Properties of Quadrilaterals

We can say 40m is the perimeter of the park 
so side of rhombus = 10 
Now $$\frac{1}{2}\times\ d_1\times\ d_2\ =\ 96$$
so we get $$\ d_1\times\ d_2\ =\ 192$$   (1)
And as we know diagonals of a rhombus are perpendicular bisectors of each other :
so $$\ \frac{d_1^2}{4}+\ \frac{d_2^2}{4}=\ 100$$
so we get $$\ d_1^2+\ d_2^2=\ 400$$    (2)
Solving (1) and (2)
We get $$d_1=12$$ and$$d_2=16$$
Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is= (12+16)(125) =3500

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC = $$\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

Let the coordinates of vertex D be (x,y)

$$\left(\ \frac{\ x+3}{2},\ \frac{\ y+4}{2}\right)=\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

x = -4 and y = 5

The answer is option D.

Given (P(-3,-2)), (Q(1,-5)), and (R(9,1)).

For parallelogram (PQRS), S = P + R - Q = (-3,-2) + (9,1) - (1,-5) = (5,4)

Diagonal SQ passes through S(5,4) and Q(1,-5).

Slope m = $$\dfrac{-5-4}{1-5} = \dfrac{9}{4}$$

Equation of SQ is  $$y - 4 = \dfrac{9}{4}(x - 5)$$

At the x-axis, y = 0: $$-4 = \dfrac{9}{4}(a - 5)$$

$$-16 = 9(a - 5)$$

$$9a = 29$$

$$a = \dfrac{29}{9}$$

So, $$a = {\dfrac{29}{9}}$$