Question

In a rhombus of side 10cm one of the diagonals is 12cm long. Find the length of the second diagonals

Answer 1

let the other diagonal is x

by pythagoras theorem,

(half of first diagonal)^2+(x/2)^2=10^2

(12/2)^2+x^2/4 = 100

6^2+x^2/4=100

x^2/4=100-36=64

x^2/4 =64

x^2=64*4

x=root 256

x=16

then length of other diagonal is 16 cm

Answer 2

Solution:-

let, Rhombus ABCD.

given:-

•The sides of a rhombus are 10cm and one diagonal is 12 cm .

let, DO = OB = ? , BD = ?

and AO = OC = 6cm, and AC = 12 cm

1) we know diagonal of rhombus are equally bisect each other and they are perpendicular to each other.

2) All sides of rhombus are equal.

so,

by Pythagoras theorem.

=> (AB)² = ( AO )² + ( OB )²

=> (10)² = (6)² + (OB)²

=> 100 = 36 + (OB)²

=> 100 - 36 = (OB)²

=> 64 = (OB)²

i.e.

=> (OB)² = 64

=> OB = √64

=> OB = 8 cm

so, we know

DO = OB = 8 cm

hence, BD = 16 cm

Area of rhombus = [(AC)×(BD)]/2

Area of rhombus = [ 12 × 16 ]/2

Area of rhombus = [192]/2

Area of rhombus= 96 cm²

Hence length of diagonal rhombus

is 16 cm and area of rhombus is

96 cm².

i hope it helps you.