Question

each side of a rhombus is 10 cm long and one of its diagonals measure 16 cm find the length of the Other diagonal and hence find the area of the Rhombus

Answer 1

Rhombus ABCD with AB=BC=CD=DA= 10cm

                                    AC=16cm

                                    AC, BD cut each other at O => OC=AC/2=8cm

Consider triangle BOC: BO= root (BC^2-OC^2) =6cm => BD=2BO=12cm

A= 0.5*AC*BD = 96 (cm^2)

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

1. consider ABCD as a rhombus .
2. AC and BD are the diagonal
3. O is the point of intersection of diagonal.

Given that,

• Each sides of rhombus be 10 cm

• AB = BC = CD = AD = 10 cm

Length of one diagonal is 16 cm ( Let AC be 16 cm )

To find :-

• Length of other diagonal which is BD.
• Area = ?

In rhombus ABCD,

AO = OC &

BO = OD (Diagonals of rhombus bisect each other at 90° )

Therefore, AO = OC = 8 cm

$\mathrm{In\:\triangle\: AOB,}$

AB = 10 cm

Angle AOB = 90°

AO = 8 cm

$\bold{by\: Pythagoras \:theorem}$

AB^2 = AO^2 + BO^2

10^2 = 8^2 + BO^2

100 = 64 + BO^2

BO^2 = 100 - 64

BO^2 = 36

BO = 6 cm

Since = BO = OD = 6 cm

Therefore, BD = 12 cm

Area of rhombus = 1/2 × (product of diagonal)

$\implies$1/2 × 16 × 12 = 96

•°• $\bold{Length\:of\:another\:diagonal=12cm}$

$\bold{Area\:=96cm}$