Question

find the area of the Rhombus whose one side measures 5 cm and one diagonal as a let ABCD be a rhombus as shown below.​

Answer 1

Step-by-step explanation:

We know that all the sides of a Rhombus are equal.

Therefore AD=AB=BC=CD

AB= 5cm

Thus, AD , BC and CD = 5cm

In this Rhombus we can see 2 triangles ADB and DCB

We also know that the diagonals of a Rhombus intersect at equal angles

Thus angle AOB , BOC , COD and AOD = 90°

To find height AO

AB^2= AO^2 + BO^2

5^2= AO^2 + 4^2

25= AO^2 + 16

AO^2 = 25-16

AO=√9

AO=3cm

Area of triangle ADB

Area= 1/2 x base x height

Area= 1/2 x 8 x 3

Area= 4x3

Area=12cm^2

Similarly

OC = 3cm(by the above method I used to find AO)

Area of triangle ACB

Area=1/2 x base x height

Area= 1/2 x 8 x 3

Area= 4x3

Area=12cm^2

Total area will be

12 +12 => 24cm^2

Hope it helps

Answer 2

Answer:

soo thanks de diye hai mene

❣️_____have a nice day____❣️