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Q). ABCD is a kite. Write equation of its diagonals. Also find its area.

Dear Student,

Please find below the solution to the asked query:

From given diagram we get :

Coordinates of A ( 0 , - 2 )  , B ( 3 , 0 ) , C ( 0 , 4 )  and D ( - 3 , 0 )

We know equation of line passing through two points : ( y -  y₁ ) = $\left(\frac{y_2 - y_1}{x_2 - x_1}\right)$ ( x  - x₁ )

For diagonal , AC we get :  x₁ = 0  , y₁ = - 2 and   x₂ = 0  , y₂ = 4

So,

Equation of diagonal AC :

$$\begin{gathered} \left(y - \left( - 2\right)\right) = \left(\frac{4 - \left( - 2\right)}{0- 0}\right)\left(x - 0\right) \\ \Rightarrow \left(y + 4\right) = \left(\frac{4 +\hspace{0.17em}2}{0}\right)\left(x\right) \\ \Rightarrow \left(y + 4\right) = \left(\frac{6}{0}\right)\left(x\right) \\ \Rightarrow \left(y + 4\right) \times 0= 6\left(x\right) \\ \Rightarrow 6 x= 0 \\ \mathbf{\Rightarrow }\mathit{x}\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)} \end{gathered}$$

And for diagonal , BD we get :  x₁ = 3  , y₁ = 0 and   x₂ = - 3  , y₂ = 0

So,

Equation of diagonal BD :

$$\begin{gathered} \left(y - 0\right) = \left(\frac{0 - 0}{- 3- 3}\right)\left(x - 3\right) \\ \Rightarrow y = \left(\frac{0}{- 6}\right)\left(x - 3\right) \\ \Rightarrow y \times \left( - 6\right)= 0\left(x - 3\right) \\ \Rightarrow -6 y= 0 \\ \mathbf{\Rightarrow }\mathit{y}\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)} \end{gathered}$$

We know area of triangle  = $\frac{1}{2}$ $\times$ Base $\times$Height  

For triangle ABC , Base = AC= 6  unit  and Height = OB = 3 unit  ( From given diagram and point ' O '  is where axis intersect each other )

So,

Area of triangle ABC  = $\frac{1}{2}$ $\times$ 6 $\times$ 3 = 3  $\times$ 3 = 9 square unit

And for triangle ADC , Base = AC= 6  unit  and Height = OD = 3 unit  ( From given diagram and point ' O '  is where axis intersect each other )

So,

Area of triangle ADC  = $\frac{1}{2}$ $\times$ 6 $\times$ 3 = 3  $\times$ 3 = 9 square unit

Therefore,

Area of kite ABCD  =  Area of triangle ABC + Area of triangle ADC = 9 + 9 =  18 square unit                              ( Ans )



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