A kite of area 40m squared has one diagonal 2m longer than the other . Find the lengths of the diagonals
Answers
Answer :
- The length of the Diagonals of kite are 8m and 10m
Given :
- A kite of area is 40m squared has one Diagonal 2m longer than the the other
To find :
- length of the Diagonals
Solution:
As we know that ,
- Area of kite = 1/2 × d₁ × d₂
where , d₁ is first diagonal , d₂ is second Diagonal and area of kite is 40
d₂ = d₁ + 2 (from Given Question)
Now putting the value :
↝ 40 = 1/2 × d₁ × d₁ + 2
↝ 40 = 1/2 × d₁² + 2 d₁
↝ 80= d₁² + 2 d₁
↝ d₁² + 2 d₁ - 80 = 0
Now apply the middle term we get,
↝ d₁² + 10d₁ - 8d₁ - 80 = 0
↝ d₁(d₁+ 10) - 8(d₁+ 10) = 0
↝ (d₁ - 8) (d₁+ 10) = 0
↝ d₁ = 8 , d₁ = - 10
Then ,
d₁ = 8
d₂ = 8 + 2 = 10
- d₁ = 8
- d₂ = 10
Hence the length of the Diagonals of kite are 8m and 10m
Answer:
Answer :
The length of the Diagonals of kite are 8m and 10m
Given :
A kite of area is 40m squared has one Diagonal 2m longer than the the other
To find :
length of the Diagonals
Solution:
As we know that ,
Area of kite = 1/2 × d₁ × d₂
where , d₁ is first diagonal , d₂ is second Diagonal and area of kite is 40
d₂ = d₁ + 2 (from Given Question)
Now putting the value :
↝ 40 = 1/2 × d₁ × d₁ + 2
↝ 40 = 1/2 × d₁² + 2 d₁
↝ 80= d₁² + 2 d₁
↝ d₁² + 2 d₁ - 80 = 0
Now apply the middle term we get,
↝ d₁² + 10d₁ - 8d₁ - 80 = 0
↝ d₁(d₁+ 10) - 8(d₁+ 10) = 0
↝ (d₁ - 8) (d₁+ 10) = 0
↝ d₁ = 8 , d₁ = - 10
Then ,
d₁ = 8
d₂ = 8 + 2 = 10
d₁ = 8
d₂ = 10
Hence the length of the Diagonals of kite are 8m and 10m
Step-by-step explanation:
Hope this answer will help you.✌️