Question

the diagonals of a rhombus are 12cm and 16cm find the area and the length of the sides of the rhombus

Answer 1

In a Rhombus ABCD 
Diagonals intersect at O
AC=12cm
​BD=16cm
area of rhombus is 1/2(products of diagonals)
therefore area of rhombus is 1/2(12*16)==96cm^2
NOW

half of AC=AO
half of 12cm=A0
​AO=6cm

half of BD=OD
half of 16cm=OD
OD=8cm

AD=side of rhombus ABCD
side2=AO​2 + OD2 (pytha. theorum)
side2=62 + 82
side2=36 + 64
side2=100
side2=102
side=10cm

 

Answer 2

$d1 = 12 \\ d2 = 16 \\ side \: = \sqrt{ \frac{ {d1}^{2} }{4} + \frac{ {d2}^{2} }{4} } \\ = \frac{1}{2} \sqrt{ {12}^{2} + {16}^{2} } \\ = \frac{1}{2} \sqrt{144 + 256} \\ = \frac{1}{2} \sqrt{400} \\ = \frac{20}{2} \\ = 10$
area of rhombus= 1/2 d1× d2
=12×16÷2
=96 square cm