies \: x = \sqrt{4√2\over \sqrt{3} }$

Answer 2

Step-by-step explanation:

given :

• yellow triangle's vertices are at the centers of the three squares

• two identical small squares and a larger
• one:

• yellow triangle's area is 4

to find :

• what's the length of the smaller square's side = ?

• what's the length of the smaller square's side = ?

according to QUESTION :

• area of AB = √x + x = 2√x

Pythogoras Theorem length of square

• = 2√x

• length of square = 2√x ( √2) = 2√2x

• centre perimeter of green line is a diagonal

• of a square is :

• centre perimeter = 2√2x /2 = x√2

• perimeter of diagonal of square = x

• centre perimeter = x√2

• so , center diagonal = x + (x√2)²

• center diagonal = √3x =x√3

• traingle area = 1/2 b×h

• = 1/2 area base × center diagonal

• = 1/2 × √2 × √3

• = 4 √4√2 /√3