Question

A kite of area 40m has one diagonal 2m longer than the other. find the lengths of the diagonals

Answer 1

A = 40 m²
d1 = d2 + 2

A = 40
1/2 × d1 × d2 = 40
1/2 × (d2 + 2) × d2 = 40
d2² + 2d2 = 40 × 2
d2² + 2d2 = 80
d2² + 2d2 - 80 = 0
d2² - 8d2 + 10d2 - 80 = 0
d2(d2 - 8) + 10(d2 - 8) = 0
(d2 - 8)(d2 + 10) = 0
d2 = 8 m (acceptable)
d2 = -10 (not acceptable)

d1 = d2 + 2
d1 = 8 + 2
d1 = 10 m

Answer 2

Answer:

The length of the diagonals of kite are 8 m and 10 m.

Step-by-step explanation:

Given : A kite of area 40 m has one diagonal 2 m longer than the other.

To find : The lengths of the diagonals?

Solution :

The area of the kite is $A=\frac{1}{2}\times d_1\times d_2$

Where, A is the area

$d_1$ is the first diagonal

$d_2$ is the second diagonal

According to question,

$d_2=d_1+2$

Substitute the values in the formula,

$40=\frac{1}{2}\times d_1\times (d_1+2)$

$40=\frac{1}{2}\times d_1^2+2d_1$

$80=d_1^2+2d_1$

$d_1^2+2d_1-80=0$

Applying middle term split,

$d_1^2+10d_1-8d_1-80=0$

$d_1(d_1+10)-8(d_1+10)=0$  

$(d_1-8)(d_1+10)=0$  

$d_1=8,d_1=-10$  

Reject negative value.

So, $d_1=8$  

$d_2=8+2=10$  

Therefore, The length of the diagonals of kite are 8 m and 10 m.