Question

find the perimeter of rhombus with diagonal 16 and 12cm

Answer 1

☆Helllo friend☆

There are two diagonals d1 and d2
So,

$\frac{1}{2} \times \sqrt{ {d1}^{2} + {d2}^{2} } \\ \frac{1}{2} \times \sqrt{ {16}^{2} + {12}^{2} } \\ \frac{1}{2} \times \sqrt{256 + 144} \\ \frac{1}{2} \times \sqrt{400} \\ \frac{1}{2} \times 20$

10cm is side if rhombus

So perimeter if rhombus=4× side
4× 10cm= 40cm

Hope it helps

☺☺☺

Answer 2

hey mate !!!


DAIGONAL of rohmbus = 16 and 12 cm


therefore , half DAIGONAL will be =>


8 cm and 6 cm .



now ,

AO ^2 + OB^2 = AB^2


6^2 + 8^2 = AB^2



AB^2 = 36 + 64


AB = 10 cm


AB is side of rohmbus .






now ,

perimeter is = 4 x side


4 x 10


=> 40 cm


hope it helps!!

$thanks$