Math, asked by ayushdeep4231, 9 months ago

Brainly ki janta, my next question is what would be portion covered by that shaded region​

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Answers

Answered by RvChaudharY50
10

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.

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Answered by Anonymous
1

Answer :D

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.

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