The area of a rhombus is 2016 sq cm and its side is 65 cm. The lengths of the diagonals (in cm) respectively are
Answers
Step-by-step explanation:
Given :-
The area of a rhombus is 2016 sq.cm
Side of the rhombus is 65 cm
To find :-
The lengths of the diagonals
Solution :-
Let the diagonals of the rhombus be
d₁ cm and d₂ cm
We know that
Area of a rhombus = ½ d₁d₂ sq.units
According to the given problem
Area of the rhombus = 2016 sq.cm
=> ½ d₁d₂ = 2016
=> d₁d₂ = 2016×2
=> d₁d₂ = 4032 --------(1)
Given that
Side of the rhombus = 65 cm
We know that
Diagonals bisects at 90° each other in a rhombus (or) Diagonals bisect each other perpendicularly.
Two diagonals divide the rhombus into four right angled triangles.
Consider a rhombus ABCD,
In ∆AOB
By Pythagoras Theorem
=> AB² = AO² + OB² --------(2)
Where, AC = d₂
=> AO = OC
=> AC = 2AO = 2OC
=> d₂ = 2AO
=> AO = d₂/2
and
d₁= BD
=> BD = BO+OD
=> BD = 2BO = 2OD
=> d₁ = 2OB
=> OB = d₁/2
Now,
Equation (2) becomes
=> 65² = (d₂/2)²+(d₁/2)²
=> 4225 = (d₁²/4)+(d₂²/4)
=> 4225 = (d₁²+d₂²)/4
=> d₁²+d₂² = 4225×4
=> d₁²+d₂² = 16900 cm---------(3)
We know that
(a+b)² = a²+2ab+b²
Now,
(d₁+d₂)² = d₁²+d₂²+2d₁d₂
=> (d₁+d₂)² = 16900+2(4032)
=> (d₁+d₂)² = 16900+8064
=> (d₁+d₂)² = 24964 cm --------(4)
=> d₁+d₂ = ±√24964
=> d₁+d₂ = ±158
Since, the length of the diagonal can't be negative .
=> d₁+d₂ = 158 cm ----------(5)
We know that
(a-b)² = (a+b)²-4ab
Now,
=> (d₁-d₂)² = (d₁+d₂)²-4d₁d₂
=> (d₁-d₂)² = 24964-4(4032)
=> (d₁-d₂)² = 24964-16128
=> (d₁-d₂)² = 8836
=> d₁-d₂ = ±√8836
=> d₁-d₂ = ±94
=> d₁-d₂ = 94 cm ---------(6)
Since ,The length of the diagonal can't be negative .
On adding (5) and (6) then
d₁+d₂ = 158
d₁-d₂ = 94
(+)
___________
2d₁+0 = 252
___________
=> 2d₁ = 252
=> d₁ = 252/2
=> d₁ = 126 cm
On substituting the value of d₁ in (5) then
126+ d₂ = 158
=> d₂ = 158-126
=> d₂ = 32 cm
Therefore, d₁ = 126 cm and d₂ = 32 cm
Answer:-
The lengths of the two diagonals of the given rhombus are 126 cm and 32 cm
Check:-
We have ,
d₁ = 126 cm and d₂ = 32 cm
Area of a rhombus = ½ d₁d₂ sq.cm
=> A = ½×126×32 sq.cm
=> A = 63×32 sq.cm
Therefore, A = 2016 sq.cm
and
AB² = AO² + OB²
=> AB² = (126/2)²+(32/2)²
=> AB² = 63²+16²
=> AB² = 3969+256
=> AB² = 4225
=> AB = ±√4225
=> AB = ±65
=> AB = 65 cm
Side of the rhombus is 65 cm
Verified the given relations in the given problem.
Used formulae:-
→Area of a rhombus = ½ d₁d₂ sq.units
- d₁ and d₂ are the diagonals of the rhombus
→ (a+b)² = a²+2ab+b²
→ (a-b)² = (a+b)²-4ab
Used Theorem:-
Pythagoras Theorem:-
"In a right angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides".
Used Properties :-
→ Diagonals bisects at 90° each other in a rhombus (or) Diagonals bisect each other perpendicularly.
→ Two diagonals divide the rhombus into four right angled triangles.
Step-by-step explanation:
Given :-
The area of a rhombus is 2016 sq.cm
Side of the rhombus is 65 cm
To find :-
The lengths of the diagonals
Solution :-
Let the diagonals of the rhombus be
d₁ cm and d₂ cm
We know that
Area of a rhombus = ½ d₁d₂ sq.units
According to the given problem
Area of the rhombus = 2016 sq.cm
=> ½ d₁d₂ = 2016
=> d₁d₂ = 2016×2
=> d₁d₂ = 4032 --------(1)
Given that
Side of the rhombus = 65 cm
We know that
Diagonals bisects at 90° each other in a rhombus (or) Diagonals bisect each other perpendicularly.
Two diagonals divide the rhombus into four right angled triangles.
Consider a rhombus ABCD,
In ∆AOB
By Pythagoras Theorem
=> AB² = AO² + OB² --------(2)
Where, AC = d₂
=> AO = OC
=> AC = 2AO = 2OC
=> d₂ = 2AO
=> AO = d₂/2
and
d₁= BD
=> BD = BO+OD
=> BD = 2BO = 2OD
=> d₁ = 2OB
=> OB = d₁/2
Now,
Equation (2) becomes
=> 65² = (d₂/2)²+(d₁/2)²
=> 4225 = (d₁²/4)+(d₂²/4)
=> 4225 = (d₁²+d₂²)/4
=> d₁²+d₂² = 4225×4
=> d₁²+d₂² = 16900 cm---------(3)
We know that
(a+b)² = a²+2ab+b²
Now,
(d₁+d₂)² = d₁²+d₂²+2d₁d₂
=> (d₁+d₂)² = 16900+2(4032)
=> (d₁+d₂)² = 16900+8064
=> (d₁+d₂)² = 24964 cm --------(4)
=> d₁+d₂ = ±√24964
=> d₁+d₂ = ±158
Since, the length of the diagonal can't be negative .
=> d₁+d₂ = 158 cm ----------(5)
We know that
(a-b)² = (a+b)²-4ab
Now,
=> (d₁-d₂)² = (d₁+d₂)²-4d₁d₂
=> (d₁-d₂)² = 24964-4(4032)
=> (d₁-d₂)² = 24964-16128
=> (d₁-d₂)² = 8836
=> d₁-d₂ = ±√8836
=> d₁-d₂ = ±94
=> d₁-d₂ = 94 cm ---------(6)
Since ,The length of the diagonal can't be negative .
On adding (5) and (6) then
d₁+d₂ = 158
d₁-d₂ = 94
(+)
___________
2d₁+0 = 252
___________
=> 2d₁ = 252
=> d₁ = 252/2
=> d₁ = 126 cm
On substituting the value of d₁ in (5) then
126+ d₂ = 158
=> d₂ = 158-126
=> d₂ = 32 cm
Therefore, d₁ = 126 cm and d₂ = 32 cm
Answer:-
The lengths of the two diagonals of the given rhombus are 126 cm and 32 cm
Check:-
We have ,
d₁ = 126 cm and d₂ = 32 cm
Area of a rhombus = ½ d₁d₂ sq.cm
=> A = ½×126×32 sq.cm
=> A = 63×32 sq.cm
Therefore, A = 2016 sq.cm
and
AB² = AO² + OB²
=> AB² = (126/2)²+(32/2)²
=> AB² = 63²+16²
=> AB² = 3969+256
=> AB² = 4225
=> AB = ±√4225
=> AB = ±65
=> AB = 65 cm
Side of the rhombus is 65 cm
Verified the given relations in the given problem.
Used formulae:-
→Area of a rhombus = ½ d₁d₂ sq.units
d₁ and d₂ are the diagonals of the rhombus
→ (a+b)² = a²+2ab+b²
→ (a-b)² = (a+b)²-4ab
Used Theorem:-
Pythagoras Theorem:-
"In a right angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides".