Question

The perimeter of a rhombus is 52 cm and one of its diagonals is 10 cm.find the length of the other diagonal and the area of the rhombus

Answer 1

perimeter of rhombus =52cm
diagonal =10cm

(as one of the property = all sides of rhombus are equal) 
each side = 52/4
each side =13cm
we can name the rhombus ABCD
AB=BC=CD=DA
AC= diagonal 1 = 10cm
draw diagonal 2= BC
(another property of rhombus = diagonals of rhombus bisect each other at 90 degree )
so, 
AO = 5 cm
OC = 5cm
we can take AOD triangle 
= (AO)^2+(OD)^2 =(AD)^2
= 5^2+(OD)^2 = (13)^2
= 25+(OD)^2 = 169
= (OD)^2 = 169-25
=(OD)^2 = 144
= OD = √144
= OD = 12 cm

now,
BO = OD 
BO= 12cm
OD= 12cm
total BD = 12+12=24cm
now we get diagonal 2=24 cm

area of rhombus = 
\begin{lgathered}= \frac{1}{2} \times diagonal1 \times diagonal2 \\ = \frac{1}{2} \times 10 \times 24 \\ = 5 \times 24 \\ = 120 {cm}^{2}\end{lgathered}=21​×diagonal1×diagonal2=21​×10×24=5×24=120cm2​ 

Answer 2

Answer:

Step-by-step explanation:

one of the properties of a rhombus is that all sides are equal so..

52/4=13cm

The length of the other diagonal:

the point at where the to diagonals meet name it O.and name the figure ABCD

length of first diagonal:10cm

choose triangle COD:

hyp*2=per*2+base*2

CD*2=CO*2+DO*2

13*2=5*2+base*2

169=25+base*2

169-25=base*2

144=base*2

$\sqrt{x}$144=$\sqrt{x}$base*2

base=12cm

length of the other diagonal:12*2

                             =24cm

area of rhombus:

b*h

=13*13

=169cm*2