Question

The perimeter of a rhomubs is 52 cm and one of its diagonals is 24 cm. The length of the other diagonal is

Answer 1

Side of a rhombus=41×52cm

=13cm

Let AC = 24cm so that 

AO=21×24cm

=12cm

∴OB2=AB2−OA2

=(169-144)=25

or OB=5 and thus BD=2.OB=2×5=10cm

Area of the rhombus =21×24×10

=120cm2

Answer 2

JioGiven :-

perimeter of rhombus =52cm

diagonal =24cm

(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 =

= \frac{1}{2} \times diagonal1 \times diagonal2 \\ = \frac{1}{2} \times 10 \times 24 \\ = 5 \times 24 \\ = 120 {cm}^{2}

hope it's help u