Question

one side of rhombus 41m,area 720m^2 then sum of diagonal of that rhombus

Answer 1

Area of rhombus = ab = $\frac{d_{1} *d_{2}}{2}$

where, a and b sides of rhombus and $d_{1}$ and $d_{2}$

are diagonals.

if, a = 41 m is a side

We, know diagonal of rhombus intersect at right angle.

hence,$d_{1}^{2} +d_{2}^{2} = 4a^{2}$ (1)

Area = 720$m^{2}$

hence, $d_{123}* d_{2} =1440$

So, $d_{2} =\frac{1440}{d_{1}}$

put this in (1)

$d_{1}^{2} + (\frac{1440}{d_{1}} )^{2} = 4*41^{2}$

on solving above equation we got $d_{1} =18 meter$

and $d_{2} = 80 meter$

hence, $d_{1} +d_{2} =98 meter$

Answer 2

Given Area of the rhombus = 720m^2.

We know that Area of the rhombus = (d1 * d2)/2

= > 720 = (d1 * d2)/2

= > 1440 = d1 * d2

------------------------------------------------------------------------------------------------------

Given Side of a rhombus = 41m.

We know that (Side)^2 = (d1/2)^2 + (d2/2)^2

= > (41)^2 = (d1/2)^2 + (d2/2)^2

= > 1681 = (d1/2)^2 + (d2/2)^2

= > 1681 * 4 = d1^2 + d2^2

= > 6724 = d1^2 + d2^2.

--------------------------------------------------------------------------------------------------------

Now,

We know that (a + b)^2 = a^2 + b^2 + 2ab

= > (d1 + d2)^2 = d1^2 + d2^2 + 2(d1 * d2)

= > (d1 + d2)^2 = 6724 + 2(1440)

= > (d1 + d2)^2 = 6724 + 2880

= > (d1 + d2)^2 = 9604

$= > (d1 + d2) = \sqrt{9604}$

$= > d1 + d2 = 98$

Therefore, the sum of diagonal of the rhombus = 98m.

Hope this helps!