Find the area of rhombus whose perimeter is 80m and its one diagonal
Answers
Correct Question:
Find the area of rhombus whose perimeter is 80 m and its one diagonal is 24 m.
Answer:
The area of rhombus is 384 m².
Step-by-step explanation:
NOTE: Refer to the attachment for the diagram.
In figure, □ABCD is rhombus.
Diagonals AC & BD intersect at point O.
AC = 24 m - - [ Given ]
P ( □ABCD ) = 80 m - - [ Given ]
Now, we know that,
Perimeter of rhombus = 4 × Sides
⇒ P ( ABCD ) = 4 × AD
⇒ 80 = 4 × AD
⇒ AD = 80 ÷ 4
⇒ AD = 20 m - - ( 1 )
We know that,
Diagonals of a rhombus bisect each other.
∴ AO = OC = ½ × AC - - ( 2 )
BO = OD = ½ × BD - - ( 3 )
Now,
AO = ½ × AC - - [ From ( 2 ) ]
⇒ AO = ½ × 24
⇒ AO = 12 m - - ( 4 )
Now, we know that,
Diagonals of a rhombus are perpendicular bisectors of each other.
∴ In △AOD, ∠AOD = 90°
∴ ( AD )² = ( AO )² + ( OD )² - - [ Pythagors theorem ]
⇒ ( 20 )² = ( 12 )² + ( OD )² - - [ From ( 1 ) & ( 4 ) ]
⇒ 400 = 144 + OD²
⇒ OD² = 400 - 144
⇒ OD² = 256
⇒ OD = √(256) - - [ Taking square roots ]
⇒ OD = 16 m
Now,
OD = ½ × BD - - [ From ( 3 ) ]
⇒ BD = 2 × OD
⇒ BD = 2 × 16
⇒ BD = 32 m - - ( 5 )
Now, we know that,
Area of rhombus = Product of diagonals / 2
⇒ A ( □ABCD ) = AC × BD / 2
⇒ A ( □ABCD ) = 24 × 32 ÷ 2 - - [ Given & ( 5 ) ]
⇒ A ( □ABCD ) = 24 × 16
⇒ A ( □ABCD ) = 8 × 3 × 8 × 2
⇒ A ( □ABCD ) = 64 × 2 × 3
⇒ A ( □ABCD ) = 128 × 3
⇒ A ( □ABCD ) = 384 m²
∴ The area of rhombus is 384 m².
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Additional Information:
1. Rhombus:
A quadrilateral with all its four sides of equal measures is called as rhombus.
2. Properties of Rhombus:
1. All sides are congruent.
2. Opposite angles are congruent.
3. Diagonals bisect each other.
4. Diagonals are perpendicular bisectors of each other.
