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