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What would be portion covered by that shaded region​

Answer 1

$\Large\boxed{\sf{\quad\dfrac{1}{16}\quad}}$

Let the length of each side of the smallest square in the fig. be $\sf{a.}$

We know that if each side of a square measures $\sf{x}$ then the length of each of its diagonal will be $\sf{x\sqrt2.}$ So,

• The length of each side of the medium square in the fig. will be equal to one of the diagonal of the smallest square, i.e., $\sf{a\sqrt2.}$

The length of each side of the largest square in the fig. will be equal to one of the diagonal of the medium square, i.e., $\sf{\left(a\sqrt2\right)\sqrt2=2a.}$

So the area of the largest square $\sf{=(2a)^2=4a^2.}$

The base of the shaded triangle can be the same as one side of the smallest square, i.e., $\sf{a.}$

Then the height of the shaded triangle is,

$\displaystyle\longrightarrow\sf{h=\dfrac{2a-a}{2}=\dfrac{a}{2}}$

due to symmetry in the fig.

Hence the area of the shaded region is,

$\displaystyle\longrightarrow\sf{A=\dfrac{1}{2}\times a\times \dfrac{a}{2}}$

$\displaystyle\longrightarrow\sf{A=\dfrac{a^2}{4}}$

So the ratio of the shaded portion to the whole portion is,

$\displaystyle\longrightarrow\sf{R=\dfrac{\left(\dfrac{a^2}{4}\right)}{4a^2}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=\dfrac{1}{16}}}}$

Answer 2

Solution :-

From image we have :-

→ ABCD is a Square .

→ EFGH are mid - Points of sides of square ABCD.

→ So, EFGH is also a square.

→ one More Square is Made inside EFGH with Mid - Points IJKL .

To Find :-

• ( Area of ∆EIJ . / Area of Square ABCD ).

Formula used :-

• Hypotenuse of Right angle isosceles ∆ = √2 * side.
• Area of Right angles isosceles ∆ = (1/2) * (side)²
• Each Angle of Square = 90°.
• Area of Square = side * side

Solution :-

Lets Assume That, Side of Each Square ABCD is 8 cm.

Than,

→ AB = BC = CD = DA = 8 cm.

So,

→ AE = AH = 4cm.

Now, in ∆AEH we have :-

→ AE = AH = 4cm.

→ Angle HAE = 90° . (Angle of Square) .

So , ∆AEH is a Right angle isosceles ∆ .

Hence,

→ Hypotenuse = √2 * 4 = 4√2 cm.

So, Each Side of Square EFGH is 4√2cm.

__________________________

Similarly , Now,

→ I is Mid - Point of EH , and J is Mid - Point of EF.

So,

→ EI = (4√2/2) = 2√2 cm.

→ EJ = (4√2/2) = 2√2 cm.

And,

→ Angle IEJ = 90° (Angle of Square).

Hence,

→ Area of Right angle isosceles ∆IEJ = (1/2) * (2√2)² = 4cm². --------------- Equation (1).

________________________

And,

→ Area of Square ABCD = (8)² = 64cm² ------ Equation(2)

So, From Equation (1) & (2), we get,

→ Area of Portion covered by that shaded region = ( Area of ∆EIJ . / Area of Square ABCD ).

→ Required Ratio = (4/64) = (1/16).

Hence, (1/16) Portion of Area is shaded.